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This integral is surprisingly difficult to evaluate, and I have looked in several references and none contain a single integral of this type. Any help would be greatly appreciated.

Evaluate $\displaystyle \int_0^\infty \frac{\sin(z)}{1 + z^2}dz$.

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    @syxiao: Evaluate $\int_{\Gamma} \frac{e^{iz}}{1 + z^2}dz$, where the contour $\Gamma$ is a large semicircle in the upperhalf plane.2011-05-31
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    @DJC: if the semicircle is where I think it is, won't the integrals over the positive and negative parts cancel? You want a quarter-circle, I think.2011-05-31
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    @DJC: The problem with using that integrand is precisely because the imaginary part is 0; or that the integral $\displaystyle \int_{-\infty}^\infty \frac{sin(x)}{1+x^2}dx = 0$. This doesn't give any information on the integral of only one side.2011-05-31
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    @Qiaochu Yuan: I couldn't figure out how to make the quarter-circle work, since the integral along the imaginary axis is also very hard to compute.2011-05-31
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    @syxiao: Not interested in accepting answers to your *other* posts?2011-05-31
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    The integral along the imaginary axis also has the added difficulty of $z = i$ being a singularity of the integrand.2011-05-31
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    @Qiaochu: Hmm. Good point. This will require some more thought then.2011-05-31
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    Ahlfors, Complex Analysis, 3rd edition, deals with $\int_{-\infty}^{\infty}R(x)\sin x\,dx$ at some length on pages 156-159, where $R(x)$ is a rational function with a zero of order at least 2 at infinity. I know this is a bit different, with lower limit of integration zero, but still it may be possible to apply what's in Ahlfors.2011-05-31
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    Similar: http://math.stackexchange.com/questions/9402/calculating-the-integral-int-0-infty-frac-cos-x1x2-without-using-c2011-06-01
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    @Qiaochu, @Syxiao: Using the quarter circle is the most reasonable approach. It allows us to rewrite the integral in a more slightly nicer form. (Specifically as $\int_0^\infty \frac{e^{-x}}{1-x^2}dx$, see answers for details) As syxiao mentions, this integral is also very hard to compute, and can only be written as a sum of exponential integrals, or equivalent such forms.2011-06-01

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