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How to evaluate the following integral?

$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$

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    $\begingroup$ In the future, please try to make the title of your questions more informative (I've done so for you now). E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. Furthermore, please do not use excessively big fonts; they come across as shouting. $\endgroup$ – Lord_Farin Jun 18 '13 at 10:38
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Differentiating with respect to $k$, one finds an easily computable integral $$-\int_0^{\infty}e^{-kx}\sin x\,dx=-\frac{1}{1+k^2}$$ Integrating back with respect to $k$ and using that for $k\rightarrow\infty$ the integral is $0$, we obtain $$\int_0^{\infty}\frac{e^{-kx}\sin x}{x}\,dx=\arctan k^{-1}.$$

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You can use Parseval's theorem, i.e.

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dw \, F(w) G^*(w)$$

where $F$ and $G$ are the Fourier transforms of $f$ and $g$, respectively. In this case:

$$f(x) = e^{-k x} \theta(x) \implies F(w) = \frac{1}{k-i w}$$

$$g(x) = \frac{\sin{x}}{x} \implies G(w) = \begin{cases}\pi & |w| < 1 \\ 0 & |w| > 1 \end{cases}$$

Then the integral is equal to

$$\frac{\pi}{2 \pi} \int_{-1}^1 \frac{dw}{k-i w} = \frac{i}{2} \log{\left( \frac{i k + 1}{i k -1}\right)} = \arctan{\frac{1}{k}} $$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty} {\expo{-kx}\sin\pars{x} \over x}\,\dd x} = \int_{0}^{\infty}\expo{-kx} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx}\,\dd q}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty} \expo{-\pars{k + \ic q}x}\,\,\,\dd x\,\dd q = {1 \over 2}\int_{-1}^{1}{\dd q \over k + \ic q} = {1 \over 2}\int_{-1}^{1}{k - \ic q \over q^{2} + k^{2}}\,\dd q \\[5mm] = &\ \int_{0}^{1/k}{\dd q \over q^{2} + 1} = \bbx{\arctan\pars{1 \over k}} \\ & \end{align}

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Note that the integral is

$$\mathcal{L}\bigg(\frac {\sin(t)}t\bigg)=F(k)$$ Where $\mathcal{L}$ is the Laplace transform operator

You can find the transformation using power series, which ultimately gives $$F(k)=\arctan(\frac 1k)$$

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One more option:$$\begin{align}\int_0^\infty\int_0^\infty e^{-(k+y)x}\sin x\mathrm{d}x\mathrm{d}y&=\Im\int_0^\infty\int_0^\infty e^{-(k+y-i)x}\mathrm{d}x\mathrm{d}y\\&=\int_0^\infty\tfrac{1}{(k+y)^2+1}\mathrm{d}y\\&=[\arctan(k+y)]_0^\infty\\&=\tfrac{\pi}{2}-\arctan k.\end{align}$$

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