Calculating residues to solve an integral

**arbolis** · August 3rd 2010, 02:32 PM

The integral is $I=\int _0 ^{+\infty} \frac{x \sin (x)}{x^2+2}dx$ . I'll probably choose a semi circle as contour since the poles aren't on the real axis. Furthermore I think that $I=\frac{1}{2} \int _{-\infty}^{+\infty } \frac{x \sin (x)}{x^2+2}dx$ .
Let $f(z)=\frac{z \sin z}{z^2+2}$ . The singularities of f are when $z=\pm i \sqrt 2$ .
Now I want to calculate the residue of f at say one singularity. But it gives infinity I think. So I believe it's a pole of order greater than 1 but I'm unsure. I tried to use the Taylor series of sin(z) in order to see what happens in $z=i \sqrt 2$ but I didn't reach anything.
So I don't know how to go further.

**Vlasev** · August 3rd 2010, 02:44 PM

There shouldn't be a problem there. Can you show me your calculation of the residue of f?

**chiph588@** · August 3rd 2010, 03:46 PM

Generally for integrals like this involving $\sin(z)$ or $\cos(z)$ , you'll need to switch them out for $e^{iz}$ . Otherwise the integrals over parts of your contour might not go to zero.

Therefore I would integrate $\displaystyle f(z)=\frac{z e^{iz}}{z^2+2}$ .

Note that $\displaystyle \int _{-\infty}^{\infty } \frac{x\sin (x)}{x^2+2}dx = \text{Im}\left(\int _{-\infty}^{\infty } \frac{ze^{iz}}{z^2+2}dz\right)$ .

**arbolis** · August 3rd 2010, 05:16 PM

Good idea chip. However I didn't have more chance.
$Res(f,i \sqrt 2 ) = \lim _{z \to i \sqrt 2} \frac{z^2e^{iz}}{z^2+2}=\frac{e^{iz}}{1+\frac{2}{z ^2}}$ . The numerator tends to $e^{-i \sqrt 2}$ while the denominator goes to 0 which is really dramatic. It happens whether or not I choose $e^{iz}$ or $\sin (z)=\frac{e^{iz} - e^{-iz}}{2i}$ since the denominator doesn't change.

**TheEmptySet** · August 3rd 2010, 05:31 PM

$\displaystyle f(z)=\frac{ze^{iz}}{z^2+2}=\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}$

Now take the limit

$\displaystyle \lim_{z \to i\sqrt{2}}(z-i\sqrt{2})\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}=\lim_{z \to i\sqrt{2}}\frac{ze^{-iz}}{(z+i\sqrt{2})}=\frac{i\sqrt{2}e^{\sqrt{2}}}{i 2\sqrt{2}}=\frac{e^{\sqrt{2}}}{2}$

**arbolis** · August 3rd 2010, 05:34 PM

Thanks TheEmptySet! I missed that despite you've done the same yesterday!

**chiph588@** · August 3rd 2010, 05:34 PM

Originally Posted by TheEmptySet

$\displaystyle f(z)=\frac{ze^{iz}}{z^2+2}=\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}$

Now take the limit

$\displaystyle \lim_{z \to i\sqrt{2}}(z-i\sqrt{2})\frac{ze^{-iz}}{(z-i\sqrt{2})(z+i\sqrt{2})}=\lim_{z \to i\sqrt{2}}\frac{ze^{-iz}}{(z+i\sqrt{2})}=\frac{i\sqrt{2}e^{\sqrt{2}}}{i 2\sqrt{2}}=\frac{e^{\sqrt{2}}}{2}$

I think the residue should be $\displaystyle \frac{e^{-\sqrt{2}}}{2}$ .

You accidently had $\displaystyle e^{-iz}$ instead of $\displaystyle e^{iz}$ .

**arbolis** · August 3rd 2010, 05:38 PM

Originally Posted by chiph588@

I think the residue should be $\displaystyle \frac{e^{-\sqrt{2}}}{2}$ .

You accidently had $\displaystyle e^{-iz}$ instead of $\displaystyle e^{iz}$ .

I was about to ask the same. I get the residue = $\frac{e^{-\sqrt{2}}}{2}$ .
Edit: I get the answer now. I hope my justification is good enough. (it's very similar to the one of Chip in a previous thread).

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