Apologies for the inteeegral business 😂😂 I was just having a bit of fun since this is the way my high school prof used to say it so I just winged it for the video😂😂. Haven't repeated it since all the comments 😂😂😂
It's a lot older than Feynman though. The general rule for differentiating an integral with respect to a parameter, where both the integrand and the endpoints vary, is referred to as the Leibniz integral rule. It includes both the fundamental theorem of calculus and differentiating under the integral sign as special cases. Feynman just made the latter method popular by discussing it in one of his autobiographical pieces. The most general form of the rule is used e.g. in the theory of weak solutions of quasilinear PDEs, for deriving the jump conditions (Rankine-Hugoniot) from conservation laws integrated across a shock discontinuity.
Another interesting property of sin(x)/x is that it's the Fourier transform of a rectangular pulse. I feel invoking FT would be somewhat a generalization of the approach presented in the video. In the conjugated space, the integral would probably quite trivial to evaluate.
I have seen many of these videos trying to get an intuitive understanding of FT. This is hands down the best. You filled in a lot of gaps I was struggling with, stuff that other presenters did not bother saying anything about (I eventually figure these out, but to hear you actually SAY it nails down my understanding). Also, presenting the solution by integrating your thought processes is extremely helpful. Kudos.
Great exercise, but I want to offer one amendment: you can do the integral at 2:15. If you introduce a regulator - say exp(-tx) (which ironically you do later anyways) - then you can regulate that derivative. The regulation parameter t will be taken to zero when computing the final answer which will indeed be π/2. So, even though the integral is divergent, forcing it to converge using regularization still allows you to use the trick for sin(ax)/x. As a MathStackExchange commenter once taught me, "convergence is overrated."
Very powerful technique, and wonderfully explained too. I could nitpick about technical details to make it more rigorous but others have already done so. You've got a new sub.
Thanks I'm actually more inclined towards physics and hence the tendency to skip the rigor in favor of "yeah that seems about right " or "that seems trivial"😂😂 But I really do love teaching math especially on UA-cam. I'm gonna keep uploading too so I'm looking forward to your feedback as well as anyone else.
I came very close just by looking at the graph. The areas cancel from negative infinity until you get to negative pi, and they cancel from pi to to positive infinity. That leaves an area above the x-axis from -pi to pi. If you draw an isosceles triangle there it will very closely approximate the shape of that area. Then, applying the area of a triangle: 1/2(base)(height) gives you 1/2(2pi)(1) = pi, which surprisingly is the exact answer. Curious.
Very clever intuition, Sin(x)/x approaches to 1 as x approaches to 0, so the triangle kinda works. As for areas beyond +/- pi, I don't have good intuition to confirm that the areas would cancel out. They indeed are a bounded sequence of decreasing quantities with alternating positive and negative signs, but I wouldn't be so sure intuitively as to whether they actually cancel out or leave some residue. Any intuition around that that I am missing perhaps? Great observation, and a great way to visualize what's going on instead of blindly following a technique.
@@sitosopti nope, what you are saying is correct. They don't actually cancel out, to convince yourself, just try a numerical integral from -pi to a very large negative value and keep making it bigger to guess the limit behavior.. it is not vanishing/ zero. What happens is the residue you refer to exactly matches the deviation of the [-pi pi] interval integral from the isosceles triangle area OP mentioned
The signed areas outside the interval I = [-pi, pi] don't cancel. If you draw your beforementioned isoceles triangle, you will see that its' area is smaller than the area between the graph and the x-axis from -pi to pi. Hence the total integral outside of this interval on the entire real axis will have to be negative in order to result in a total value of pi for the generalized integral from -inf to inf. Even if you don't know the result pi beforehand, you can also see that the areas don't cancel from noting that the function f(x) = sin(x)/x is asymptotically approaching zero when |x| --> inf. So the crests and troughs of the graph become damped in amplitude, while the period remains unaltered. The biggest of these crests and troughs are throughs on the adjacent sides of the central peak and the interval I (with negative values). Hence they will give the biggest contribution (negative value) to the original integral aside from the positive contribution from the central peak that is slightly larger than pi in area. (Integral f(x) from -2pi to 2pi would have a value less than pi, but integrating "outwards" from the origo the area will approach pi as the limit, from altenating sides for every extra period of the function)
This is a shorter path to solve this integral using comlex analysis: integral(-inf, inf) dx sin(x)/x=integral(-inf, inf) dx Im(exp(ix))/x (using Euler's formula) As 1/x is real and integration is linear we can bring the Imaginary part in front of the integral. = Im(integral(-inf, inf) dx exp(ix)/x Now let's assume +/- infinity as lim R->inf of R) and add some semicircle with radius R to the path of integration leading to a closed path of integration. In the limit as R goes to inifinity, the interal along the semicircle vanishes and we have added zero to the original integral. Now using the residue theorem we can easily evaluate this integral as pi*i*exp(ix)|x=0 (The factor p*i equals the integral along an inifitely small semicircle along the pole at x=0) Plugging x=0 into this expression leads to: pi*i, from which we have to take the imaginary part being pi.
Why do I get answer 2pi instead of pi following your way. Closed integral, couchy: int (f(z)/(z-0))=2pi*i*f(0)=2pi*i On the other hand, closed integral is a sum of: 0 (R->inf) + integral of e(ix)/x along the x axis. Infinitizemal smal semi circle path integral around 0 is zero, right? What am I missing? Small semi circle goes around and BELOW zero, right? Tnx for reply
@@meeehc The correct integration path is the following: Suppose R being a large radius and r being a small readius. Integrate from -R to -r, make a semi circle with radius r around pole z=0, integrate from r to R and make a big semi circle with radius R back to -R. As there is no pole within the path of integration, this path integral will be equal zero. In the limit R-> infinity and r -> 0 the straght paths conicide with the goal integral and the big circle vanishes. Consequently the goal integral + the integral along the small semi circle must be equal 0. It is important to distinguish between real and complex poles. If the pole is real as in the present problem, you need a semi circle to complete the closed path. On the other hand, if you have a complex pole, you have to make the following path: Assume the pole is x + iy: Integrate from -R to x, integrate from x to x+ i(y-r), integratge a full circle with radius r surrounding pole, integrate from x+i(y-r) to x, integrate from x to R... You will see that both integrals in the imaginary direction are eqal with opposite signs and therefore cancel each other. The remaining difference is that the path around the pole will be a full circle. I think your problem is that you do not start with a correct Cauchy path. Make a drawing of the desired path using finite values for r and R nearly comprising the goal integral and correctly surrounding all poles. This path consits of straight lines, semi cicles and full circles only. If you do so, you will see that you will need a semicircle instead of a full circle. In a second step you can apply the limits to it.
@@manfredwitzany2233 with your help I found the simplest, sborteat way. We write TWO closed curve Couchy integrals. One is big semi circle (R-->inf) + our integral along x axis + small circle integral (semi circle goes ABOVE zeoro, to avoid it) = 0 (no poles inside) Second statement is a FULL small circle around zero with one pole = 2pi*i. Once you take into account, that value of small circle integral from first statement is half of full small circle integral, because symmetry and even property.... You just sum both statements together and that number "2", that I was missing before comes from half value of full small circle integral. Very easy with drawings and simple. It was frustrating afternoon:)
Nice video. A slightly technical note: the argument that you give for why the int_0^inf e^(-ax) sin(x)/x dx must tend to zero as a tends to infinity (namely, that for each fixed x the integrand tends to zero as a tends to infinity) isn't quite complete. As a simple example, consider the function f(a,x) which is defined to be 1 if a < x < a + 1 and zero otherwise. Then the integral of f(a,x) dx is always 1 (for a > 0) but for any fixed x, f(a,x) tends to zero as a tends to infinity. The trick to do this rigorously is to observe that |e^(-ax) sin(x)/x| =< e^(-ax) =< e^(-x) for a > 1 and x > 0, since |sin(x)| =< |x|. Then since int_0^inf e^(-x) dx = 1 we have that the whole sequence is dominated by a integrable function, and so now the argument goes through by the Dominated Convergence Theorem. This is indeed a large part of the reason why negative exponentials work so well for this trick. Some of the details here are of course a bit technical, but it might be good to mention in spots like this where technical details are being skipped over, to give the viewer somewhere to look if they are interested in the details. EDIT: Just realised that the Dominated Convergence Theorem is overkill, there is actually a very nice simple argument along the same lines: | int_0^inf e^(-ax) sin(x)/x dx | =< int_0^inf |e^(-ax) sin(x)/x| dx =< int_0^inf e^(-ax) dx = 1/a so as a tends to infinity, int_0^inf e^(-ax) sin(x)/x dx must tend to zero.
Loved the explanation and yes I agree that there are areas where the explanation can be more rigorous. However being primarily a physics teacher I tend to skip over a bit of rigor in favour of "yeah that seems about right"😂 In my head it went like sinx is a bounded function anyway so it all comes down to a "race" between an exp function and a polynomial (x) and the exp function would get to zero much faster than the x in the denominator would....yeah not much rigor but seems to work😂
OMG this is so much easier using contour integration in the plane! Cauchy principal value and you're done. I love integration by differentiation, but this example is a heck of a lot of work!
Very impressive, however 10:18 - "0 Times something is zero", fortunately in that case it is indeed, since both sin and cos oscillates. However, it's good to mention that we have to make sure this something does not approach to infinity:)
there is still a problem in this moment: he says that with x-->inf exp(-ax) goes to 0, but later he uses result he gats from this statement while integrating with a-->0, so basically he multiplies 0 by inf at some point
That would have been cool to point out for sure! I know there is some people not used to these arguments that would really need some clarification on the rigorousity in integration. It's not widely spread on the internet, but it's so important in order to get to higher mathematics 😬
@Jasper Antonelli how about the fact that it gets you the correct answer? That’s enough for physics Edit: I feel like I overstated the point a bit too much here, but nevertheless I think it's safe and pedagogically correct to sometimes separate the physics from the more rigorous parts of the mathematics. It's the same reason we don't have to derive everything from scratch every time we learn something new.
@Jasper Antonelli you do need to check that but not by checking the convergences and all that as mathematicians would do. That’s how physicists can do many integrals that mathematicians can’t because those are ill defined mathematically. Physicists check the the results make sense by checking it against a set of rules but they are all physics motivated, causality for example is typically used. Ideally, they should spend some time to make their formulae more rigorously defined mathematically but that is not happening. Nor is it clear if it’s entirely possible. That’s the nasty reality about QFT these days.
There’s a better way Improper integrals from 0 to infty where the integrand takes the form f(x)/x simplify to L{f(x)} where L is the Laplacian. Thus the integrand becomes 1/(1+x^2) which evaluates to atan(infty)-atan(0) which is π/2
when it comes to integrating from 0 to +inf the function sin(x)*e^-ax instead of using integration by parts (or DI method with a table as shown by blackpenredpen which is less prone to error i think) I believe it's easier to compute it as the imaginary part of the integral from 0 to +inf of e^(-a+i)x which happens to be especially nice here becomes you don't have any terms that remain in the exponent (because from 0 to +inf) and it saves quite a lot of time here.
The integral over e^(-ax) sin x is easier to solve: instead of twice IBP, use 2 sin x = e^(ix) - e^(-ix) and you get a simple exponential to integrate.
an interesting formula I have come across recently that could work is called the Dirichlet Lobachevsky formula, where then the integral from 0 to inf of sinx/x dx becomes the integral from 0 to pi/2 dx which yields pi/2, multiply that by two since we are integrating an even function from -inf to inf, you get pi
Very cool technique, thanks! Two things I noticed could be more rigorous: sinx/x is not defined at 0, but we can replace it with its limit at 0 to make a new function to integrate, and to evaluate I(inf) at the end I believe you also have to check when x goes to inf as well, whether that changes anything (it doesn't).
@@tupoiu not that, u have to verify that after differentiating under the integral sign with respect to the dummy variable, that the domain for that variable is valid when u sub back in at the very end, and if not u have to use limits to justify it. For example, if u set a change of variables involving a partial fraction 1/a-1 as the intermediary step, and then substitute a = 1 at the final stage to determine your integral, this is a prod pen because of the hole at a = 1. We instead have to deduce that I(a) is continuous and the limit as x->1 I(x) exists. Same story with functions going off to +_ infinity under the new variable
To avoid integration by parts, write the integral as the imaginary part of exp(-(a-i)x) integrated from 0 to infinity. So that gives you Im (a-i)^{-1} = 1/(1+a^2) much more quickly. The other way to do this integral elegantly is by contour integration, but Feynman's claim was that he could always avoid contour integrals by finding an alternative trick.
the fact that he had the urge to do this is real ( the mathness screaming inside his head) but it's too early for the general yt audience to get a second existential crisis in the same video which they have thought they know to solve. 😊
Richard Feynman was a genious in Physics. As for Maths, his tricks allow to solve interesting problems but not this one. The integral of the sinc function was deciphered 2 centuries before him: Laplace, Fourier, Residues, etc.
@@MSloCvideos "When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration." From the wiki article you linked. No one likes a pedantic ass.
Residues ? There's no circulation path on which to integrate here ....Just minus infinite , plus infinite 1D .... Sure you could integrate on half an infinite circle and find a way to find the value of the curved part to retract it ...But still ,very cumbersome way of solving that ( i've done a few of those a while back ) , a change of variable , or some kind of substitution trick is always welcome ...
I think the IBP part of the calculation can be simplified by using sin(x) = Im exp(ix), which reduces the integral to that of an exponantial function with complex argument. But the given proof also works if the students are not familiar with complex numbers yet.
When reduced to e^(-ax)*sinx you can convert sinx to Im(e^ix), integrate with respect to dx and be left with (i-a) as a denominator,multiply by its conjugate (i+a) both top and bottom, separate the real and imaginary parts and bingo!!
Sorry, but I don't get the part from 3:00 to 5:07. You can't just throw an extra function into an integral and multiply it by the existing one. Unless the extra function is number 1, but here it's not. It's not the same function any more. Yes, it approaches zero at infinity, but it still changes the values before infinity, so it's still a different function.
Don't think of it as the same function Think about it as a different function altogether that we're solving for a more "general" solution and the integral we we're given is a special case of our general answer
You do not need to have infinity in integration limit of the integral with respect of a. It can arbitrary, so just as well make it the most convenient.
Using a laplace transform: f(t) = integral[t,inf](sin(x)/x)dx L(f(t)) = (1/s)(integral[s,inf](1/(1+s^2))ds = (1/s)(pi/2-arctan(s)) f'(t) = -sin(t)/t L(f'(t)) = L(f(t))s - f(0) = pi/2 - arctan(s) - f(0) = L(sin(t)/t) = arctan(s) - pi/2 pi - f(0) = 0 f(0) = pi therefore the integral[-inf,inf](sin(x)/x) = 2*pi ...now to watch the video :P (and yes, I'm aware that I didn't actually solve for f(t) but that's because it's not an elementary function anyway)
Why is this called Feynman's technique? It's a classic technique for integrals like this, and with some grain of salt in exchanging limits it can be tought in high school. However, it's very beautiful when you encounter this the first time. Feynman's technique usually refers to calculate path integrals by regularized determinants and zeta functions.
Well this method was actually popularised by Feynman. However its actually the Leibniz integral rule. Its normally called Feynman's trick rather than technique.
@@maths_505 Referring this integral to Feynman would be appropriation (and Feynman would like not claim this). However, Wikipedia seems to wrongly refer to this as Feynman's Trick, too). This integral is historically called Dirichlet-Integral, who solved this by Laplace transformation which is very close to your presentation ("Sur la convergence des séries trigonométriques qui servent à représenter une fonction arbitraire entre des limites données", 1829), without dirctly spelling out the ODE for J. However, that might we a wrong referrel already, as first consideration already go back to Euler. Hardy ("Mathematical Gazette 5, p98-103)", 1909) calls it a classical proof.
when u have taken the factor exp(-ax) then it became as Laplace integral transform. So, there may be another type of integral transform if u take log(+ or - ax) , sin(ax), ax , etc. but on the condition that the integral should exist when we do so ?
@@maths_505 Yes indeed. Point taken. Feynman saves the day for many such tricky ones. It is just that many videos at this part move to IBP (incl BPRP one) and i always wondered if i was missing a nuance at that point. Thank you!
I might like to make math videos--or tutor math online--but I find myself stymied by an inability to draw mathematical symbols on a keyboard. How are you allowing your own drawing to be turned into a video?
@@briansauk6837 Matter of taste. In my opinion Laplace is the most elegant way to find the Dirichlet integral (sinc x = sin x /x over R) : Laplace transform of sin x , then Laplace transform of f(x)/x, and final value theorem. Now, if you play around with the Fourier transform of e.g. the triangle function you get indeed additional results : the integral of the sinc squared is also pi/2, and Bessel-Parseval tells you that the integral of sinc to the power 4 is pi/3. These results can also be guessed with IBP of the Dirichlet integral.
@@briansauk6837 Yes I know, problem 1 in my 101 Fourier course. 1 lign with Fourier, 2 ligns with Laplace. But the result by Pierre-Simon Laplace is deeper. Not surprising since the Fourier transform is just a special instance of the Laplace transform (p = ik).
@@laurentthais6252 Agreed. The only small point I was making is that the approach in video was directly matching the Fourier steps (but without making that connection). I do appreciate your insights.
Well You can integrate indefinitely but the definite integral treatment seems more elegant and I didn't have to worry about figuring out the value of C
@@jshadow1988 well I(0) is what I wanted to find and I knew the behaviour of I(a) in the limit a approaches infinity...so it made the calculation convenient That's what the Feynman technique is all about...its more about getting creative and making things work to your liking rather than just some systematic algorithm.
@@maths_505 yes, that's right now! But in the video seems like you integrate alpha from zero to infinity just because the first integral in x is from zero to infinity. Maybe my english is not really good, but thats what I've understood.
@@jshadow1988 yeah that's sort of an accidental motivation stemming from my choice of the alpha function The original integral prompted my choice of the function and so it provided me with information that came in handy doing the last integration w.r.t a
could you please explain why at the end, I(infinity)-I(0)=-pi/2, we didn't accept the answer -pi/2, but had to find what is I(infinity) and get to the answer pi/2
I'm just thinking why the upper limit wasn't emphasized in his evaluation of the derivative of "I wrt a" much same way he emphasized while evaluating the lower limit. I'm worried because: if he does; the result will be an indeterminate of the kind (0×infinity). We all know this result isn't zero.
If we take 1/ x as a log x and another sinx fun as it is and then if we solve this by integration by parts the ans will be right ...but why are u doing this extra ?
The sine limits the numerator to +-1, and the 1/x limits to zero. The function is odd because sin(x)/x = sin(-x)/(-x). Therefore the answer is a non zero positive value.
Apologies for the inteeegral business 😂😂 I was just having a bit of fun since this is the way my high school prof used to say it so I just winged it for the video😂😂. Haven't repeated it since all the comments 😂😂😂
I yearn for the inteeeegral formulation of Calculus. 🥺
@@daddy_myers SUIIIIIIIIIIIIIIIIII
Nah its genius I'm gonna start saying inteegral too lmao
Thought you were an AI.
Feynman's integration is such a fascinating and beautiful way to solve inteegrals.
It's a lot older than Feynman though. The general rule for differentiating an integral with respect to a parameter, where both the integrand and the endpoints vary, is referred to as the Leibniz integral rule. It includes both the fundamental theorem of calculus and differentiating under the integral sign as special cases. Feynman just made the latter method popular by discussing it in one of his autobiographical pieces. The most general form of the rule is used e.g. in the theory of weak solutions of quasilinear PDEs, for deriving the jump conditions (Rankine-Hugoniot) from conservation laws integrated across a shock discontinuity.
Another interesting property of sin(x)/x is that it's the Fourier transform of a rectangular pulse. I feel invoking FT would be somewhat a generalization of the approach presented in the video. In the conjugated space, the integral would probably quite trivial to evaluate.
I have seen many of these videos trying to get an intuitive understanding of FT. This is hands down the best. You filled in a lot of gaps I was struggling with, stuff that other presenters did not bother saying anything about (I eventually figure these out, but to hear you actually SAY it nails down my understanding). Also, presenting the solution by integrating your thought processes is extremely helpful. Kudos.
Great exercise, but I want to offer one amendment: you can do the integral at 2:15.
If you introduce a regulator - say exp(-tx) (which ironically you do later anyways) - then you can regulate that derivative. The regulation parameter t will be taken to zero when computing the final answer which will indeed be π/2. So, even though the integral is divergent, forcing it to converge using regularization still allows you to use the trick for sin(ax)/x.
As a MathStackExchange commenter once taught me, "convergence is overrated."
Very powerful technique, and wonderfully explained too. I could nitpick about technical details to make it more rigorous but others have already done so. You've got a new sub.
Thanks
I'm actually more inclined towards physics and hence the tendency to skip the rigor in favor of "yeah that seems about right " or "that seems trivial"😂😂
But I really do love teaching math especially on UA-cam. I'm gonna keep uploading too so I'm looking forward to your feedback as well as anyone else.
@@maths_505 its in-teg-ral not in-TEE-gral
@@yongdetao7005 keep crying
I came very close just by looking at the graph. The areas cancel from negative infinity until you get to negative pi, and they cancel from pi to to positive infinity. That leaves an area above the x-axis from -pi to pi. If you draw an isosceles triangle there it will very closely approximate the shape of that area. Then, applying the area of a triangle: 1/2(base)(height) gives you 1/2(2pi)(1) = pi, which surprisingly is the exact answer. Curious.
Very clever intuition, Sin(x)/x approaches to 1 as x approaches to 0, so the triangle kinda works. As for areas beyond +/- pi, I don't have good intuition to confirm that the areas would cancel out. They indeed are a bounded sequence of decreasing quantities with alternating positive and negative signs, but I wouldn't be so sure intuitively as to whether they actually cancel out or leave some residue. Any intuition around that that I am missing perhaps?
Great observation, and a great way to visualize what's going on instead of blindly following a technique.
@@sitosopti nope, what you are saying is correct. They don't actually cancel out, to convince yourself, just try a numerical integral from -pi to a very large negative value and keep making it bigger to guess the limit behavior.. it is not vanishing/ zero. What happens is the residue you refer to exactly matches the deviation of the [-pi pi] interval integral from the isosceles triangle area OP mentioned
@@jjukjkjiok7782 Thanks, I was guessing that might be the case.
The signed areas outside the interval I = [-pi, pi] don't cancel. If you draw your beforementioned isoceles triangle, you will see that its' area is smaller than the area between the graph and the x-axis from -pi to pi. Hence the total integral outside of this interval on the entire real axis will have to be negative in order to result in a total value of pi for the generalized integral from -inf to inf.
Even if you don't know the result pi beforehand, you can also see that the areas don't cancel from noting that the function f(x) = sin(x)/x is asymptotically approaching zero when |x| --> inf. So the crests and troughs of the graph become damped in amplitude, while the period remains unaltered. The biggest of these crests and troughs are throughs on the adjacent sides of the central peak and the interval I (with negative values). Hence they will give the biggest contribution (negative value) to the original integral aside from the positive contribution from the central peak that is slightly larger than pi in area.
(Integral f(x) from -2pi to 2pi would have a value less than pi, but integrating "outwards" from the origo the area will approach pi as the limit, from altenating sides for every extra period of the function)
@ 6:21 That integral can also be computed by re-writing in terms of complex exponentials
This is a shorter path to solve this integral using comlex analysis:
integral(-inf, inf) dx sin(x)/x=integral(-inf, inf) dx Im(exp(ix))/x (using Euler's formula) As 1/x is real and integration is linear we can bring the Imaginary part in front of the integral.
= Im(integral(-inf, inf) dx exp(ix)/x Now let's assume +/- infinity as lim R->inf of R) and add some semicircle with radius R to the path of integration leading to a closed path of integration.
In the limit as R goes to inifinity, the interal along the semicircle vanishes and we have added zero to the original integral. Now using the residue theorem we can easily evaluate this integral as
pi*i*exp(ix)|x=0 (The factor p*i equals the integral along an inifitely small semicircle along the pole at x=0) Plugging x=0 into this expression leads to:
pi*i, from which we have to take the imaginary part being pi.
What if comments in YT could illustrate Latex formula...
@HB Net Yes my way shown above is much shorter.
Why do I get answer 2pi instead of pi following your way. Closed integral, couchy: int (f(z)/(z-0))=2pi*i*f(0)=2pi*i
On the other hand, closed integral is a sum of: 0 (R->inf) + integral of e(ix)/x along the x axis.
Infinitizemal smal semi circle path integral around 0 is zero, right? What am I missing? Small semi circle goes around and BELOW zero, right? Tnx for reply
@@meeehc The correct integration path is the following: Suppose R being a large radius and r being a small readius. Integrate from -R to -r, make a semi circle with radius r around pole z=0, integrate from r to R and make a big semi circle with radius R back to -R. As there is no pole within the path of integration, this path integral will be equal zero. In the limit R-> infinity and r -> 0 the straght paths conicide with the goal integral and the big circle vanishes. Consequently the goal integral + the integral along the small semi circle must be equal 0. It is important to distinguish between real and complex poles. If the pole is real as in the present problem, you need a semi circle to complete the closed path. On the other hand, if you have a complex pole, you have to make the following path: Assume the pole is x + iy: Integrate from -R to x, integrate from x to x+ i(y-r), integratge a full circle with radius r surrounding pole, integrate from x+i(y-r) to x, integrate from x to R... You will see that both integrals in the imaginary direction are eqal with opposite signs and therefore cancel each other. The remaining difference is that the path around the pole will be a full circle.
I think your problem is that you do not start with a correct Cauchy path. Make a drawing of the desired path using finite values for r and R nearly comprising the goal integral and correctly surrounding all poles. This path consits of straight lines, semi cicles and full circles only. If you do so, you will see that you will need a semicircle instead of a full circle. In a second step you can apply the limits to it.
@@manfredwitzany2233 with your help I found the simplest, sborteat way.
We write TWO closed curve Couchy integrals.
One is big semi circle (R-->inf) + our integral along x axis + small circle integral (semi circle goes ABOVE zeoro, to avoid it) = 0 (no poles inside)
Second statement is a FULL small circle around zero with one pole = 2pi*i.
Once you take into account, that value of small circle integral from first statement is half of full small circle integral, because symmetry and even property....
You just sum both statements together and that number "2", that I was missing before comes from half value of full small circle integral.
Very easy with drawings and simple. It was frustrating afternoon:)
Nice video. A slightly technical note: the argument that you give for why the int_0^inf e^(-ax) sin(x)/x dx must tend to zero as a tends to infinity (namely, that for each fixed x the integrand tends to zero as a tends to infinity) isn't quite complete. As a simple example, consider the function f(a,x) which is defined to be 1 if a < x < a + 1 and zero otherwise. Then the integral of f(a,x) dx is always 1 (for a > 0) but for any fixed x, f(a,x) tends to zero as a tends to infinity.
The trick to do this rigorously is to observe that |e^(-ax) sin(x)/x| =< e^(-ax) =< e^(-x) for a > 1 and x > 0, since |sin(x)| =< |x|. Then since int_0^inf e^(-x) dx = 1 we have that the whole sequence is dominated by a integrable function, and so now the argument goes through by the Dominated Convergence Theorem. This is indeed a large part of the reason why negative exponentials work so well for this trick.
Some of the details here are of course a bit technical, but it might be good to mention in spots like this where technical details are being skipped over, to give the viewer somewhere to look if they are interested in the details.
EDIT: Just realised that the Dominated Convergence Theorem is overkill, there is actually a very nice simple argument along the same lines:
| int_0^inf e^(-ax) sin(x)/x dx | =< int_0^inf |e^(-ax) sin(x)/x| dx =< int_0^inf e^(-ax) dx = 1/a
so as a tends to infinity, int_0^inf e^(-ax) sin(x)/x dx must tend to zero.
Loved the explanation and yes I agree that there are areas where the explanation can be more rigorous.
However being primarily a physics teacher I tend to skip over a bit of rigor in favour of "yeah that seems about right"😂
In my head it went like sinx is a bounded function anyway so it all comes down to a "race" between an exp function and a polynomial (x) and the exp function would get to zero much faster than the x in the denominator would....yeah not much rigor but seems to work😂
IN-teg-ral
Not in-TEE-gral
And the way he says differentiate too
the way he says integration right but integral wrong
He explained in a comment; This is a joke
"in-TEE-gral" - are we being trolled?
Yes, he is joking, he knows that’s super wrong
And I was pronouncing it inte-GRAAAL all the time
Not to mention "differenseeate" and "swaydo".
@@chrischiesa609 yeah away do that took me a while ...
It’s a method to get comments
OMG this is so much easier using contour integration in the plane! Cauchy principal value and you're done. I love integration by differentiation, but this example is a heck of a lot of work!
Very impressive, however 10:18 - "0 Times something is zero", fortunately in that case it is indeed, since both sin and cos oscillates. However, it's good to mention that we have to make sure this something does not approach to infinity:)
there is still a problem in this moment: he says that with x-->inf exp(-ax) goes to 0, but later he uses result he gats from this statement while integrating with a-->0, so basically he multiplies 0 by inf at some point
Laplace makes it even better
1:28 You cannot just shift differentiation operator inside the integral. You need to check some convergence theorems in order for that to work.
That would have been cool to point out for sure! I know there is some people not used to these arguments that would really need some clarification on the rigorousity in integration. It's not widely spread on the internet, but it's so important in order to get to higher mathematics 😬
It's a physics channel
@Jasper Antonelli how about the fact that it gets you the correct answer? That’s enough for physics
Edit: I feel like I overstated the point a bit too much here, but nevertheless I think it's safe and pedagogically correct to sometimes separate the physics from the more rigorous parts of the mathematics. It's the same reason we don't have to derive everything from scratch every time we learn something new.
The proof of dominananted convergence, I knew I'd see this eventually in the comments. A Dirichlet class function will always have this issue.
@Jasper Antonelli you do need to check that but not by checking the convergences and all that as mathematicians would do. That’s how physicists can do many integrals that mathematicians can’t because those are ill defined mathematically. Physicists check the the results make sense by checking it against a set of rules but they are all physics motivated, causality for example is typically used. Ideally, they should spend some time to make their formulae more rigorously defined mathematically but that is not happening. Nor is it clear if it’s entirely possible. That’s the nasty reality about QFT these days.
I've never heard anyone say integral like that my entire life and I've had math professors from many different countries
InTEEgral
Wanna smack him lol
Are you looking for the derivation?
There’s a better way
Improper integrals from 0 to infty where the integrand takes the form f(x)/x simplify to L{f(x)} where L is the Laplacian.
Thus the integrand becomes 1/(1+x^2) which evaluates to atan(infty)-atan(0) which is π/2
Why work so hard? Directly apply Lobachevsky's Dirichlet integral formula where f(x) =1 and get the answer pi/2 in just one step.
Cool. Can you explain i havent heard about it
@@rishabhsemwal4180 Integral from 0 to infinity f(x). (sin x)/x dx is same as integral from 0 to pi/2 f(x) dx. In this case, f(x) = 1.
Very nice, especially the part with reasoning behind using e^-ax and not another function.
Thank you!
It's the most fun part of using the technique.
when it comes to integrating from 0 to +inf the function sin(x)*e^-ax instead of using integration by parts (or DI method with a table as shown by blackpenredpen which is less prone to error i think) I believe it's easier to compute it as the imaginary part of the integral from 0 to +inf of e^(-a+i)x which happens to be especially nice here becomes you don't have any terms that remain in the exponent (because from 0 to +inf) and it saves quite a lot of time here.
At 6:27, it does not really matter which one. You get the same answer either way. But this does not work always. Thank you.
The integral over e^(-ax) sin x is easier to solve: instead of twice IBP, use 2 sin x = e^(ix) - e^(-ix) and you get a simple exponential to integrate.
2i sin x
@@jerebenitez8542 yes
an interesting formula I have come across recently that could work is called the Dirichlet Lobachevsky formula, where then the integral from 0 to inf of sinx/x dx becomes the integral from 0 to pi/2 dx which yields pi/2, multiply that by two since we are integrating an even function from -inf to inf, you get pi
Yeah that's a pretty cool formula. I made a proof video on that and used it in a few integrals as well.
Feynman used this technique but he got it from a book by Woods.
Very cool technique, thanks! Two things I noticed could be more rigorous: sinx/x is not defined at 0, but we can replace it with its limit at 0 to make a new function to integrate, and to evaluate I(inf) at the end I believe you also have to check when x goes to inf as well, whether that changes anything (it doesn't).
I think this is actually one of the most crucial steps in this argument, and is unfortunately just overlooked too much
You don't need a function to be defined at its endpoints to be able to integrate it. sinx/x is not defined at infinity.
@@tupoiu not that, u have to verify that after differentiating under the integral sign with respect to the dummy variable, that the domain for that variable is valid when u sub back in at the very end, and if not u have to use limits to justify it. For example, if u set a change of variables involving a partial fraction 1/a-1 as the intermediary step, and then substitute a = 1 at the final stage to determine your integral, this is a prod pen because of the hole at a = 1. We instead have to deduce that I(a) is continuous and the limit as x->1 I(x) exists. Same story with functions going off to +_ infinity under the new variable
I've been looking for this since a long time. Thank you so much ❤
To avoid integration by parts, write the integral as the imaginary part of exp(-(a-i)x) integrated from 0 to infinity. So that gives you Im (a-i)^{-1} = 1/(1+a^2) much more quickly.
The other way to do this integral elegantly is by contour integration, but Feynman's claim was that he could always avoid contour integrals by finding an alternative trick.
the fact that he had the urge to do this is real ( the mathness screaming inside his head)
but it's too early for the general yt audience to get a second existential crisis in the same video which they have thought they know to solve. 😊
Beautiful, no words can define the beauty.
Thx for sharing. Love that you tried an approach at the beginning so we see why it fails (not just saying it doesn't work)
Thanks for the appreciation....makes a depressed math teacher's day😂😂😂
Richard Feynman was a genious in Physics. As for Maths, his tricks allow to solve interesting problems but not this one.
The integral of the sinc function was deciphered 2 centuries before him: Laplace, Fourier, Residues, etc.
I was gonna come into this comment section yelling something something laplace transform but you beat me to it 😅
This is an illustration of Feynman's trick, rather than an illustration of the solution of the integral.
@@neilgerace355 But it's not even his trick. This is called the Leibniz integral rule ( en.wikipedia.org/wiki/Leibniz_integral_rule )
@@MSloCvideos "When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration." From the wiki article you linked. No one likes a pedantic ass.
Residues ? There's no circulation path on which to integrate here ....Just minus infinite , plus infinite 1D .... Sure you could integrate on half an infinite circle and find a way to find the value of the curved part to retract it ...But still ,very cumbersome way of solving that ( i've done a few of those a while back ) , a change of variable , or some kind of substitution trick is always welcome ...
Great video about differenciating inteegrals.
Ehehehe
This is just the Fourier transform of sinc(x) function at 0. So it is pi*rect(0)=pi
I think the IBP part of the calculation can be simplified by using sin(x) = Im exp(ix), which reduces the integral to that of an exponantial function with complex argument. But the given proof also works if the students are not familiar with complex numbers yet.
Exactly
I wanted a video that even a high school calculus student could understand
When reduced to e^(-ax)*sinx you can convert sinx to Im(e^ix), integrate with respect to dx and be left with (i-a) as a denominator,multiply by its conjugate (i+a) both top and bottom, separate the real and imaginary parts and bingo!!
Sorry, but I don't get the part from 3:00 to 5:07. You can't just throw an extra function into an integral and multiply it by the existing one. Unless the extra function is number 1, but here it's not. It's not the same function any more. Yes, it approaches zero at infinity, but it still changes the values before infinity, so it's still a different function.
Don't think of it as the same function
Think about it as a different function altogether that we're solving for a more "general" solution and the integral we we're given is a special case of our general answer
just, wow! fascinating video, great explanation I could follow every step!
Basically Sin(t)/t = Sa(t), and Integrating x(t) from -inf to +inf = Area[x(t)], Area[Sa(t)] = pi
7:03 According to the rule "ILATE", u = sinx and dv = e^(-ax)dx right?
Edit: Can anyone explain what he did at 7:00
You’re an excellent teacher.
You do not need to have infinity in integration limit of the integral with respect of a. It can arbitrary, so just as well make it the most convenient.
Ofcourse we don't "need" it ....
However we can be creative in our choices to give us answers more efficiently
This Integral is damnnn.............really loved it!!!!
It took me 5 minus to realize this is made on a phone. Very interesting video
There is a problem of convergence of dI/da at a=0.
Just a special case of Fourier transform pair
You could just use Laplace transform to solve exp(-ax)sinxdx. I think it's much easier.
A beauty, and very well shown!
In-tee-gral?
Using a laplace transform:
f(t) = integral[t,inf](sin(x)/x)dx
L(f(t)) = (1/s)(integral[s,inf](1/(1+s^2))ds = (1/s)(pi/2-arctan(s))
f'(t) = -sin(t)/t
L(f'(t)) = L(f(t))s - f(0) = pi/2 - arctan(s) - f(0) = L(sin(t)/t) = arctan(s) - pi/2
pi - f(0) = 0
f(0) = pi
therefore the integral[-inf,inf](sin(x)/x) = 2*pi
...now to watch the video :P
(and yes, I'm aware that I didn't actually solve for f(t) but that's because it's not an elementary function anyway)
No wonder physics is exhausting. It took over 18 minutes to solve this single problem.
What is the difference between Feynman integration with Laplace transform?
Instead use Forurier transform technique to find result within seconds
Why is this called Feynman's technique? It's a classic technique for integrals like this, and with some grain of salt in exchanging limits it can be tought in high school. However, it's very beautiful when you encounter this the first time. Feynman's technique usually refers to calculate path integrals by regularized determinants and zeta functions.
Well this method was actually popularised by Feynman. However its actually the Leibniz integral rule. Its normally called Feynman's trick rather than technique.
@@maths_505 Referring this integral to Feynman would be appropriation (and Feynman would like not claim this). However, Wikipedia seems to wrongly refer to this as Feynman's Trick, too). This integral is historically called Dirichlet-Integral, who solved this by Laplace transformation which is very close to your presentation ("Sur la convergence des séries trigonométriques qui servent à représenter une fonction arbitraire entre des limites données", 1829), without dirctly spelling out the ODE for J. However, that might we a wrong referrel already, as first consideration already go back to Euler. Hardy ("Mathematical Gazette 5, p98-103)", 1909) calls it a classical proof.
@@timtaler2435 the title of my video doesn't refer to this integral as Feynman's....the title says I'm solving it using Feynman's trick
@@maths_505 It's not Feynman's Trick.
@@timtaler2435 okay cool...the important thing its damn beautiful 😂
The inteeeegral of a circles circumference is it’s area….
I've been asked to solve this integral in the ICTS-TIFR interview. Fortunately I was able and got selected 🥳
Using contour integral ..was easy.
Great work sir. It's really wonderful. Can you please reply to what's the software you are using...?
Samsung notes on a smartphone for this video
You are so kind. Thank you Sir
You should create a video on the indefinite integral of sinx/x
A piece of art (when "a" positive) !
when u have taken the factor exp(-ax) then it became as Laplace integral transform. So, there may be another type of integral transform if u take log(+ or - ax) , sin(ax), ax , etc. but on the condition that the integral should exist when we do so ?
I had said yet that it is not the Feynman's method, it is just the method of integration by parameter.
Wow, very nice problem and wonderful explanation!👍
Dude !!!
Your math videos are so cool !!!
Thank you so much your recognition really means so much !!!
@@maths_505 🙏
You should justify the limit integral interversion, it's incomplete.
could you also show technque to solve using Residue thm ?
Nice video. At 6:17, can’t we just use euler’s formula for sine? I find it faster and maybe one less “thing” to think about.
Ofcourse we can
However this video was just a demonstration of how beautiful Feynman's technique is when it comes to solving tricky integrals
@@maths_505 Yes indeed. Point taken. Feynman saves the day for many such tricky ones. It is just that many videos at this part move to IBP (incl BPRP one) and i always wondered if i was missing a nuance at that point. Thank you!
But yes I do agree that one should skip the IBP for something more sophisticated like complex analysis
Great! Easy understanding for french people.
I might like to make math videos--or tutor math online--but I find myself stymied by an inability to draw mathematical symbols on a keyboard. How are you allowing your own drawing to be turned into a video?
The Laplace transform gives you the result in 2 ligns. The video would be short.
or Fourier - with his 'a' as a pure real, he really did just derive it by Fourier. Regardless, you are absolutely correct.
@@briansauk6837 Matter of taste. In my opinion Laplace is the most elegant way to find the Dirichlet integral (sinc x = sin x /x over R) :
Laplace transform of sin x , then Laplace transform of f(x)/x, and final value theorem.
Now, if you play around with the Fourier transform of e.g. the triangle function you get indeed additional results : the integral of the sinc squared is also pi/2, and Bessel-Parseval tells you that the integral of sinc to the power 4 is pi/3.
These results can also be guessed with IBP of the Dirichlet integral.
@@laurentthais6252 Integral of f from - inf to +inf is simply F(0), where F is Fourier transform of f. Sinc -> Rect. Rect(0) is pi. Done.
@@briansauk6837 Yes I know, problem 1 in my 101 Fourier course. 1 lign with Fourier, 2 ligns with Laplace.
But the result by Pierre-Simon Laplace is deeper.
Not surprising since the Fourier transform is just a special instance of the Laplace transform (p = ik).
@@laurentthais6252 Agreed. The only small point I was making is that the approach in video was directly matching the Fourier steps (but without making that connection). I do appreciate your insights.
Excwllent video...which software are you using?
It's the default samsung notes software on my note 9 and I'm using the s pen for writing
Please sir, I want to know the technology you use in making your videos, do you use graphic tablet?
Thanks 🙏
8:24 is me 30s into solving any calculus problem
That's brilliant! Awesome
What does Antigua have to do with calculus?
Why you integrate dI/da from 0 to infinity? I was expecting an indefinite integral.
Well
You can integrate indefinitely but the definite integral treatment seems more elegant and I didn't have to worry about figuring out the value of C
@@maths_505 but why a goes from zero to infinity?
@@jshadow1988 well I(0) is what I wanted to find and I knew the behaviour of I(a) in the limit a approaches infinity...so it made the calculation convenient
That's what the Feynman technique is all about...its more about getting creative and making things work to your liking rather than just some systematic algorithm.
@@maths_505 yes, that's right now!
But in the video seems like you integrate alpha from zero to infinity just because the first integral in x is from zero to infinity. Maybe my english is not really good, but thats what I've understood.
@@jshadow1988 yeah that's sort of an accidental motivation stemming from my choice of the alpha function
The original integral prompted my choice of the function and so it provided me with information that came in handy doing the last integration w.r.t a
This seems like a lot of hand waving that would be unnecessary if you would just introduce an integration factor, a la Dirac, et al.
nice Integral! Which App do u use to present this solution? "Nebo"?
It's just the default samsung notes app
could you please explain why at the end, I(infinity)-I(0)=-pi/2, we didn't accept the answer -pi/2, but had to find what is I(infinity) and get to the answer pi/2
Great video, thank you. It’s funny to see mathematicians in the comments annoyed that Feynman gets his name on this technique 😆
😂😂😂
Well technically it is Leibniz's integral rule but Feynman's the one who popularised it
just use the residue theorem, then you will get the answer in 1sec.
Yes indeed but the purpose of this video is to the show the beauty of the Feynman trick and how to tackle problems with it.
The way he says integral pissed me off sm i couldn't finish the video
Oh that was just a joke. I didn't even take this video seriously....the rest of the content on my channel avoids this
I'm just thinking why the upper limit wasn't emphasized in his evaluation of the derivative of "I wrt a" much same way he emphasized while evaluating the lower limit.
I'm worried because: if he does; the result will be an indeterminate of the kind (0×infinity). We all know this result isn't zero.
Please can someone help explain; maybe I'm twisting things here.
Thanks
My gosh, that is beautiful
@ 7:48 I would have computer this integral using the fact sin x is the imaginary part of e^ix
Very nice Sir !! ... Even if using Laplace transform techniques it would have been done even quicker. My Regards.
But e^(-ax) is exactly the factor of Laplace transform
@@Retroist2024 so ????
Look my mobile keyboard doesn't have all those signs but 1 can be written as e^-(0x).... Next you know how it can be done.
Can you give a good reference for the derivation of the Feynman Technique that is easy to understand?
You could search for the derivation under the name Leibniz's integral rule and that should turn up something useful.
why not simply using integration by part method !?
How is this apply to every day life?
you can use complex variable easily to solve this
Great video! I don’t mind however you say “integral”, the concepts are explained well.
You use Laplace transform and solve this problem with in 4-5 steps ... anyway it is wonderful integration problem...
Indeed
But the purpose of this video was to demonstrate the beauty of Feynman's trick
If we take 1/ x as a log x and another sinx fun as it is and then if we solve this by integration by parts the ans will be right ...but why are u doing this extra ?
How do you justify that you can indeed derivate I(a)?
Heh. I remember a little from all those math classes 40+ years ago. A very little lol.
I have absolutely no idea what's going on, but it looks cool!
Well
I'll give you full marks for the energy
@@maths_505 One day I’ll understand, I promise. Remind me in 3 years :)
@@ecIipsed you'll understand much sooner than 3 years ....cuz you're curious and really want to learn this stuff
@@maths_505 I’m only a sophomore in High School and wont take physics for another year. I assume this is more college-level calculus?
@@ecIipsed it's quite possible to understand it when you're done with high school calculus as well
"in-TEE-gruhl" ?
Just a bit of fun 😂
Mari es una profesora encantadora.
what's with writing Pi like that?
Everybody jumps over the int(1/(1+x^2)) dx = arctan(x) step. Doesn't seem obvious to me at all.
The x cancels out. The answer is that you’ve committed sin.
Could use a sprinkle of rigor here and there but I'll give you full marks for elegance and simplicity
@@maths_505 it's what I do best professor 😎Calculus 1 ain't got nothing on me.
Great video man! Thank you!
The sine limits the numerator to +-1, and the 1/x limits to zero.
The function is odd because sin(x)/x = sin(-x)/(-x).
Therefore the answer is a non zero positive value.
The Fourier transform gives you the answer in 1lign. Joseph Fourier was better than Pierre-Simon Laplace.