Need help in complex integration

  • Thread starter krindik
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  • #1
Can somebody help me with this.

Homework Statement

[tex]\int_{-\infty}^{\infty} \, \frac{\sin{x}}{x} \, dx[/tex]
Could u pls advice me with the procedure to follow not only the answer?

The Attempt at a Solution

1. Use complex numbers as there is a pole of order=0 at x=0

\int_{-\infty}^{\infty} \! f(x) \, dx = 2\pi\, i \sum_{res\, upper\, hp} {f(x)} \, + \pi\, i \sum_{res\, real\, axis} {f(x)}

which give 0 as the answer

2. Expand by sin(x) by Taylor series around 0 and multiply by x this gives a divergent series

Couldn't figure out which is correct?


Homework Statement

Homework Equations

The Attempt at a Solution


Answers and Replies

  • #2
have you drawn the contour for this problem? the fact that there is a simple pole at x=0 (i.e. on the real axis) means you'll need a small semi-circular indent to take care of this.

also you can use Cauchy's residue theorem to get the integral round the whole contour then your probably going to need some sort of analysis with Jordan's lemma to eliminate the contributions from the parts of the contour not on the real line...

oh it will be easier to consider [itex]\int_{-\infty}^{\infty} \frac{e^iz}{z} dz[/itex] and then take the imaginary part of this at the end.
  • #3
I tried the contour as in image" [Broken]

taking the pole to z=0.

I used the following theorem


\textbf{Principal Value}\left\{\int_{-\infty}^{\infty} \! f(x) \, dx \right\} = 2\pi\, i \sum_{res\, upper\, hp} {f(x)} \, + \pi\, i \sum_{res\, real\, axis} {f(x)}

But there are no poles in upper half plane so the first sum is zero.

There's a pole at x=0 on real axis. Residue at x=0 is zero.

I'd really like to how u came to the below conclusion
\int_{-\infty}^{\infty} \frac{e^iz}{z} dz[/itex]

is it by experience or is there any standard procedure u followed?
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  • #4
Science Advisor
Homework Helper
If you just look at sin(z)/z on that contour you can't control the size of the function on the large semiarc. Use sin(z)=(exp(iz)-exp(-iz))/(2i) and split the integral into two parts. Now exp(iz) is small if z has a positive imaginary part and vice versa for exp(-iz). So close the outer part of the contour in the upper half plane for the first and the lower half plane for the second. Now pick out which one of those integrals contains the pole and work it out. Pay attention to the orientation of the contour.

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