Complex Analysis: Integral of (x-sin(x))/x^3

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Today, we use contour integration to evaluate the integral of (x-sin(x))/x^3 from -infinity to infinity.

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it's nice how you got the integral on the gamma path, but you could calculate it as (θ2-θ1) i Res(f(z), z=0), we must consider that it is triple pole so it would give something like (1/2!) times second derivative of z³*(1+iz-e⁽iz)/z³)). some like that: (0-Pi)*i*(1/2) and the second derivative of 1+iz-e⁽iz is equal to 1.

hectore.garcia

That was amazing !!! I love this series. I'll definetly use your playlist to lean complex analysis once i'm ready.

daniellindner

Nice Integral with another beautiful sollution! 💚
I think an equally interesting film on your part would be a video with Hankel contour.
It would be amaizing to see Riemann's Zeta function and Euler's Gamma function in that form...

nightmareintegral

I've evaluated this one before with complex analysis, but I think I ended up using Feynman's technique on the sine term and remove one factor of x from the numerator and denominator. This allowed me to take the real part instead of imaginary part and prevented the contour avoiding the origin from being divergent.

TheRandomFool

Very nice! As The1RandomFool said, this is a prime candidate for Feynman Integration with f(t)=(tx-sin(tx))/x^3. Taking 2 derivatives wrt t will generate a well known(to all of us) value. Then re-integrate twice and done!

theblainefarm

I found two things absolutely brilliant, because they definitely were not immediately clear to me:

1. Your clever selection of the complex contour integral
2. Your use of Taylor series in solving for little gamma.

I am going to try solving this myself now, without cheating by watching your video again. Thank you, your methods are clear, articulate and in this case truly brilliant. Your solution was awesome.

fordtimelord

Elegant as always! loved that small trick! but I have a question:

I do get why we added a +1, but Why didn't we just applied contour integration on the imaginary part of e^iz/z^3? since separating the function would give us a trivial 1/x^2 and sin(x)/x^3?

manstuckinabox

Can u do sinx /x^3 integral from - infinity to infinity

bhavyajain

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