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$$\int_0^\infty \frac{\sin(2\pi x)}{x(x^{2}+3)}$$

I looked at $\frac{e^{2\pi i z}}{z^{3}+3z}$, also calculated the residues, but they don't get me the right answer. I used that $\int_{-\infty}^\infty f(z)dz = 2\pi i (\sum \operatorname{Res} z_{r}) + \pi i Res_{0}$, but my answer turns out wrong when I check with wolframalpha.

Residue for $0$ is $1$, for $z=\sqrt{3}i$ it's $-\frac{e^{-2\pi}}{2}$ . . .

In a worse attempt I forgot $2\pi$ and used $z$ only (i.e. $\frac{e^{iz}}{z^{3}+3z}$) and the result was a little closer, but missing a factor of 2 and and $i$.

Can anyone see the right way? Please do tell.

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    How do you get the residue at 0 to be 1?2011-11-20
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    Once you calculated the residue of $f(z) = \frac{e^{2\pi i z}}{z(z^2+3)}$ at $z = 0$ and $z = i\sqrt{3}$, you will get $$\int_{-\infty}^{\infty} = 2\pi i \operatorname{Res}_{z=i\sqrt{3}} f(z) + \pi i \operatorname{Res}_{z=0} f(z).$$ Then the actual answer must be $\frac{1}{2i} \int_{-\infty}^{\infty} f(z) \; dz$. I guess this is what you have missed.2011-11-20
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    Hey sos440, I do get the right answer with $\frac{1}{2i}$, the $\frac{1}{2}$ comes from the integration boundaries, and the $\frac{1}{i}$ because sine is the imaginary part of $e^{iz}$, right?2011-11-20
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    @VVV , what contour are you taking for your calculations?? Also, after the limit when $\,R\to\infty\,$ (or whatever) is done, we still remain with the real part of the complex function ($\,cos 2\pi x\,$ instead of $\,\sin 2\pi x\,$, which does *not* converge on $\,(0,\infty)\,$... Also, you seem to be taking a contour that contains only the positive imaginary pole and not the negative one.2012-07-07
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    @sos440 , what contour are you taking for your calculations? and why the integral is multiplied by $\,1/2i\,$ and not by $\,1/2\pi i\,$ ?2012-08-20
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    @DonAntonio, The contour I used is a semicircular contour with small indent at the origin, see [here](http://en.wikibooks.org/wiki/File:ContourSinzz.gif). Also, we are not calculating the residue of $f(z)$, which means that there is no need to divide the integral by $2\pi i$. The factor $\frac{1}{2i}$ arises because our goal is to evaluate $$\Im \int_{0}^{\infty} f(z) \; dz = \frac{1}{2i}\int_{-\infty}^{\infty}f(z)\;dz.$$2012-08-20
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    @sos440 , thanks for the comment. Would you mind writing down an answer for this (and thus rescue this poor question from the unanwered questions' dragon)? I understand that Jordan's lemma gives us that the integral on the indentend part tends to zero when epsilon does, and on the large arc also the same whe $\,R\to\infty\,$, yet I still cannot get the correct result which according to WA is $$\frac{\pi}{6}\left(1-e^{-2\sqrt 3 \pi}\right)$$. I can't get that one inside the parentheses.2012-08-20

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