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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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    $\begingroup$ @syxiao: Evaluate $\int_{\Gamma} \frac{e^{iz}}{1 + z^2}dz$, where the contour $\Gamma$ is a large semicircle in the upperhalf plane. $\endgroup$
    – JavaMan
    May 31, 2011 at 21:32
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    $\begingroup$ @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. $\endgroup$ May 31, 2011 at 21:51
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    $\begingroup$ @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. $\endgroup$
    – syxiao
    May 31, 2011 at 21:54
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    $\begingroup$ @syxiao: Not interested in accepting answers to your other posts? $\endgroup$
    – Did
    May 31, 2011 at 22:03
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    $\begingroup$ Similar: math.stackexchange.com/questions/9402/… $\endgroup$
    – Aryabhata
    Jun 1, 2011 at 0:38

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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    $\begingroup$ Eric: Your first step (where you replace sin by sinh) lacks justification. As a matter of fact (unless a miracle occurs in this specific case, which you should explain), this step is false. $\endgroup$
    – Did
    May 31, 2011 at 23:05
  • $\begingroup$ @Didier: You are completely right. I edited to provide a complete solution. I don't know why, but when I was looking this over I thought I saw using $\sinh(z)$ as a shortcut.... $\endgroup$ Jun 1, 2011 at 1:07
  • $\begingroup$ Hmmm... Either you are supremely confident in your contour integration skills, or the revised version goes too fast. For example, you could write precisely which contour you use (i.e. a closed path avoiding the pole at i (or else...), but how does it avoid this point), explain why the |z|=R and |z-i|=epsilon parts of the contour do no contribute to the end result (if indeed they do not), and avoid writing integrals which simply do not exist (or in a very specific sense which you should explain), like int(exp(-t)/t,t=-1..+infty). .../... $\endgroup$
    – Did
    Jun 1, 2011 at 6:04
  • $\begingroup$ .../... Note in particular that Gompertz constant appears in the first integral, the one where you divide by u+1, and certainly not in the second integral, which does not exist. Sorry. :-) $\endgroup$
    – Did
    Jun 1, 2011 at 6:05
  • $\begingroup$ @Didier: I am supremely confident in my contour integration skills. Partly joking, but yes this version does go fast, I leave out many steps, and the fact that some of these integrals are principle values. Some details: The integral clearly dies on the circle radius $R$ by Jordan's lemma. The contour does a half circle at $z=i$ and this contributes by the fractional residue theorem. Because we are taking limits to approach $i$, this integral is then a Cauchy principle value. Since $Ei(z)$ incorporates principle values by definition, I did not mention it. $\endgroup$ Jun 1, 2011 at 15:28
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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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    $\begingroup$ 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. $\endgroup$ May 31, 2011 at 22:46

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