Integral of 1/(x^2+1) from -inf to inf, Contour Integral

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In this freestyle, right on the spot lecture, Dr. Peyam explains how to integrate 1/(x^2+1) from -inf to inf by using contour integration from complex analysis. This is a great starter for the people who are interested in this topic!

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100/(1-x)

[CLeonard]

Best explanation of contour integral I have seen. Hopefully, y'all continue with a series of more complex integrals.

[gsniteesh]

Please make a video on complex analysis :)

[victorpaesplinio]

This video summarizes my calc 4 exam for next month on residue theorem.
Thank you very much!

[whyit]

Marvel: Infinity War is the most ambitious crossover event!


Bprp and Dr. Peyam: Hold my markers

[Vampianist]

This is the first time I know that a lot of Peyam’s videos were taken in his office. Always liked the picture beside the board.

[habouzhaboux]

Finally I found a worth watching video about contour integrals

[tobyinsley]

Honestly this is the most intuitive and clear explanation I've found of the process of finding bounds with triangle inequality etc - thanks Dr. Peyam, this was very useful!

[itachi]

Is it me or is this a lot more interesting than the usual videos? It gives an insight to how you'd work in university.

[FractalMannequin]

To compute the residue just notice that 1/(z^2+1) = 1/(z-i) * 1/(z+i) and the second factor has no poles at z=i, so its Laurent expansion coincides to its Taylor expansion. The -1-th coefficient of 1/(z^2+1) now must be 1 (i.e., 1/(z-i)'s -1-th coefficient) times the 0-th coefficient of 1/(z+i), which is its value at z=i, namely 1/2i. So the residue is 1/2i (or -i/2).

[Erik]

Complex analysis is *really* cool but for people to understand it from the ground and then people don’t have to assume it’s magic :)

Basically the reason why the residue is important can be explained by the Taylor series of the (analytic, meaning it has a complex derivative) function, breaking it up into a sum of different powers of z.

Every term has a nice primitive function [remember integral of x^n=(x^(n+1))/(n+1) ] *except* the 1/z term! (What’s kind of interesting in real analysis is really important in complex analysis here.) It’s the derivative of the complex logarithm which is multi valued.

If we integrate on a closed curve, are start and endpoint is the same and so those terms with a primitive function integrate to zero! The exception is 1/z and famously the (complex) logarithm increases by 2*pi*i in one lap around the origin.

So the only thing we care about when integrating on a closed curve is where the function has any singularities (else the 1/z term coefficient is necessarily 0 as the function is “nice”) and in those singularities, what the coefficient of 1/z is, and we call that the residue!

(Note: some singularities have residue 0, like 1/z^2 in z=0, and do not contribute to the integral.

It also does not have to be a simple “pole”, you can also integrate e^(1/z) around z=0 for example. This is an essential singularity where the function approaches infinity and 0 in the same point but it does not matter!)

[egillandersson]

Hokusai's Great Wave and Van Gogh's Starry Night : Peyam is definitely the Artist of mathematics !
Well, about contour integral, I have to work a bit more :-(

[kcldnx]

I just had complex analysis last semester and saw the thumbnail. I immediately thought residue theorem and it was super fun to see you use it

[atrath]

This video brought back some memories from graduate school... I really miss those times...

[nishan]

Love dr. P. Very approachable person and loves his subject.

[cbbuntz]

Peyam's enthusiasm for math is infectious

[RapidScience]

Wow this is amazing, he made some amazing points throughout the video

[IoT_]

Hello, Dr. πm) I would like to say thank you that you invited blackpenredpen to your lecture because they asked very useful questions for novices. As for me, I was graduated from the engineering faculty of the russian university and surprisingly I had all these stuff with proofs (surprisingly because even mathematicians have less mathematics than us- engineers ). Anyway, thank you that you reminded me of the best subject - Complex Analysis) . By the way, Residue in Russian means "Вычет" 😁

[bandamkaromi]

wow!! Dr. Peyam. Brilliant try. Thank you. BlackPenRedPen!!

[manfredwitzany]

Calcualtion of the residuum is much easier than demonstrated. As the pole is of first order, you can use the following identity for residuum calculation:

\textstyle \operatorname {Res}_{a}f=\lim _{{z\rightarrow a}}(z-a)f(z)
The term (z-a) cancels out and the remaining function value is equal to the residuum. I have integrated some even more complex functions. This method is pretty easy.

[chris-hjqd]

Man, how beautiful is it to see a culture that appreciates learning. So wonderful.