Complex Analysis: Gaussian Integral

[@konchady1]

What makes this proof elegant is also that you used the general rule of making contour integrals useful. That is, if you don't have poles or singularities, add the appropriate ones to get the result. This is itself an art.

[@elahehghanati4011]

Thanks a lot. In my doctorate research I use Gaussian integrals as I work on laser beam propagation. Your solution now will be part of my thesis appendix and I will cite your name. I feel really better knowing this integral solution process. Do you know what will be the result for exp(-z*x^2), where z is complex number?

[@konchady1]

Brilliant !! For most of my life, I was looking for a countour proof that didn't involve that annoying parallelogram contour. You cleverly changed it into something much more elegant and that is commendable.

[@digxx]

Did you come up with it yourself, or where is it from? Btw your argument about ib=tau being complex technically defines a parallelogram can be made a bit more rigorous for every finite R (large enough so that the pole is inside), by noting that you can deform the contour without crossing any singularity.

[@dijkstra4678]

The end was incredibly satisfying

[@datsmydab-minecraft-and-mo5666]

Absolutely amazing!

[@arielfuxman8868]

Awesome. Huge appreciation!

[@mustafaunal1834]

Thank you! This method is ingenious! Who discovered the idea?

[@nasim09021975]

Bravo! A real tour de force! 😊👍

[@Rundas69420]

What journey man :)

[@Calcprof]

The value of τ makes the size of the "rectangle" slightly scew, so that the upper integral is not precisely from R to -R, but its only off a little, and in the limit it down't matter

[@mihaigabrielbabutia4595]

When calculating the limit of the integral over gamma_2, do we have a guarantee that the limit and the integral can be interchanged for the argument? I was thinking of the dominated convergence theorem for Lebesgue integration here, that may be the key here. Very interesting approach, I only saw the method of evaluation by double integral, but this is far more elaborate and pleasant to follow!

[@qncubed3]

Yes, most of the time I assume DCT with the interchange of limits. Usually it's quite clear, but you can always bound the integrand further.

[@mihaigabrielbabutia4595]

@@qncubed3 thank you. Everything else was very clead and detailed enough to answer any possible question on-the-fly

[@reinaldomaiasilvafilho1754]

Such a nice integral

[@dfk1991]

Hi! What about integrals INVOLVING Gaussian functions? I know by default it is not suitable for a contour enclosed at infinity, as the Gaussian diverges in both directions of the imaginary axis, but any tricks on that?

[@arielfuxman8868]

30:10 Is it formal? Interchanging of an integral and a limit

[@qncubed3]

You would have to use DCT to show it properly, but it's quite clear in this example. You could also factor e^(-R^2) out of the integral (which goes to 0) and it's clear the remaining integral is finite.

[@bonelesspizza6311]

I think there might be a mistake at 19:18. how does dividing by sqrt(pi*i) on both sides give sqrt(pi*i)? you gave tao/2 but it would be pi*i/(2)(sqrt(pi*i) and then the odd integer. could I get why we got that fraction instead?

[@aamirshaikh474]

ipi = τ², and sqrt(ipi) = τ , τ²/ τ leaves us with a τ term on the right side

[@geremiasjunior138]

Hello, how do you define the vertices of the rectangle, for example, I'm solving an integral whose denominator is cosh (pix / 2), how can I get the upper vertices, would it be something like R + ib, -R + ib?

[@qncubed3]

It depends on what the integral is. In most cases you will not need a rectangular contour

[@geremiasjunior138]

@@qncubed3 the integral I'm calculating needs, it is of type 1 / cosh (ax) as -inf to inf, but is there any method to define this vertex?

[@qncubed3]

@@geremiasjunior138 The cosh function has singularities at every odd multiple of i*pi. For this function, you can choose b=pi so that your horizontal contour paths lie in the middle of each singularity. Hopefully this helps!

[@geremiasjunior138]

@@qncubed3 it helped a lot, thanks, so the question depends on your uniqueness, I love your channel, I watch many of your videos, keep it up.

[@qncubed3]

@@geremiasjunior138 thanks :) I may consider doing a similar integral in the future as it's a nice application of rectangular contours, which are not common.

[@ian731]

When do I use a rectangle as a contour?

[@qncubed3]

When the imaginary part of the poles are periodic

[@ian731]

@@qncubed3 how i see this in this gaussian integral

[@qncubed3]

@@ian731 The function we are using to integrate has periodic poles

[@ian731]

@@qncubed3 Would have like to see first the function, like look and already realize that the countour is rectangular

[@ian731]

@@qncubed3 I say only see f(z) could have an idea if the countour would be rectangular, if yes which specific functions usually have this behavior

[@ld1ego_733]

How would it be resolved if it were e^-(ax^2)?🦊

[@qncubed3]

You can make a substitution for sqrt(a)*x and you'll be left with the integrand we had in the video.

[@hectore.garcia2244]

What would justify your assumption that g(z)=g(z+tao), what leads you to think that this is actually the case?

[@user-wu8yq1rb9t]

I want to learn complex analysis, you think this example (Gaussian integral) is good place to start or you have another suggestions for me? (And I should tell you I'm familiar with this kind of thing (complex analysis), and I just to practice more and ... ) Thank you

[@qncubed3]

Take a look at my contour integration playlist on my channel, I've arranged my complex analysis videos from basic examples to more complicated ones

[@user-wu8yq1rb9t]

@@qncubed3 Thank you so much

[@Serghey_83]

Wow😮

[@yateendrasihag3319]

Hey thanks for the video but can you tell me how do you write "exp(sqrt(i))=exp(i*pi/4)" at 23:13 ?

[@qncubed3]

sqrt(i)=i^(1/2)=(e^(i*pi/2))^(1/2)=e^(i*pi/4)

[@yateendrasihag3319]

@@qncubed3 got it, Thank you very much, I appreciate this!!

[@zakirreshi6737]

interesting :) 👍🏻👍🏻

[@tubevortex]

Oooff nice

[@aweebthatlovesmath4220]

Hey can't you say integral over gamma 2 is equals to mines integral over gamma 4 so integral over gamma 4 is negative zero so zero.

[@qncubed3]

Although they are similar, they are two different paths in the complex plane and we will have to check both of them

[@shuewingtam6210]

Why f(z+ib) but not f(z-ib)? On 7:28

[@qncubed3]

Let z=w+ib Then w=z-ib For the bounds: w_start=-R w_end=R Then the integrand: f(z)=f(w+ib) And dz=dw Then I just replaced w with z so that both integrals are with respects to the same variable.

[@dynastieeyala4236]

Si seulement. Il expliquait tout ça en français j'aillais peut-être comprendre

[@c_ornato]

En gros l'intégrale Gaussienne est complètement Zehef par sa méthode de dingo et il utilise une division bien Lacoste TN bien yeux jaunes pour que la somme de deux constituantes du contour soit égale à l'intégrale et pour que les deux autres soient à 0