There is some 'complex' stuff happening here...

A nice integral related to the Gaussian integral.

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[tomeknowak]

10:30 The fact that we can integrate over half-line from 0 to infinity, instead of half-line from 0 to alpha times infinity is a consequence of Cauchy integral theorem. In my opinion doing it the same way, as we would in real analysis course, without any explanation is a certain abuse.

[matthiasbergner]

10:28 : There is something I do not understand: You say that u is just a real variable. But x is already a real number and u = alpha * x where alpha is a complex number with nonzero imaginary part, so u cannot be a real number. Technically, the integral in the u-variable is not along the positive real axis but along some straight ray in the complex plane. I am not sure whether this integral has the same value as the wellknown Gaussian integral.

[tharunsankar]

The frickin integral screamed: RESIDUE AND JORDAN’S LEMMA to me when I saw this monster lol

[TheRandomFool]

I used Feynman's technique with the extra parameter in the sine function. Differentiating allowed me to eliminate x^2 in the denominator. I then complexified the cosine and turned it into a contour integral in the complex plane. My substitution is beta=sqrt(1+a*i), where a is my extra parameter. Michael Penn's substitution is sketchy here. He makes a leap to a real-valued integral without showing why that's the case. With the proper application of complex analysis, it can be shown it does. Much of my work is the same as his here, but at the end, I have an additional integral to evaluate.

[shanestrickland]

15:57 : The moment I realized you were going to use the 1/2 formula that way, set the calculation god knows how many steps b4, and busted out laughing like a madman. I trade stocks and options for a living now adays... My God I miss solving these problems! Some days, watching your channel is what gets me through day of life in a cold dark world that has abandoned logic, math, and physics. Seriously, thank for your dedication!

[yt]

Somebody pointed at the back of our professor with laser. Euler maybe? 😛

[clementcoine]

At 9:17 this really requires some justifications using Cauchy's formula. But well, the result is correct :)

[cernejr]

Wolfram Alpha:, approx. 0.81, about 90% of the value without the sin(x^2)/x^2 which is sqrt(pi)/2 as mentioned in the video.

[vaibhavcm]

I think you can easily solve it, by the substitution x²=t, and then use the complex definition of sinx and then some Laplace transform.

[aproductions]

Representing sin and cos as real and imaginary parts of a complex exponential is a nice little trick -- I always expand them out to their full form as a sum of complex exponentials and it ends up being a real mess for a little bit before I can collapse it back down.

[thierrychenevier]

Nice integral, thanks !! Two small comments : first, as we are talking about integration among real numbers, the "trick" to use complex notations might be valid, and might not. As already legitimately noted by a few other comments, this is not as obvious as "goes to infinity so everything is ok"... and a bit of an additionnal justification might be useful. Second, when you say that cos (pi/8) - sin (pi/8) is equal to the square root of its square, you still need to check whether this number os positive, since the square root of a square is the absolute value of the number, not always the number.

[hudson]

Here is one idea.

After the first simplification, one needs to calculate the integrals of sin(x^2)e^(x^2) and cos(x^2)e^(x^2), which are the imaginary and real parts of the integral of e^{ix^2}e^{x^2} = exp((i-1)x^2). This is a real function with values in R^2 (or C) and we integrate each coordinate individually.

Next, define F(y) = \int exp((iy-1)x^2) dx, for real values of y. If we can differentiate inside the integral, F'(y) is F(y) divided by a complex like -i(1-yi). The solution of this complex ODE will be a F(0) times a term with the sqrt of (1-iy). After that, one just calculates F(1).

The advantage of this method is that one doesn't need residuals nor limits on C. Also, it is not hard to see we can really differentiate inside the integral using the exponencial decay and uniform convergence (a la Weiestrass).

Hopefully we get the same answer.

[yoantardy]

Great video ! I want to say that I really enjoy watching these videos that are really original and kind of "high level" compare to other maths videos. I have an alternative proposition for the step of the integration by part : maybe putting a parameter t in the exponential and calling the all fonction phi (t), see it as an parameter integral and derive it with respect to t. You have a simplification beween the square of x that divide the expression and the square of x that arrive because of the derivation and you have directly a calcuable expression of the derivative of phi (t) which is calculable by the same way you are doing in the end of the video.

[victoriakurt]

Why does he write the infinity symbol like that?

[mihaipuiu]

Very interesting integral and, ...of course very nice proof.

[_skyslayer]

I don't agree that at 9:17 the boundaries of integration do not change. integral_0^oo is a completely different thing from integral_0^(oo*i). In this example I believe you should end up with an integral_0^(oo + oo*i), which you should split into a sum of integral_0^oo + integral_0^(oo*i). I bet the second part of this sum tends to 0, but you didn't show it!

[viktoryehorov]

btw, it could be solved using Ramanujan's master theorem

[mrminer]

Improper at x = 0. Basically you're looking at sin x approximated by x, divided by x^2, so it blows up at x =0

[immolator]

Can someone explain to me why we can assume that u is a real number (10:20), although it comprises of alpha which is a complex number?

[antoninnguyen]

Sorry I don't understand at 10:00 how can Re(exp(-u^2))/α is equal to Re(α)× exp(-u^2)