One of the most beautiful calculus results you'll see! Solved using the Laplace transform 

Beautiful interplay of the two techniques! This result was marvelous.

nice. however there is an elementary solution in one line using the residue theorem applied to the function Re(e^(iz)/(z^2+1)).

Very nice method! Really liked it. Thank you🙏🙏

Can somebody tell me how all of this is used in the real world

Another aprroach is to use Fourier -transforms:Consider the integral of Exp[i x y]/(1+x^2) ,which corresponds to √(2*π) * Fourier transform of 1/(x^2+1). The FT of 1/(x^2+1) is √(π/2)*Exp[- | y |]. For y=1 you get π/e . To see that the Fourier transform of 1/(x^2+1) is given by the term written before , one can just use the inverse FT of Exp[- | y |] which is easy to calculate.

Absolutely gorgeous one ❤️‍🔥

Wow! Great result and techniques for solving! Thank you👍

which software you use to write like these?

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But what if replace the cos(x) with sin(x) ? It surely converge (seems converge to 0.64...). , but I can't work out. I get it equal to infnity*0

The technique used is by no means simple. I would assume that someone who is able to do that also knows residue theory .Use that you may take the real part of Exp[ i x] and use that 1/(x^2+1)= 1/2 i (1/(x-i)-1/(x+ i) ,which leads to π * i *(-i *Exp[-1] = π/e

Are U kidding us?

Very nice.

Beautiful! This is a combination of the Feynman Technique and the Laplace Transform right?

Why should I use such a long method? There are much shorter methods

❤❤❤❤❤❤ Beautiful interplay of techniques!!

wow... very nice, stunning result

Let's say I integrate from -infty to +infty the function x, wrt x. Can I use the fact that x is an odd function in order to say that the I definite integral is zero? The same way here, how do we know the parity on an indefinite integral applies ? Because the function you're integrating is bounded maybe? In that case why is it a valid argument?

such an underrrated math channel

Love from India 👍

beautiful solution.

Thank you very much!

Nice video along a touristic route of mathematical techniques. Contour integration in the complex plane makes this a one-liner.

awesome

Nice!

Beaufiful result. Can also be accomplished using complex analysis.

Hi You skipped too many core concept steps :(

I'm not very sure on 0:30 passage: function is even ok, but doubling the integral and behalfing the integration domain in case of integrand even function works only when extremes of integration are finite. If extremes are not finite we don't know how each extreme approaches its infinity (+inf and -inf respectively) and so we don't necessarily have a symmetric integration area. I think the passage is true only in the principal value of the integral sense.

I've always found this result mind-boggling. I mean who asked for that e? 😂Made even more strange by the introduction of cosx into the integrand of all things! Results like this seem to be hints to a higher-order generationalization of e and trigonometric functions via complex analysis.