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I would like to evaluate the integral $$\int_{-\infty}^\infty \frac{\sin x}{x-i} dx,$$ which I believe should be equal to $\frac{\pi}{e}$. However, I cannot reproduce this result by hand. My work is as follows: first, we evaluate the indefinite integral.

\begin{align*} \int \frac{\sin x}{x-i} dx &= \int \frac{\sin(u+i)}{u} du \text{ where }u=x-i \\ &= \int \frac{\sin u \cos i + \cos u \sin i}{u} du \\ &= \mathrm{Si}(u) \cos i + \mathrm{Ci}(u) \sin i \\ &= \mathrm{Si}(x-i) \cosh 1 + i\mathrm{Ci}(x-i) \sinh 1 \\ \end{align*}

Then, we insert the bounds.

\begin{align*} \int_{-\infty}^\infty \frac{\sin x}{x-i} dx &= \mathrm{Si}(\infty-i) \cosh 1 + i\mathrm{Ci}(\infty-i) \sinh 1 \\ &\phantom{=}-\mathrm{Si}(-\infty-i) \cosh 1 - i\mathrm{Ci}(-\infty-i) \sinh 1 \\ &= \frac{\pi}{2}\cosh 1 + 0 +\frac{\pi}{2} \cosh 1 + \pi \sinh 1 \\ &= \pi (\cosh 1 + \sinh 1) \\ &= \pi e \end{align*}

I assume I have made some mistake in manipulating the complex sine and cosine integrals, since the result would be correct if $\mathrm{Ci}(-\infty-i)$ were evaluated to $-i \pi$. However, I cannot pinpoint the error.

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    $\begingroup$ $u=x-i$ but you have sustituted it by $x+i$. I think that's the problem. $\endgroup$
    – S -
    Commented Jul 25, 2014 at 7:07
  • $\begingroup$ @Adolfo Thanks, I've fixed the signs on $i$. However, this doesn't seem to change anything. $\endgroup$ Commented Jul 25, 2014 at 7:11
  • $\begingroup$ @Adolfo No, I take that back--$\mathrm{Ci}$ appears to have a branch cut across the negative real axis, which is the source of the issue. Thanks! $\endgroup$ Commented Jul 25, 2014 at 7:18
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    $\begingroup$ Try to use contour integration and see what you get. $\endgroup$ Commented Jul 25, 2014 at 7:44

4 Answers 4

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Write the sin-function as the difference of two exponentials and obtain the corresponding integrals by contour integration. This gives the result: \begin{eqnarray*} I &=&\int_{-\infty }^{+\infty }dx\frac{1}{x-i}\sin x=\int_{-\infty }^{+\infty }dx\frac{1}{x-i}\frac{1}{2i}\left\{ e^{ix}-e^{-ix}\right\} \\ \int_{-\infty }^{+\infty }dx\frac{1}{x-i}e^{ix} &=&2\pi ie^{-1} \\ \int_{-\infty }^{+\infty }dx\frac{1}{x-i}e^{-ix} &=&0 \\ I &=&\frac{1}{2i}2\pi ie^{-1}=\frac{\pi }{e} \end{eqnarray*}

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  • $\begingroup$ Suppose you used the Residues Theorem in the second step. $\endgroup$
    – MathArt
    Commented Mar 16 at 15:26
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The mistake is simple--$\mathrm{Ci}$ has a branch cut across the negative real axis, so $\mathrm{Ci}(-\infty - i)$ should indeed evaluate to $-i \pi$ rather than $i \pi$.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\begin{align} \color{#f00}{\int_{-\infty}^{\infty}{\sin\pars{x} \over x - \ic}\,\dd x} & = \int_{-\infty}^{\infty}{x\sin\pars{x} \over x^{2} + 1}\,\dd x = \Im\int_{-\infty}^{\infty}{x\expo{\ic x} \over x^{2} + 1}\,\dd x \\[5mm] = & \ \Im\pars{2\pi\ic\,{\ic\expo{\ic\ic} \over \ic + \ic}} = \color{#f00}{{\pi \over \expo{}}} \end{align}

The integration was performed along a semi-circle in the upper half complex plane by using the Residues Theorem.

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  • $\begingroup$ Quote: "However, I cannot pinpoint the error." How does this pinpoint the error? $\endgroup$
    – Did
    Commented May 19, 2016 at 8:23
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It is also possible to use integral transforms of Laplace and Fourier.

\begin{align} \int_{-\infty}^{\infty}{\sin{x} \over x-i}\,\mathrm{d}x&= 2\int_{0}^{\infty}{x\sin{x} \over x^{2} + 1}\,\mathrm{d}x\\ &=2\int_{0}^{\infty}\mathcal{L}^{-1}\left[{s \over s^{2} + 1}\right](x)\mathcal{L}\left[\sin{t}\right](x)\,\mathrm{d}x\\ &=2\int_{0}^{\infty}\frac{\cos(x)}{x^2+1}\mathrm{d}x\\ &=\frac\pi e. \end{align}

The last step is not easy either to do in a fundamental approach but the generalized one is given here. How can I demonstrate the following equality: $\int_0^\infty \frac{\cos(yx)}{y^{2}+\lambda^{2}}\, dy = \frac{\pi}{2\lambda}e^{-\lambda x}$ $$\frac{\pi}{2}\mathrm{e}^{-|t|} = \int_{0}^{+\infty}\frac{\cos(xt)}{x^2+1}\mathrm{d}t.$$

Just found as well here: What is wrong in this calculation $\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx$?.

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