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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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    @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

2 Answers 2

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Edit: Since the definition of the exponential integral incorporates Cauchy principle values, I did not previously write which integrals were in fact principle values. This has now been changed.

We can evaluate it as a sum of exponential integrals by integrating along the quarter circle contour:

Consider $\int_0^\infty \frac{e^{iz}}{1+z^2}dz.$ The residue at $x=i$ is $\frac{\pi}{e}$ which in particular tells us the value of the related integral $\int_0^\infty \frac{\cos(x)}{1+x^2}dx=\frac{\pi}{2e}.$ However, we care about the imaginary part, not the real part. Lets integrate on the contour which is the quarter circle of radius $R$ in the right half plane which avoids the point $z=i$ by going around a half circle of radius $\epsilon$. Then, in the limit as $\epsilon\rightarrow 0$ and $R\rightarrow \infty$ we have that $\int_0^\infty \frac{e^{iz}}{1+z^2}dz=i\left(p.v.\int_0^\infty \frac{e^{-z}}{1-z^2}dz\right)+\frac{2\pi}{e}.$ The portion on the circle of radius $R$ goes to zero by Jordans lemma, and the $\frac{\pi}{2e}$ comes from the fractional residue theorem.

Looking at the imaginary parts of both sides we conclude $\int_0^\infty \frac{\sin(z)}{1+z^2}dx=\int_0^\infty \frac{e^{-z}}{1-z^2}dz$ Split this up using partial fractions to get $\frac{1}{2}\left(\int_{0}^{\infty}\frac{e^{-u}}{u+1}du-p.v.\int_{0}^{\infty}\frac{e^{-u}}{u-1}du\right).$ Lets turn each of these into an exponential integral by shifting the lower limit of integration to $0$. We then have $\frac{1}{2}\left(e\int_{1}^{\infty}\frac{e^{-t}}{t}dt-e^{-1}\left(p.v.\int_{-1}^{\infty}\frac{e^{-t}}{t}dt\right)\right).$ This last line is then equal to $\frac{e^{-1}Ei(1)-eEi(-1)}{2}$ by definition, and we have evaluated the integral in terms of known functions.

It is highly unlikely that you can write this in a more satisfying way without in turn implying relations about exponential integrals. Also, the term $-eEi(-1)$ is a constant called Gompertz Constant.

Hope that helps,

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    @Didier: I think you are being too critical of the revised version, most likely because of lack of confidence stemming from the completely **blatant error** I had before with $\sinh (z)$ . The decisions for the contours were straightforward, and understandable from "quarter circle." The fact that the integrals were principle values followed from the context of exponential integrals, and the evaluation of the integral, (such as on $|z|=R$) was also straightforward. In any case, I added these details to make the answer more complete.2011-06-01
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Wolfram Alpha (and so I suppose Mathematica) gives for the indefinite integral

$\int \frac{\sin(x)}{1+x^2} dx = $

$ \frac{(e^2 - 1) (Ci(i+x) + Ci(i-x)) + i (e^2 + 1) (Si(i+x) + Si(i-x))}{4 e} + \text{constant}$

where $Ci(x)$ is the Cosine integral and $Si(x)$ is the Sine integral.

Looking at the graph of this, the imaginary part seems to be constant for real $x$ and the real part seems to tend to $0$ for large $x$, and if so the the answer to original question is the negative of the real part of this expression when $x=0$, i.e. about $0.64676112277913$.

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    Remark: When we evaluate at $x=0$, by using the fact that $Ei(-x)=\text{ci}(ix)+i\left(\frac{\pi}{2}+\text{si}(ix)\right)$ it is possible to rewrite this expression in the form given in my answer.2011-05-31