A beautiful result in calculus: Solution using Feynman integration ( Integral cos(x)/(x^2+1) )
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- Опубліковано 21 вер 2024
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Leibniz integral rule: • The Leibniz rule for i...
Dirichlet 1: • Doing double Feynman I...
Dirichlet 2: • DESTROYING THE DIRICHL...
Dirichlet 3: • A COMPLEX BOI! Integra...
Haha :D Quite some tounge slipperinos in this video XD Especially in the end, it's meant to be ,,Patreon" not,,video" :D
Nvm, I hope you guys are enjoying this great technique of solving this integral using feynman integration :) Turning an integral into a second order ordinary homogeneous differential equation is such a nice way of aproaching a problem!!! :)
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How fast does a cow fall when thrown from the top of a building?
Engineers : pi squar.. I mean g.
Physicists : okay first let's assume the cow to be a perfect sphere in a vaccum...
Mathematicians : what's a cow... Or a building?
Physicists would probably be the only ones to give you an answer😂😂
Btw we manage to solve problems way more complicated than this
Like Feynman said „Mathematicians care about the general case“.
So they‘d ask how fast x mass would fall when thrown from n meters.
@@obinator9065 an n dimensional cow
Physicist stole n dimensions mths
Thrown requires an initial non zero velocity and is not the same as dropped.
Dude, isn't pi/e just 1 according to the Fundamental Theorom of Engineering
@@PapaFlammy69 ≈ = =
@@arnavanand8037 approximately
No
Yo do engineers actually do this?
Yes, all the time
This is a very elegant proof. Really enjoyed. The Leibniz method of differentiating under the integral sign always seemed like magic, but it's a very powerful technique. I think the benefit this method has over your first method is it can actually be used to solve an infinite number of variations of the integrand by just varying the value of t you're trying to solve for. Good job ^_^
A technique you cannot apply here, since I(t) is not derivable at zero...
While working on this, I just noticed: How can we verify the result? We can verify that a division was done correctly by performing a multiplication. Or an exponentiation was done correctly by taking the root. That goes up until indefinite integrals, where you can differentiate the result to see if the original comes out. But for definite integrals? If you went the antiderivative route, we can verify the antiderivative is correct, and the subtraction is correct, but you didn't go that route here.
Because I have a problem with something right before the end: I'(0) = -pi. But that's only if we take the last expression for I'(t). If we take any expression before that, sin(tx) was a factor of all of them, so wouldn't they all get a zero integrant?
Now, I know your approach is correct, because you get the correct result (rule of consistence; if multiple methods get the same result, that result is probably correct), but I still wonder how that works.
simple. papa flam' should have done a limit and not directly put the t value. look.
step 1: dance
step 2: substitute to the integrals of I'(t) tx=u
step 3: NOW put the limit as t aproches 0
step 4: *realisation*
I knew about Feynman technique and Liebnez, but had never seen Feynman technique used twice to get a differential equation... my eyes lit up as soon as I saw the second derivative equal to the original I(t)! Excellent video!
clicked for terence tao stayed for the math memes
I was blown away when the second part of the sum reduced to a linear second order ODE. Great video !
The magic of trig functions and their invertible derivatives.
''take a leak into the describtion''
_with pleasure_
You have to be careful with I'(0) since I'(0) is actually undefined. What you actually had was I'(0+)=-pi, if you take the limit on the other side you get I'(0-)=+pi. (This happens because the Dirichlet integral changed sign when its argument changes sign.)
Thank you I thought I missed something
But wolfram alpha agreed with him
I started praying too right as you started calculating and Shazam i did the whole equation in my head and knew the answer before you finished. Finally i am as smart as master flammy. I knew i was going to learn how the big boys do it as i subscribed to your channel.
Can't imagine how amazing mathematics is; i saw your video about doing the same problem, and that was, with the use of laplace transform... And guess what, both methods revealed the same result. 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
the virgin cauchy vs the chad dirichlet
At 9:40, I don't understand how plugging in the 0 into I'(t) makes the second integral 0, but not the first one. Why does the first one remain as -π? Wouldn't the top also become sin((0)x) = sin(0) = 0, making the integral 0?
Substitute z=t*x.
This integral can be done in fewer steps by contour integration. Replace cos(x) by Re exp(ix) and make the substition x -> z, in the integrand, where z is a complex number:
cos(x)/(x^2+1) -> Re exp(iz)/(z^2+1).
The denominator has poles at z=+i and z=-i. The numerator goes to zero along the positive imaginary axis and diverges along the negative imaginary axis. So we can close the (-infinity, +infinity) line with an arc in the upper halve plane that encloses the pole at +i. Then
Integral = Re [ 2 pi i exp(iz)/(z+i) | (evaluated at z=i)] = Re [ 2 pi i exp(-1)/(2i)] = Re (pi /e) = pi/e
which is a check on the result found in the video.
So |exp(ix)| is bounded on upper half plane. So you can create a contour to get int(e(ix)/(1+x^2)). From residue theorem this is immediately 2pi i exp(i i)/2i = pi / e.
I have never seen a simple way of solving such a intergral but I must say that dividing the equation is somehow did not create a problem even though the limit was moving from -inf to + inf. Usually, dividing an equation by X creates a problem. Cool video, thanks..
What makes it good is that you move quickly, you use chalk and write clearly as well as clear thought.
I love your way of explaining and having fun with mathematics 😍😍😍😍
I've done this integral in my complex analysis class. Not gonna lie, this is by far the tastiest way of solving this integral I've seen.
Yeah it was truly beautiful..
when u put t = 0 for initial conditions for I'(t) on one hand u have - pi which won't be affecting by plugging in 0 but on the other hand if u consider it to be sin(tx)/x in -inf,inf it is 0 .plz clarify.
I used a mclaurin series to aproximmate cos (x) only two functions of the series. Then I took out a numerator constant. Then I add a zero using +1-1. Then resulting a direct integral. Afterward just evaluate the limits😀
I've scrolled comments and didnt find a convincing answer to the problem of the derivative. Here's my take on it.
One should first note that the derivative of the integrand cannot fulfill the domination condition. Thus, the classic theorem for derivating under the integral doesn't apply.
That's why he gets nonsensical result with I'(0). In fact, one can show that I(t) is not differentiable in 0.
How can we then know that it gives the correct derivative for t in ]0,1[ ?
In my opinion, the way he did it, we can't. At least, I dont know any derivation theorem that would justify it.
But, by studying I(t+h)-I(h) /h, one can quite easily show that the derivative given in the video is correct for t non zero. For t=0, it doesn't converge when h->0
In conclusion, it's likely that he got lucky, unless if he knows some less known theorem. But in this latter case, he would have mentionned it in the comment, which he did not
One of the most amazing integral I have ever seen
@FlammableMaths The actual value of I(t) is π×exp(-|t|), which is not differentiable at 0.
Idk what you're doing, but ik that i suscribed to be fascinated by your work.
10:10 When solving for t=0 for I’(t) you used the result earlier where we had that integral - pi but that relied on the dirichlet results for sin(tx)/x which only is true for t > 0. When t = 0 the dirichlet evaluates to 0 so I’(0) = 0. What you need to use is the limit as t approaches 0 from the positive side.
I(t) is actually pi*e^(-|t|) which makes more sense since if you look at the original definition of I(t) you can see that it’s an even function.
And that also explains why it is continuous but not differentiable at t = 0
I’(t) = -sgn(t)*pi*e^(-|t|)
I have a question. What is the story behind this channel? Like, do you just love maths and memes and think “I should put them together”? Anyway, great video. Es war sehr interessant 👌🏼
Normally, Id use the contour integral for this but this is awesome.❤
I can support you with my appreciation only, so thanks for sharing this video.
I did this in my head using complex analysis just now
Singularity is at z=i
Find the corresponding residue.
The integral over the semi circle is zero by Jordon's lemma.
Contour integral equals 2πi times residue
You end up with π/e
Unbelievably Beautiful! Or as my Phys-Teacher would say in a very nice and comfortable way "Wonderfuuuuul"
great tempo! thanks for the video. You really let me appreciate the Feynman integration technique.
Serious question, papa flammy. I've recently self-studied feynman integration technique, but I have one question to ask. How do you train picking the parameter in order to use Feynman Integration technique, it makes sense in all regards except for that parameter part, what are the guidelines or how should you approach this tricky question?
Jack Vernian This.
I suppose it's something that comes naturally with experience. If you see what kind of parametrizations work in which situations, you somewhat get a feeling of what might be the right one. In other words, this statement isn't helpful
+Flammable Maths
so in essence, it's similar to picking comparison convergence criterion for infinite series, you just practice and get a feel for what is bigger or smaller and at the same time converges or diverges. Thanks.
The way you do it is wrong when you took the 1st derivative of the integral using Leibniz rule, it fails as the corresponding integral doesn’t have any upper bound that converge (which you omit to check)
Hi Flammable Maths, great video! Could you try solving this using Complex Analysis? I think it would be pretty cool to see
Note that the calculation of I(t) is correct for t > 0 only (sufficient to calculate the wanted result). The solution valid for all t is I(t) = π exp(-|t|).
yas!
I’m confused. I plugged in I(-a) = I(a) because cos (ax) is an even function, but the result I got was wrong. I don’t understand this because the function should be even and the result u got pie^-a is not even
something isnt right.
considering I(t) is an even function, since cos is even, then we should rather get I(t) = pi.cosh(t)
and when we substitute with 1 we get the wrong result.
in the first place, differentiating under the integral requires a few conditions which are not met.
There's something that i really didn't catch: at 9:35 he plugged t=0 in the derivative and he said that the second expression is equal to 0, but that also happens in the first expression, so, shouldn't it be equal to 0 instead of -pi? the one that he wrote is the actual case of t=1
You can't quite plug in t=0 in that expression. I needs be to continuous and differentiable on [0, 1] for the logic to hold up. The first I' expression is approximately equal to integral sin(tx)/x, but the integral of 1/x diverges. So as sin(tx) approaches 0 (Due to t approaching 0), you could still get any real number for that integral. Even if t=0 algebraically gave you 0 there, what you actually care about is the limit as t goes to 0 (Otherwise I'(0) is at best undefined, but still not 0). Taking the limit gives you the actual slope of I around 0 (Which is what you need to integrate I' from 0 to 1 and get I(1)). The latter expression is approximately sin(tx)/x^3, which actually goes to 0 as t goes to 0, so there the actual slope of I has been discovered.
There are two things to note here:
1) The 1st integral evaluates to -pi. You can verify that by letting u = tx.
2: In doing so, you realise that this integral is independent of t, and thus by evaluating this integral with respect to u, you find the limit -pi, and the fact that this limit does not contain t.
As for the other integral, using the same substitution, you realise that this integral is indeed depending on t such that the limit of this integral contain t in its limit.
Consider:
I’(t) = -pi + int[-inf;inf]sin(tx)/(x(x^2+1))dx =>
I’’(t) = 0 + int[-inf;inf]cos(tx)/(x^2+1)dx ( = I(t) )
Now, let t = 0, then:
I’(0) = -pi + 0
I’’(0) = pi ( = I(0) )
Further more assume:
I(t) = ce^at => I’(t) = cae^at => I’’(t) = ca^2e^at
then:
I’’(t) - I(t) = ca^2e^at - ce^at = ce^at(a^2 - 1) = 0
for which we find a = +/- 1, then:
I(t) = (c1)e^t + (c2)e^-t => I’(t) = (c1)e^t - (c2)e^-t
Let t = 0, then:
I(0) = c1 + c2 = pi
I’(0) = c1 - c2 = -pi
Thus we find:
I(0) + I’(0) = c1 + c2 + c1 - c2 = pi - pi = 0 c1 = 0
and
I(0) - I’(0) = c1 + c2 - (c1 - c2) = pi - (- pi) = 2pi 2c2 = 2pi
@@kennethkramer9615 The function I is even, so if it had a derivative at 0, that derivative would be zero. Hence the whole thing is wrong. That's what happens when you commute limits without checking that you can apply the standard theorems. You have to check your hypotheses first. That's not just a mathematician's whim, you'll get erroneous results if you're not careful.
@@npip99 What derivation theorem did you use to conclude that it gives the right derivative on ]0,1[ ?
The thing he did is rigorously false, as I has no derivative in 0. But it happens to give the correct answer on ]0,+inf[. Still, I dont find any derivation theorem for this to be proven
It can be done using residue theorem too ,right?
you can calculate it in your head if you know the theorem and have a little bit of experience ;)
Tf is that
@@ryanjagpal9457 complex analysis
@@MrMrPomki Oh ok
I don’t understand how people know these things such as calculus, does calculus even require 150 iq or is it non-sensical knowledge
@@ryanjagpal9457 that's bcoz of interest and deep questioning of logics.. U may wonder on it but these small things helps to discover the secrets of universe ( physics) in more logical manner and it works.. Even communication systems which helps us to exchange our opinion works based on lots and lots of integrals like these
This is your best video. Well done!
Thanks Phil :)
The I’(t) calculated at 2:14 doesn’t have the same I’(t=0) as I’(t) at 4:21. At 2:14 I’(t=0) is equal to 0, while at 4:21 it instead is -π, however the only way to obtain the π/e by this method would be to use the -π rather than 0, but what’s the reason for this ?
The answer to your question that I can come up with is that when you have x in the numerator at 2:41, at the limits of infinity you will get an indeterminate results. because 0 times infinity is indeterminate. So, you can't readily conclude that the integral at 2:14 is zero.
I agree with Assaad, but I'd like to provide my own insight: at 2:14, the integrand is asymptotically proportional to x^-1 (or at least its envelope is), whereas at 4:21, it is asymptotically proportional to x^-3. It's well known that the integral of x^-1 diverges as the integration bounds grow to infinity, so setting t=0 does give an indeterminate form of 0 * infinity. Of course, no such scruples present themselves in the second case.
I think the problem is that the integral of sin(tx)/x is pi only if t is not zero, if t=0 then that integral is zero, so we would get zero anyway. I think that the issue has to do with the continuity and differentiability of I(t), but i dont know how to study continuity and differentiability of this kind of functions.
Peter Stasiak: I think thats a really interesting question and I'd like to try to answer this, altough I'm not sure this is right.
The two I'(t) are defined over an integral. So if you want to determine it's value at t=0, the correct formulation would be:
lim of t->0 of I'(t) = lim t->0 of Integral from -inf to inf of sin(tx)/(x^2+1) OR sin(tx)/(x*(x^2+1)), depending on which formulation you are interested in. The problem now is that you cant just "put the limit into the integral"; there are some conditions.
One condition here is the so called Dominated Convergence Theorem (dont know if thats the right translation):
if lim i->inf of f_i = f and if there is a integrable function g such that |f_i(x)| < |g(x)| for all i and x, then f is integrable and lim i->inf of integral of f_i dx = integral of f dx.
So for the first version at 2:14, the integrand is sin(tx)/(x^2+1) ==> you can find a function g for the DCT and you can get the limit into the integral, so this evaluates to 0. BUT at 4:21, the additional x in the denominator makes your life hard, so you cant just put the limit into the integral, and so you cant say this term vanishes.
I hope thats correct; pls papa have a look and correct me if smth is wrong :D
In the other videos one can see why the first integral on the right has the value -pi. (That integral is called "Dirichlet Integral", as papa Flammy says in the video). If you simply plug 0 in the place of "t" in that integral, you would obtain 0 (no -pi), but you would be wrong. Why? See the video about the Dirichlet integral (it goes from 0 to infinity, but with that result it is not difficult to calculate it from minus infinity to infinity).
YOUR HAND WRITHING IS TOO GOOD😑😑
Normally we might justify differentiating under the integral sign to compute I'(t) by showing for all t in some interval, the partial derivative of the integrand with respect to t is dominated by some positive function whose integral with respect to x converges. But the integral of |(-sin tx)*x /(x^2+1)| with respect to x from -infinity to infinity diverges for any fixed t>0. So how do we justify differentiating under the integral sign here? I would really like to know this.
I thought it was very interesting how you solved it using a second order differential equation. But when you plug in the initial conditions at 10:30, I was wondering whether you could say that I(-alpha) = I(alpha) due to the symmetry of cos(alpha * x). Then I(1) = I (-1). However, when I plug this in, I end up with A = B which is incorrect. I dont get why this does not work
Don't you get that the integral of sin(xt)/x gives pi*sgn(t) as opposed to just pi? So that doesn't vanish when you differentiate the second time with respect to t. It does vanish everywhere that t is non-zero though.
Does I'(0)=0 then
"Ansatz" sounds better than "assumption" 'cause Deutsch
Good to hear that at least in the field of mathematics Germans influence the English speaking world instead of the other way around.
Grüße aus Deutschland!
Ansatz that
That is fantastic, mate! I loved the way that you solved.
:)
You are missing the dependence on t when you insert the dirichlet integral near 4:30. Note e.g. the case t=0 where the integral becomes zero. Not that it matters where you’re interested (t>0)
Edit: spoken too soon. You actually insert zero. I guess there’s a continuity argument that allows you to do this, but you’re glossing over some difficulties here. OTOH you could use t=1 as your initial condition.
The function I(t) is clearly even, so how can I'(0) be non-zero? And does anybody ever check for convergence on you tube?
I think that there is a slight mistake in the general answer: Since I(t) is an even function (of t). So I(t)=pi*exp(-t) can't be correct. Rather it should be pi*exp(-abs(t)). Anyways.... substituting t=1 gives the same answer in either case:). (But dare I substitute t=-1 in your answer?;)
Right. Meaning, I is not differentiable at zero. Meaning part of the reasoning is false (it is at zero). You need to be much more careful than that when you derive an integral depending on a parameter. Since the differential equation is false at zero, you can't just plug in the value t = 0. You need a continuity argument at zero. Besides, you need to find a proper way of computing the left derivative of I at zero (and even show that it exists). This is all much more complicated than shown in the video.
Why can we multiply by x/x? Are we assuming x is not 0? Are we just doing this because it is nicer to cancel out the x^2 + 1?
This is the same (by u sub) as integral of cos(tan(x)) from -pi/2 to pi/2
Beitrag zu noch perfekteren Videos:
"... ist nichts anders als..." ➞ "... is nothing else..." ✘
"... ist nichts anders als..." ➞ "... is just..." ✔︎
I can not understand why I'(0)=pi on 9:43. I'(t) is divided into two terms, both of them are the function of t. The first term equals to pi only when t=1. When t=0, both terms should vanish if the second term equaling to 0 is justified. Can anyone explain me where do I make a mistake?
You didn't
The hypothesis to apply the derivation theorem are not fulfilled
His reasoning can be saved though but it needs a bit more justification
"once again we are back at this bad boy again " !!!.. that my friends is priceless !!!
These videos make me realise how OP residue theorem is.. I got the same answer in just 2 lines
Hail Papa Flammy!
I’ll pray for you any day papa flammy 🙏🙏😫👅💦
Great Papa ...
I just enjoyed (and of course, learned).
Thank you ... Thank you 💞
I have a doubt if we take first expression for I'(t) we get I'(0)=0 but if we take last expression we get I'(0)=-π.......I can't understand this part...
Really nice video! Thanks a lot, but I have a doubt about the solution, I have gone through the same process but when we computed I'(0) it gave us 0 because in that case the Dirichlet integral is the integral of sin(0*x)/x, which I'm pretty sure is 0, but the strange thing is that this gives a different and wrong(we've computed some good aproximations) answer
I've had the same doubt, any solutions?
I check my solution and I get π÷e that is correct answer or 1.15... is right answer my solution also right answer
That's actually a really surprisingly nice answer lol.
Take eye cornea as a two points of ellipses and point of nose as third point for elpsoid
And make equation 😁
It was amazing man, i apreciate very much your work.
Great content in every single video. Thx a lot man!
I(t) is obviously an even function, so it cannot possibly be just Pi * e(-t). The correct solution is Pi * e(-|t|). You missed the fact that I'(t) is not continuous or even defined at 0. What you called I'(0) is in fact I'(0+).
I do not know if the first differentiation under integral sign is correct. Because the integrand that results is not absolutely integrable
On the seventh day, God created Eddie Woo, and on the eighth day, he sent this flammable boi just in case.
I like your accent man. You British accent is so beautiful.
He isn't british
@@sagoot But his accent sounds like British.
Beautiful solution I am glad I found this
i want you to be a friend of mine! never seen such a nice guy!
In my view you can not use Feynman trick for improper integrals!
Smoothly and lovely solution thanks
Sorry sir
Can you tell me that integral is from whish level ?
I saw Terence Tao. WHY PAPA FLAMMY WHY
there is a mistake.at 10:13 .how did you write
I'(0)= -pi. it is equal to zero.please check..
@Flammable, I think there is an error in the step where you state that int((sin(tx)/t)) is equal to pi. This integral depends on t and the the result is an expression in t. It is only equal to pi when t = 1
It's pi for t strictly positive
Very nice, as usual. I love your german accent too, ("...and ten we are tan"). Keep going!
However... why is there a Terry Tao image in the video's preview?
The integral of sin{tx}/x from minus infinity to infinity isn't π/t and not just π?...So the final equation is
I''(t)-I(t)=π/t^2...I Love you videos btw :)
Sir are brilliant
Nice:)
You reminded me another one brilliant method to solve this integral: alpha-representation. Have you already made video on this? If not, I can make an example of using this method for you:)
*somebody misses his Hagoromo Chalk and his Hagoromo Chalk misses him too!*
When you apply the second boundary condition, you take sin(0x)=0 and say that the second term will vanish, but the first term also is zero, because of the same argument and not -pi. Someone can explain this better for me ? I don't understand this passage
Superb bro ,super genious 👍👍
Couldn't the answer also be pi * e since c1 or c2 could be 0 making the other pi? So suppose c2 is 0, then I(t=1)=pi * e^1 = pi * e.
hi flammable maths why did you take I'(0)=-Pi when both the integrals of I'(t) was having sin(tx) leading to Zero as the result of I'(0)? kindly explain
It looks like I=pi/e^abs(t) (clearly I(t)=I(-t) ). I think this is because the integral from -inf to inf of sin(tx)/x is -pi (not pi) when t is negative, so c1=pi and c2=0 in this case. So the first derivative is discontinuous at zero. But you could use the limit of the first derivative as t->0+ and I think the dirichlet integral remains equal to pi and everything is good.
The memetician is at it again
That was syk flammy
Loved it...Would totally be a patreon if i ...had money....or was older than 15... lol
ah smart, not gonna get into trouble at all for that
PS: sweet sweater; looks great on you
I don't know how to ingerate but it's sign looks cool
Why I'(0)=-π,and not I'(0)=0? (at the start of I'(t)=..sin(tx)…)
What was the need to solve that differential? The answer of I''(t)=I(t) was so obvious!
Because it has two answers
Wow, that was really something man!
[pray for Jens]
I don't know how to use the emoji lol
I did it using residue integration!
I like that you use German words
Uh, the first term in I'(t) doesn't depend on t? How is that possible if change of variables would imply a linear dependence?
I have a question about the final result. According to the original function, I(t) should be equal to I(-t), because cos function is even function. But the pi*e^(-t) is not.
That means we will get two different result when t=-1
It has been 2 years but I may still answer. When we were evaluating I'(0), we were actually taking the limit as t approaches to 0 from right (we are looking at how the function acts on the right of 0, we are not dealing with what is it at 0 because it is discontinuous there), so assuming t>0. If we were to do what you say, we would take the limit as t approaches to 0 from left, which would result in the first term in I'(t) to equal π instead of -π. You can then see that taking I'(0) = π, the answer is π/e as expected.