Teacher, I follow the lessons you post on UA-cam I am studying in the college of maths and physics number whats app please tell me you can help me with maths
I like this more than the polar coordinates method because it is far easier to understand. No Jacobian, no different kind of coordinate system, just one substitution (really the second one you could do without the substitution, just by inspection).
There is no Jacobian probably because substitution is done in iterated integral only (he changed only one variable at the time) Another approach is Gamma function with reflection formula
This is a computationally brilliant method for sure. However, compared to the polar approach the appearance of pi as a result comes across as the outcome of a somewhat artificial-looking substitution process. In contrast, the polar approach relies on the circular symmetry of the "Bell surface" about the z axis, which makes a pi-related value of the integral fairly obvious. Besides, one can relate the square root in the result to the Gaussian curve being a cross-section of that surface, whilst the squaring of the integral in Laplace's approach looks like a mere trick based on nothing but the factorizability of the exponential. For those who are interested, Dr Peyam in his channel does around 20 different derivations of this result!
@@holyshit922 That might actually be a line of circular reasoning, since many of the proofs of Gamma function integral rely on the Gaussian integral result!
@@Xoque551 not necessarily circuloar reasoning Reflection formula can be derived from product representation of Gamma function and Euler's product for sine
It is great seeing this integral done in cartesian coordinates. All the textbooks I used so far either used the approach over polar coordinates or just used the result. Thank you for this video!
Interestingly, this quite related to polar coordinates but avoids using them :). From the 2D point of view in the first quadrant, we used the coordinates (x,t) where (x,y) = (x,tx). t = y/x is the slope of the line starting from the origin passing through (x,y), in other words t = tan(theta) where theta is polar angle. Cool video :)
As a stats guy, this is a beautiful detour from the more popular solution. I enjoyed the detailed steps and exploits of the symmetry inherent in the integral. Whilst the polar coordinates approach is easier to explain to anyone who has done trig, this solution is elagant af too.
I remember doing this integral shortly after learning about the jacobian. There is so much joy in doing this integral for the first time, thank you Prof. Steve!
@@MohammadIbrahim-sq1xn i shouldve clarified but my integral was from 0 to x and the intial variable was t. Once you integrate the series you get Σ(-1)^n * x^(2n+1)/((2n+1)(n!)) which is the expansion of sin(x), cheers.
I watched 10 minutes of this stuff and we went from a simple formula to a bunch of mumbo jumbo. Can one of you calculus students explain to me what in the world this stuff is useful for? And be specific and have you found a job to where you are actually crunching out formulas to prove they are right or wrong for a living. Can't use teacher or professor either. Some other profession please.
In 1995 the internet arrived. And this channel is the only good thing worth watching (just a long time waiting :)). Love your style and you teach like someone who understands rather than repeats. Thanks for your hard work!
Beautiful :) I did'nt know this trick. BTW when you switch to double integral and polar coordinates, you don't need Jacobian. You just can calculate the volume under the function as sum of volumes of cylinder shells. (Height of shell is e^(-x^2-y^2) = e^-r^2; length is 2 pi r and thickness is dr, so you integrate e^-r^2 . 2 pi r . dr from 0 to infinity.)
Plus I don't you understand his y = xt where his translation derivatives doesn't equal in dy where dy = x dt + a missing in his formulas t dx term: dy = x dt + t dx I thought? 🤔
@@lawrencejelsma8118 In the substitution y = xt in the inner intergral is x in the role of constant. (Like "for given fixed x from the outer integral we shall calculate this inner integral..."). So the substitution y = xt is just recepee for transition from y to t. This is the reason why dy = x.dt.
@@aninob ... Yeah your correct. He should have written it out dy = xdt + tdx but the tdx term equals zero just like t can translate through the integral during summation. I was just use to "two space" partial differentiation mathematics of ut (ut)' = (u')t + u(t'). The recommended solution of real only or even when considering imaginary numbers are always polar solutions of rcosx + jrsinx solutions because trigonometric integral tables solutions are easier to understand.
@@MrPoornakumar 3B1B just made a beautiful video about it (plus some mathematical sugar on the top I never knew about). ua-cam.com/video/cy8r7WSuT1I/v-deo.html
Franchement très intéressant. Je ne connaissait que la méthode avec passage en polaire. Et j'étais persuadé que c'était la seule méthode possible ! Merci pour cette brillante présentation.
Love the aspect of switching the order of integration, I was wondering how it could be done. But then realised that it is ofcourse summation of product. At the start was confused as to how you could make the function 2* (0 to Infinity) of the f(x), but i had missed that this is a converging function. made math interesting again.. Thanks!
This is a cool integral to know in that you can use integrals used by probability distributions to simply rewrite the integral in terms of it's probability distribution and then if they go from -inf to +inf they just become 1
Can you solve this problem? Q. If f[{x + √(1 + x^2)}/x] = x^2. Then find f(x); domain & Range of f(x) =? Video link:- ua-cam.com/video/YVboBTlmUo8/v-deo.html
My own technique also has value: create an infinite alternating series through iterative integration by parts. After creating the first few terms the series rule can be inferred and condensed into series summation nomenclature ( Σ). It gives rise to a very nice little "sum i" with multiples of e^-2x^2 in the numerator and a fun little concatenation "π" product in the denominator.
Hi, I worked on this for years when I was young, until I found the polar solution in a book. But I'm glad to see that there is a method that avoids polar coordinates. Thanks a lot for this 👍
I really doubt polar coordinates since I don't know how it works. With this method, I believe now that the integral of e^-x^2 from -inf to inf is equal to sqrt of pi. Amazing.
You can imagine the coordinate and variable switch in the polar coordinate trick as allowing one to sweep from 0 to infinity in a circular way all at once (radially with r from 0 to infinity and circularly with theta from 0 to 2pi). Since the function is actually radially symmetrical from the origin and there is only an infinity (imagine it in 3D), with polar coordinates you do not need to split it up in two (-+inf,0] halves.
Start with the original integral: Integral e^(-x^2) dx Square it: (Integral e^(-x^2) dx)^2 = double integral e^(-x^2) * e^(-x^2) dx dx Change one of our variables of integration to y: double integral e^(-x^2) * e^(-y^2) dx dy Using properties of exponents, the product of the two exponential functions becomes e^(-x^2 - y^2): double integral e^(-x^2 - y^2) dx dy In polar coordinates, r^2 = x^2 + y^2.. The differential area dx*dy is equivalent to r dr dtheta. Thus: double integral e^(-r^2) r dr dtheta The limits on this integral are the full domain of x and y. In polar coordinates, it is 0 to infinity for r, and 0 to 2*pi for theta. This generates the derivative of the inside function, so we can use u-substitution. Work with the inner integral first, and theta isn't involved. integral r*e^(-r^2) dr Let u = -r^2. Thus du = -2*r dr, and dr = du/(-2), and the integral becomes: -1/2*integral e^u du, which evaluates to -1/2*e^u + C. In the r-world, it becomes: -1/2*e^(-r^2) + C We'd like to evaluate this from r=0 to r=infinity: (-1/2*e^(-inf^2)) - (-1/2*e^0) = 1/2 The outer integral is trivial, it's just integral 1 dtheta, which is theta + C. Evaluate from 0 to 2*pi, which is 2*pi. Multiply with the r-integral result, which gives us the result: [integral e^(-x^2) dx from 0 to infinity]^2 = pi Since we originally squared the integral, take the square root to get the original integral we want: integral e^(-x^2) dx from 0 to infinity= sqrt(pi)
Nice to see a different way, I really liked the polar method when it was explained in multivariable calc. I also liked the way the Fourier transformation is used to find the improper integral of sin(x)/x. Could be nice to see an alternative way to do that one.
Question at 7:36 : why is (1+t^2) constant? If we’re integrating over all x and all y, and t is defined as y/x, and the function in question isn’t linear, how could t be a constant?
It's constant with respect to the inner integral, not in general. Once you make the change of variables you no longer have to deal for these complications since the dependency has been accounted for. You can treat it as a proper variable from then on.
Sir , at 6:32 , i didnt understood the part where u divided y by x saying it is a positive number , so ♾️ /+ve = ♾️, but the value of x ranges from 0 to ♾️ and then we would get two sets of different integral forms ie ♾️/0 and ♾️/♾️ , can you pls explain
When you changed variables around 10:00, why is it assumed that t is independent of x (constant) even though t=y/x. I would understand if it was written that y=tx for whatever x in 0 to inf
Oh my god you are so amazing. I just loved it. The way you make it so easy for us is commendable. You are incomparable. Thank u so much. Because of you I am able to solve it without remembering polar coordinates typical method.
when you define y=xt therefore t=y/x, how can you say that 1+t^2 is constant in the x world? t=y/x so it varies with x, it doesn't seem constant to me. What am I getting wrong? thank you!
Good point. It wasn't a constant at first in the dt-integral right after substitution. But it does become a constant for the dx-integral after application of the Fubini-Tonelli theorem that changes the order of integrations. Now the inner integral runs over x values (x ist the integration variable) and t is a constant here. Only when evalutating the outer dt-integral, t is the integration variable and cannot be considered constant.
Question: If you are defining t=y/x, then x cannot be equal to zero. But you are letting that x be the same as the x in the other integral which is requires x=0 on the integral limits. How is this reconciled?
Yea. I think the function can be broken up when there are only a finite number of discontinuities and you can prove the absolute value after substitution still converges approaching the now asymptomatic discontinuity. It’s probably one of those theorems you prove using real analysis techniques. Not that I remember much of that.
This is amazing. Now I won't forget integral of e^(-x²) from 0 to ∞. I just have to learn the equation but knowing the proof just makes it easy to remember.
y = xt , when y goes to 0, the value of t does not have to be always 0. because in y = xt , the value of y also depends on x. in the integral we can see that x varies from 0 to infiniti. For example if x = 0, then t can have many values while xt remains 0. Can you explain why we assume when y = 0 that t = 0.
The value at some isolated point never effects the integral as a whole. The substitution is okay if still have convergence approaching zero. Polar coordinates has the same problem as theta can be anything at the origin.
That's a good question. @marshallsweatherhiking1820 already gave you the reason why the the substitution is still valid/correct. If you change the integrand of the Riemann integral at only a finite number of points, the integral stays the same. So the value of the integrand at x=0 doesn't matter and you can ignore it. You can also think of this as taking the integral on the *open* interval (0, +inf) so that x is always positive. Note that you're actually doing something similar on the upper limit of the integral: inf is not a real number that you can divide by x. Here, too, the upper limit itself is not actually included in the computation of the integral; a limiting process is used instead.
If you substitute t= y/x then you have substituted the tan of the angle for polar coordinates. Also the substitution for u is minus the square of the radius in polar coordinates. So you have used polar coordinates, it is just disguised.
I would love to see you use the Feynman trick with some rigorous explanations (uniform convergence) for the swap between the derivative and the integral! Keep up the good work 😉
The Feynman trick has nothing to do with uniform convergence though, you prove it using the dominated convergence theorem - it essentially requires to dominate the integral of the partial derivative
Very neat and direct. For those interested in a simple way, using polar coords, ( and simple to understand) I suggest you refer to Prime Newtons' video- it is nice for beginners.😊
Not apparent that use of iterated improper integrals as double improper integral and the interchange of integration in the iterated improper integrals behave this way. An explanation with improper integrals as limiting values would be lengthy. But with I(t) = (integral of e^(-x^2)) from 0 to t)^2 and g(t) =integral from 0 to 1 of e^(-t^2(x^2+1))/(1+x^2) with respect to x, we can show that I(t)+g(t)= is a constant = g(0) = integral of 1/(1+x^2) from 0 to 1 and is Pi/4. We can use differentiation under the integral sign here. Letting t tends to infinity you get your answer since g(t) tends to 0.
This was quite easy to follow - which is weird, I'm very bad at integrals. Honestly the only point I tripped up was in the end where 2*int[0,inf] (1/1+t^2) * dt became 2*tan^-1 (t) [0,inf]. But that is probably because I don't remember formulas for anti-derivative of 1/(1+x^2) and didn't know derivative of tan^-1 (x). My big fail in trigonometry is remembering all the formulas that can be used. I only remember that sin^2+cos^2 = 1. Very entertaining and informative video - thanks!
Very clever, I love it! I have a little problem with it though. I could be wrong, but it seems to have a dimensional mismatch between the two sides of the equation. The left-hand side, I, represents an area under the curve, which is 2-d, a number to the power of 2. The right-hand side is the square root of a double integral, which represents a volume-a number to the third power. Taking the sqr. root of that number produces a power of 3/2, which is not 2. Am I seeing it wrong?
Your definition of dimension is bizzare and certainly doesn't apply to integrals that flexibly. How many dimenisons cylinder have, knowing that it's integral is defined by DOUBLE integral in 2D polar coordinates multiplied basically by a constant? How many dimensions integral of sphere has, knowing that it's integral is TRIPLE integral? Inconsistent.
Also double integral came not from a 'natural observation' of a geometrical object, but was a product of multiplication of two usual integrals, which would rather double the 'dimensionality' of I becoming I^2 rather than adding 1 to it
Nothing says the dimensions have to match up, I think it was the squaring of an area and then labeling that extra area with a new variable that caused your confusion
e^(-x^2)->e^(-x^2-y^2) r=sqrt(x^2+y^2) tan(theta)=y/x ->e^(-r^2) So imagine spinning the e^(-x^2)=y graph around the y axis through your screen in the 3d dimension
Geometrically speaking, the area under 2/(1+x^2) from 0 to infinity is exactly the square of the area under e^-x^2 as is the case with pi*x*e^-x^2 from -infinity to infinity
He did hit on this method, and he later defines the Laplace Transform in the same book. So maybe it was the other way around, that thinking about this and also generating functions led him to transforms? Someone would have to go through his unpublished papers and see what he was working on, because he doesn't tell you how he came to something useful, just the thing he came to. (Usually after saying it can be easily seen....)
@@honourabledoctoredwinmoria3126 Thank you Sir for your prompt and informative reply. I was complimented by Professor Leif for how well I learnt Laplace and Inverse Laplace Transform to solve differential equations. I was a student in her ODE class at KCC(CUNY).
I havent gone to Uni yet so i don't know about calculus II, but, at the point where you derive Y = XT, since we count Y, X and T as variables, shoul the derivative of RHS be y+ x*dt? Since when you derive the product between two variables you usually do, por example: xy--> dx*y +x*dy
I think Laplace must have figured this by trial and error. He doesn't say how he got to the substitution. He actually mentions a few methods that work. The first is using a rather strange u substitution to turn the gaussian integral into an integral of the square root of a logarithm. The other is to calculate using this double integral with y = xt. I think he kept fiddling with it, knowing that the expression had no closed form antiderivative, but thinking he could use the properties of exponentials to get something he could solve. This was in the context of working out the properties of the normal distribution, so maybe he hit on using two variables that way.
There is an intimate connection between the circle and the complex exponential so this isn't too surprising. Honestly think of the complex numbers as the circle numbers for this reason.
Substitute s=x^2 and you get the gamma function for 1/2. Gamma(1/2)*Gamma(1-1/2)= pi/sin(pi/2) (Gamma reflection formula). Reason is that zeros of sine(pi*x)/pi function match the poles of this product of gamma functions , think of factorials. This is where pi comes from.
It's cute, but you cant hide from the trig on this one. You end up with the arctan anyway. Thats kinda obvious because the answer has pi in it, but still fun to watch you sweep it under the rug for as long as possible. I like different approaches, but I still like the move to polar better at the end of the day.
Apart from this awesome video supplied by ur easy explanation . Actually, Like the Gamma function, since there is no explicit integral to f(X) = e^-x² , another special function can easily be defined to be the distribution function F(X) = the integral of f(X) between -∞ and x. Obviously it is positive , strictly increasing and limited. and actually , since ex is equal to the Taylor series that has the terms 1/n! .X^n . Then e^-x² is equal has the Taylor series with terms : 1/n! . (-x²)^n . Thus F(X) can easily be obtained by integrating of that series. I think This can be helpful in numerical analysis as the you can have an approximation to F(X) by studying a polynomial of sufficiently large degree and dropping the rest of the power series.
I really love this video the only part that confused me was the y=xt part that felt a little out of no where but after I thought about it the function is just saying that some number y is equal to some number x times some scaler or constant of t i think you could of explained that it helped the rest of treating (1+t^2) as a constant seem less arbitrary.
I really love the way you solve this problem. But I quite don't understand why y and x have linear relationship? Why we can put y=xt but not other transformations?
I feel great, even though I have studied calculus only on my own, not in school yet, I was able to follow along with what's happening. And oh boi, it was beautiful
It is true that tan^-1 (inf) = pi/2 but arctan is a multivalued function. Any odd positive multiple of pi/2 is also inf. What is the rigorous reason we only consider pi/2 and not (2*n + 1)pi/2?
The bell shape created by e^(-x^2) is so beautiful as that created by other base constants instead of 'e'. At excel drawing ,eg, 12^(-x^2), gives that shape, with less variance, centered at 0 and same maximum value 1. Sure integrating this with the same tricks showed by Blackpenredpen, will obtain other finite razonable constant . So is very interesting explain why Gauss uses base 'e' instead of any other number (except the obvious not useful like base 1 or 0.5 or other between 1 and 0). Tradition ? Because 'e' is the "natural" base although irrational? Advantages of this? ...
When you do this same integral with base ‘a’, ie a^(-x^2) the result is sqrt(pi/log(a)), and with the natural log function popping into the result, you can already see how e is present in the result. By using base e, the result becomes sqrt(pi/log(e))=sqrt(pi). Aside from just being the nicest result, I’d assume this is the easiest to work with in stats and whatnot, although I’m not far enough into my studies to say for sure.
Learn more calculus from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)
Teacher, I follow the lessons you post on UA-cam
I am studying in the college of maths and physics number whats app please tell me you can help me with maths
sir from where I will get more videos of definite integration
ua-cam.com/channels/oLMpMr0JTdLZz4LPdvOf3A.html
Lol I don't have much time to go to brilliant when we have enough brilliant content on this channel.
Teacher may I know your telegram? I want to ask you something.
this truly is one of the most integral of all time
Truly. This integral is, in fact, an integral.
an integral integral
Indeed
The most integral?
This integral is indeed an integral
I like this more than the polar coordinates method because it is far easier to understand. No Jacobian, no different kind of coordinate system, just one substitution (really the second one you could do without the substitution, just by inspection).
There is no Jacobian probably because substitution is done in iterated integral only (he changed only one variable at the time)
Another approach is Gamma function with reflection formula
This is a computationally brilliant method for sure. However, compared to the polar approach the appearance of pi as a result comes across as the outcome of a somewhat artificial-looking substitution process. In contrast, the polar approach relies on the circular symmetry of the "Bell surface" about the z axis, which makes a pi-related value of the integral fairly obvious. Besides, one can relate the square root in the result to the Gaussian curve being a cross-section of that surface, whilst the squaring of the integral in Laplace's approach looks like a mere trick based on nothing but the factorizability of the exponential. For those who are interested, Dr Peyam in his channel does around 20 different derivations of this result!
@@FleuveAlphee what's an "artificial" looking substitution process?
@@holyshit922 That might actually be a line of circular reasoning, since many of the proofs of Gamma function integral rely on the Gaussian integral result!
@@Xoque551 not necessarily circuloar reasoning Reflection formula can be derived from product representation of Gamma function and Euler's product for sine
It is great seeing this integral done in cartesian coordinates. All the textbooks I used so far either used the approach over polar coordinates or just used the result. Thank you for this video!
Never knew you could solve this without using polar coordinates... excellent video!
I didn't know you could solve it with polar coordinates so I guess we balance the universe out somehow. 😆
@@abebuckingham8198 It's very quick to solve with polar coordinates, but there is also a 3rd geometric way to solve this integral.
They're the two proofs outlined in Wikipedia.
there are other several ways to prove this remarkable fact.
@@jacoboribilik3253 yes, Dr Peyam presents 12 of them in his video collection here ua-cam.com/play/PLJb1qAQIrmmCgLyHWMXGZnioRHLqOk2bW.html
I love this integral! Funnily enough in all the physics exams it is always just given 😅
😆
keep up great work sir
Because in physic we just use... In math it depends of the subject
In physics the solution to this integral is an intuitive truth.
You're luckyy in engineering my profs made us do it
Interestingly, this quite related to polar coordinates but avoids using them :). From the 2D point of view in the first quadrant, we used the coordinates (x,t) where (x,y) = (x,tx). t = y/x is the slope of the line starting from the origin passing through (x,y), in other words t = tan(theta) where theta is polar angle. Cool video :)
Thanks!!
Wow never thought about this, very interesting thanks!
As a stats guy, this is a beautiful detour from the more popular solution. I enjoyed the detailed steps and exploits of the symmetry inherent in the integral. Whilst the polar coordinates approach is easier to explain to anyone who has done trig, this solution is elagant af too.
I remember doing this integral shortly after learning about the jacobian.
There is so much joy in doing this integral for the first time, thank you Prof. Steve!
I had this on my calc 2 exam!! I approximated it using the mclaurin series for e to the x and then integrating that summation!! Cool video
I am getting 2 Σ(-x)^(2n+1)/((2n+1)(n!)) [0, inf]
isn't this diverging? I know I have done a mistake somewhere but I am not able to spot it
@@MohammadIbrahim-sq1xn i shouldve clarified but my integral was from 0 to x and the intial variable was t. Once you integrate the series you get Σ(-1)^n * x^(2n+1)/((2n+1)(n!)) which is the expansion of sin(x), cheers.
@@fooddrive8181 I have a doubt where did you get the (-1)^n from (this might be very silly but I can't seem to figure it out)
@@MohammadIbrahim-sq1xn the (-x^2) is (-1) (x^2) so in my expansion i applied the n to both from (a^m)^n = a^(m*n)
I watched 10 minutes of this stuff and we went from a simple formula to a bunch of mumbo jumbo. Can one of you calculus students explain to me what in the world this stuff is useful for? And be specific and have you found a job to where you are actually crunching out formulas to prove they are right or wrong for a living. Can't use teacher or professor either. Some other profession please.
As a student who hasn't taken Calc 3 but still uses parts of it, I really appreciate this video and seeing this explanation!
In 1995 the internet arrived. And this channel is the only good thing worth watching (just a long time waiting :)). Love your style and you teach like someone who understands rather than repeats. Thanks for your hard work!
Beautiful :) I did'nt know this trick. BTW when you switch to double integral and polar coordinates, you don't need Jacobian. You just can calculate the volume under the function as sum of volumes of cylinder shells. (Height of shell is e^(-x^2-y^2) = e^-r^2; length is 2 pi r and thickness is dr, so you integrate e^-r^2 . 2 pi r . dr from 0 to infinity.)
aninob
Yes. That is more elegant.
Plus I don't you understand his y = xt where his translation derivatives doesn't equal in dy where dy = x dt + a missing in his formulas t dx term:
dy = x dt + t dx I thought? 🤔
@@lawrencejelsma8118 In the substitution y = xt in the inner intergral is x in the role of constant. (Like "for given fixed x from the outer integral we shall calculate this inner integral..."). So the substitution y = xt is just recepee for transition from y to t. This is the reason why dy = x.dt.
@@aninob ... Yeah your correct. He should have written it out dy = xdt + tdx but the tdx term equals zero just like t can translate through the integral during summation. I was just use to "two space" partial differentiation mathematics of ut (ut)' = (u')t + u(t'). The recommended solution of real only or even when considering imaginary numbers are always polar solutions of rcosx + jrsinx solutions because trigonometric integral tables solutions are easier to understand.
@@MrPoornakumar 3B1B just made a beautiful video about it (plus some mathematical sugar on the top I never knew about). ua-cam.com/video/cy8r7WSuT1I/v-deo.html
The beauty of the mathematics lies not in the destination but in the elegance of the paths. Thanks for illustrating it.
Franchement très intéressant. Je ne connaissait que la méthode avec passage en polaire. Et j'étais persuadé que c'était la seule méthode possible ! Merci pour cette brillante présentation.
Google translates _la méthode avec passage en polaire_ as "the fleece method."
Love the aspect of switching the order of integration, I was wondering how it could be done. But then realised that it is ofcourse summation of product. At the start was confused as to how you could make the function 2* (0 to Infinity) of the f(x), but i had missed that this is a converging function. made math interesting again.. Thanks!
This integral is great. It's amazing how there are several connections between the Exponential function and pi, complex numbers and this at least.
I'm of the whacky opinion that complex numbers hide (or reveal) a door into other universe, Or universes. No evidence. Just a "gut feel".
@@starpawsy that doesn't mean anything
@@amineaboutalib Nope. Not a thing.
@@starpawsy Holy crankometer Batman, it's a kook!
@@holliswilliams8426 Oh look, I fully recognize the wackiness. That's ok. Im old enough not to care.
I have seen over 10 different proofs of this result, I don't think I have seen that one before. Great job!
Thank you! Cheers!
Hey, then you'll now have to make a video about the other 9 methods, to let the rest of us in on them! ;-)
Fred
This is a great video Steve; I've seen this integral solved before, but only with the usual polar coordinates method. Thanks!
This is a cool integral to know in that you can use integrals used by probability distributions to simply rewrite the integral in terms of it's probability distribution and then if they go from -inf to +inf they just become 1
Can you solve this problem?
Q. If f[{x + √(1 + x^2)}/x] = x^2. Then find f(x); domain & Range of f(x) =?
Video link:- ua-cam.com/video/YVboBTlmUo8/v-deo.html
My own technique also has value: create an infinite alternating series through iterative integration by parts. After creating the first few terms the series rule can be inferred and condensed into series summation nomenclature ( Σ). It gives rise to a very nice little "sum i" with multiples of e^-2x^2 in the numerator and a fun little concatenation "π" product in the denominator.
Nice route to solving a tricky integral. Great videos ... keep it up!
Thanks
Thank you!
Hi,
I worked on this for years when I was young, until I found the polar solution in a book.
But I'm glad to see that there is a method that avoids polar coordinates. Thanks a lot for this 👍
The way he swicthes pens is no less than a magician.
I didnt know this approach. Thank you for the very clear and instructive presentation.
how good are your tutorials? I passed a calculus 2 course with 70% with little to no help from my professor. keep up the great work man!
This has to be one of the most beautiful integrals out there
3:15 I was like “yeah I get it” 💪
No polar? Something against polar?
Glad I finally found someone doing this without going to polar coordinates.
Polar killed my father.
@@chitlitlah
Noooo!!!
change of variables would be calculus 3 ;)
Too COLD!
I really doubt polar coordinates since I don't know how it works. With this method, I believe now that the integral of e^-x^2 from -inf to inf is equal to sqrt of pi. Amazing.
You can imagine the coordinate and variable switch in the polar coordinate trick as allowing one to sweep from 0 to infinity in a circular way all at once (radially with r from 0 to infinity and circularly with theta from 0 to 2pi). Since the function is actually radially symmetrical from the origin and there is only an infinity (imagine it in 3D), with polar coordinates you do not need to split it up in two (-+inf,0] halves.
Start with the original integral:
Integral e^(-x^2) dx
Square it:
(Integral e^(-x^2) dx)^2 = double integral e^(-x^2) * e^(-x^2) dx dx
Change one of our variables of integration to y:
double integral e^(-x^2) * e^(-y^2) dx dy
Using properties of exponents, the product of the two exponential functions becomes e^(-x^2 - y^2):
double integral e^(-x^2 - y^2) dx dy
In polar coordinates, r^2 = x^2 + y^2.. The differential area dx*dy is equivalent to r dr dtheta. Thus:
double integral e^(-r^2) r dr dtheta
The limits on this integral are the full domain of x and y. In polar coordinates, it is 0 to infinity for r, and 0 to 2*pi for theta.
This generates the derivative of the inside function, so we can use u-substitution. Work with the inner integral first, and theta isn't involved.
integral r*e^(-r^2) dr
Let u = -r^2. Thus du = -2*r dr, and dr = du/(-2), and the integral becomes:
-1/2*integral e^u du, which evaluates to -1/2*e^u + C. In the r-world, it becomes: -1/2*e^(-r^2) + C
We'd like to evaluate this from r=0 to r=infinity:
(-1/2*e^(-inf^2)) - (-1/2*e^0) = 1/2
The outer integral is trivial, it's just integral 1 dtheta, which is theta + C. Evaluate from 0 to 2*pi, which is 2*pi.
Multiply with the r-integral result, which gives us the result:
[integral e^(-x^2) dx from 0 to infinity]^2 = pi
Since we originally squared the integral, take the square root to get the original integral we want:
integral e^(-x^2) dx from 0 to infinity= sqrt(pi)
Very cool to see someone so passionate about a topic that so many people wrongly think of as boring
Nice to see a different way, I really liked the polar method when it was explained in multivariable calc. I also liked the way the Fourier transformation is used to find the improper integral of sin(x)/x.
Could be nice to see an alternative way to do that one.
How beautiful the result and the way to solve it , thanks
I am so happy to live in a world where bprp exists! greetings from Brazil!!
Thank you
It is so satisfying to watch you explain the math.
(The first thing that catch my eye is the 荼果 doll under the e)
I have always loved your enthusiasm !! Also, nice way to solve the integral
Never mentioned Laplace. Would be interesting to hear just a little about when and how these guys came up with these things. We kinda owe them.
Question at 7:36 : why is (1+t^2) constant? If we’re integrating over all x and all y, and t is defined as y/x, and the function in question isn’t linear, how could t be a constant?
It's constant with respect to the inner integral, not in general. Once you make the change of variables you no longer have to deal for these complications since the dependency has been accounted for. You can treat it as a proper variable from then on.
You are one of the best maths teacher on internet
Quite impressive in terms of your presentation, well done
Sir , at 6:32 , i didnt understood the part where u divided y by x saying it is a positive number , so ♾️ /+ve = ♾️, but the value of x ranges from 0 to ♾️ and then we would get two sets of different integral forms ie ♾️/0 and ♾️/♾️ , can you pls explain
Thanks for your videos! It's fun to watch your process!
6:02 Slopping over possible divergence at x = 0. 8:10 Circular argument.
When you changed variables around 10:00, why is it assumed that t is independent of x (constant) even though t=y/x. I would understand if it was written that y=tx for whatever x in 0 to inf
Because of this video, now I understand how upper and lower bounds of integral change due to it's variable change. Thank you so much⭐
Oh my god you are so amazing. I just loved it. The way you make it so easy for us is commendable. You are incomparable. Thank u so much. Because of you I am able to solve it without remembering polar coordinates typical method.
when you define y=xt therefore t=y/x, how can you say that 1+t^2 is constant in the x world? t=y/x so it varies with x, it doesn't seem constant to me. What am I getting wrong? thank you!
√(1+t) the correct translation
When you change the variable X to U
Yes, can someone please explain the process
Good point. It wasn't a constant at first in the dt-integral right after substitution. But it does become a constant for the dx-integral after application of the Fubini-Tonelli theorem that changes the order of integrations. Now the inner integral runs over x values (x ist the integration variable) and t is a constant here. Only when evalutating the outer dt-integral, t is the integration variable and cannot be considered constant.
Question: If you are defining t=y/x, then x cannot be equal to zero. But you are letting that x be the same as the x in the other integral which is requires x=0 on the integral limits. How is this reconciled?
Yea. I think the function can be broken up when there are only a finite number of discontinuities and you can prove the absolute value after substitution still converges approaching the now asymptomatic discontinuity. It’s probably one of those theorems you prove using real analysis techniques. Not that I remember much of that.
Thank you,I’ve been thinking about a method to do it without polar coordinates cuz I didn’t learn them,great job ❤
Simple and brilliant, never occurred to me!
This is amazing. Now I won't forget integral of e^(-x²) from 0 to ∞. I just have to learn the equation but knowing the proof just makes it easy to remember.
I’m a masters in ML but I love math and calculus, love these videos❤
さとみくんの｢まず｣って言い方好きなの分かりますか？！🥺
youtubemn.com/watch?v=zZt0708hbPO
お母さんに言われた時は天国へのカウントダウンしよっかなって思っちゃったけど
nice, but what is ML??
@@janami-dharmam Machine Learning
@@zannyrt I see; these are rather infant sciences. Math is well established.
y = xt , when y goes to 0, the value of t does not have to be always 0. because in y = xt , the value of y also depends on x. in the integral we can see that x varies from 0 to infiniti. For example if x = 0, then t can have many values while xt remains 0. Can you explain why we assume when y = 0 that t = 0.
The value at some isolated point never effects the integral as a whole. The substitution is okay if still have convergence approaching zero. Polar coordinates has the same problem as theta can be anything at the origin.
That's a good question. @marshallsweatherhiking1820 already gave you the reason why the the substitution is still valid/correct. If you change the integrand of the Riemann integral at only a finite number of points, the integral stays the same. So the value of the integrand at x=0 doesn't matter and you can ignore it. You can also think of this as taking the integral on the *open* interval (0, +inf) so that x is always positive. Note that you're actually doing something similar on the upper limit of the integral: inf is not a real number that you can divide by x. Here, too, the upper limit itself is not actually included in the computation of the integral; a limiting process is used instead.
if y ranges from 0-infinity how can t do the same, ie if y is infinity then surely t wont be infinity aswell? 6:30
I dont understand calculus one bit, but something about your explanation style just drives me towards your videos
I´m calculus 2 student from Argentina and I understand it so well, I´ll believe in Chen Lu
If you substitute t= y/x then you have substituted the tan of the angle for polar coordinates. Also the substitution for u is minus the square of the radius in polar coordinates. So you have used polar coordinates, it is just disguised.
I would love to see you use the Feynman trick with some rigorous explanations (uniform convergence) for the swap between the derivative and the integral!
Keep up the good work 😉
The Feynman trick has nothing to do with uniform convergence though, you prove it using the dominated convergence theorem - it essentially requires to dominate the integral of the partial derivative
Dhanyawad bhaiyaa 🙏🏻🙏🏻.
Love from BHARAT 🇮🇳
When he was doing the first integral I wondered how the π would show up. Then he wrote down 1/2(1+t^2) and it hit me!
Very neat and direct. For those interested in a simple way, using polar coords, ( and simple to understand) I suggest you refer to Prime Newtons' video- it is nice for beginners.😊
this is just realy original, congratulations!
Hi bprp, thank you for the comprehensible and clear derivation, I am now practicing Laplace calculus.
Great job! Good pace and explanation, an alternative to polar conversion method. Thank You!
Not apparent that use of iterated improper integrals as double improper integral and the interchange of integration in the iterated improper integrals behave this way. An explanation with improper integrals as limiting values would be lengthy. But with I(t) = (integral of e^(-x^2)) from 0 to t)^2 and g(t) =integral from 0 to 1 of e^(-t^2(x^2+1))/(1+x^2) with respect to x, we can show that I(t)+g(t)= is a constant = g(0) = integral of 1/(1+x^2) from 0 to 1 and is Pi/4. We can use differentiation under the integral sign here. Letting t tends to infinity you get your answer since g(t) tends to 0.
This was quite easy to follow - which is weird, I'm very bad at integrals. Honestly the only point I tripped up was in the end where 2*int[0,inf] (1/1+t^2) * dt became 2*tan^-1 (t) [0,inf]. But that is probably because I don't remember formulas for anti-derivative of 1/(1+x^2) and didn't know derivative of tan^-1 (x). My big fail in trigonometry is remembering all the formulas that can be used. I only remember that sin^2+cos^2 = 1.
Very entertaining and informative video - thanks!
Very clever, I love it! I have a little problem with it though. I could be wrong, but it seems to have a dimensional mismatch between the two sides of the equation. The left-hand side, I, represents an area under the curve, which is 2-d, a number to the power of 2. The right-hand side is the square root of a double integral, which represents a volume-a number to the third power. Taking the sqr. root of that number produces a power of 3/2, which is not 2. Am I seeing it wrong?
Your definition of dimension is bizzare and certainly doesn't apply to integrals that flexibly. How many dimenisons cylinder have, knowing that it's integral is defined by DOUBLE integral in 2D polar coordinates multiplied basically by a constant? How many dimensions integral of sphere has, knowing that it's integral is TRIPLE integral? Inconsistent.
Also double integral came not from a 'natural observation' of a geometrical object, but was a product of multiplication of two usual integrals, which would rather double the 'dimensionality' of I becoming I^2 rather than adding 1 to it
Nothing says the dimensions have to match up, I think it was the squaring of an area and then labeling that extra area with a new variable that caused your confusion
∫ydx=∫∫dydx
What changed, really?
You could also try x=rcosa and y=rsina for solving double integral
Masterpiece . Thanks a lot for your great efforts ,Sir. 💖💖💖
Can someone explain why you can just "put the e^-x^2" inside the red integral? (4:00)
Because it's considered a constant according to the red integral
Nice vid!
Question: Geometrically, what does mean to square the integral?
probability? wrt y
e^(-x^2)->e^(-x^2-y^2)
r=sqrt(x^2+y^2)
tan(theta)=y/x
->e^(-r^2)
So imagine spinning the
e^(-x^2)=y graph around the y axis through your screen in the 3d dimension
Oh no, you watch Whatifalthist? 😂😂😂 The guy that says the scientific method is sh!t, and scientists are wrong?
Nothing at the first place. The gemetrical meaning only comes when you transform it into double integral.
Thank you so much You are the greatest teacher in the world🤩🤩🤩
Geometrically speaking, the area under 2/(1+x^2) from 0 to infinity is exactly the square of the area under e^-x^2 as is the case with pi*x*e^-x^2 from -infinity to infinity
Is this the method Laplace used to solve the Gaussian Integral? Was the Laplace Transform or Inverse Laplace Transform used in this method?
He did hit on this method, and he later defines the Laplace Transform in the same book. So maybe it was the other way around, that thinking about this and also generating functions led him to transforms? Someone would have to go through his unpublished papers and see what he was working on, because he doesn't tell you how he came to something useful, just the thing he came to. (Usually after saying it can be easily seen....)
@@honourabledoctoredwinmoria3126 Thank you Sir for your prompt and informative reply. I was complimented by Professor Leif for how well I learnt Laplace and Inverse Laplace Transform to solve differential equations. I was a student in her ODE class at KCC(CUNY).
You are better than 99% of calc teachers!
I havent gone to Uni yet so i don't know about calculus II, but, at the point where you derive Y = XT, since we count Y, X and T as variables, shoul the derivative of RHS be y+ x*dt? Since when you derive the product between two variables you usually do, por example:
xy--> dx*y +x*dy
Is there any rule to choose y=xt at the beginning ? Such that comes out 1/(1+t^2) and is just the intergrand of tan^-1 ? how brillant this choice !
I think Laplace must have figured this by trial and error. He doesn't say how he got to the substitution. He actually mentions a few methods that work. The first is using a rather strange u substitution to turn the gaussian integral into an integral of the square root of a logarithm. The other is to calculate using this double integral with y = xt. I think he kept fiddling with it, knowing that the expression had no closed form antiderivative, but thinking he could use the properties of exponentials to get something he could solve. This was in the context of working out the properties of the normal distribution, so maybe he hit on using two variables that way.
Yes because x²+y² is homogeneous so we put y= xt
Really still. Love how pi just comes out nowhere when we are doing something related to exponentials
There is an intimate connection between the circle and the complex exponential so this isn't too surprising. Honestly think of the complex numbers as the circle numbers for this reason.
Substitute s=x^2 and you get the gamma function for 1/2. Gamma(1/2)*Gamma(1-1/2)= pi/sin(pi/2) (Gamma reflection formula). Reason is that zeros of sine(pi*x)/pi function match the poles of this product of gamma functions , think of factorials. This is where pi comes from.
It's cute, but you cant hide from the trig on this one. You end up with the arctan anyway. Thats kinda obvious because the answer has pi in it, but still fun to watch you sweep it under the rug for as long as possible. I like different approaches, but I still like the move to polar better at the end of the day.
His face in the end saying "i love this one so much" is priceless. Greetings from Brazil
Well done. You made this very easy to follow. Thanks.
I don’t even know basic functions, except for linear functions, yet I still watch his videos as If I understand something.
Beautiful problem, balckpen! Thank you for sharing :)
Apart from this awesome video supplied by ur easy explanation . Actually, Like the Gamma function, since there is no explicit integral to f(X) = e^-x² , another special function can easily be defined to be the distribution function F(X) = the integral of f(X) between -∞ and x. Obviously it is positive , strictly increasing and limited. and actually , since ex is equal to the Taylor series that has the terms 1/n! .X^n . Then e^-x² is equal has the Taylor series with terms : 1/n! . (-x²)^n . Thus F(X) can easily be obtained by integrating of that series. I think This can be helpful in numerical analysis as the you can have an approximation to F(X) by studying a polynomial of sufficiently large degree and dropping the rest of the power series.
I really love this video the only part that confused me was the y=xt part that felt a little out of no where but after I thought about it the function is just saying that some number y is equal to some number x times some scaler or constant of t i think you could of explained that it helped the rest of treating (1+t^2) as a constant seem less arbitrary.
Thank you so much. This is the best explanation of this ive ever seen
OMGGGGGG Thanks you so much, i dont have words 😍😍😍
BPRP, can you integrate √(x+√x) with respect to x? Yes, there is an elementary answer :D
amazing sir👍today I had learnt little something,... and understood that, there is a lots of yet to learn ...
13:31 "And there's no +c"
CRIES OUTTA HAPPINESS
I really love the way you solve this problem. But I quite don't understand why y and x have linear relationship? Why we can put y=xt but not other transformations?
I really don't get why in 5:16 I can set y=xt, why would y be a linear function of x? Isn't that arbitrary to the problem?
No more words need to be said. Just... Elegant.
I feel great, even though I have studied calculus only on my own, not in school yet, I was able to follow along with what's happening. And oh boi, it was beautiful
It is true that tan^-1 (inf) = pi/2 but arctan is a multivalued function. Any odd positive multiple of pi/2 is also inf. What is the rigorous reason we only consider pi/2 and not (2*n + 1)pi/2?
The bell shape created by e^(-x^2) is so beautiful as that created by other base constants instead of 'e'.
At excel drawing ,eg, 12^(-x^2), gives that shape, with less variance, centered at 0 and same maximum value 1.
Sure integrating this with the same tricks showed by Blackpenredpen, will obtain other finite razonable constant .
So is very interesting explain why Gauss uses base 'e' instead of any other number (except the obvious not useful like base 1 or 0.5 or other between 1 and 0). Tradition ?
Because 'e' is the "natural" base although irrational? Advantages of this? ...
When you do this same integral with base ‘a’, ie a^(-x^2) the result is sqrt(pi/log(a)), and with the natural log function popping into the result, you can already see how e is present in the result. By using base e, the result becomes sqrt(pi/log(e))=sqrt(pi). Aside from just being the nicest result, I’d assume this is the easiest to work with in stats and whatnot, although I’m not far enough into my studies to say for sure.
This is one of the youtube videos of all time
how are you able to treat the "t" as constant in the first integral when it depends on x and the integral is with respect to x?
10:31 / 15:00 but -00 is least than 0 so it must be integral -00 to 0