Evaluating $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$ How to evaluate the following integral?
$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$
 A: 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
$$\int_0^{\infty}\frac{e^{-kx}\sin x}{x}\,dx=\arctan k^{-1}.$$
A: 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
$$\frac{\pi}{2 \pi} \int_{-1}^1 \frac{dw}{k-i w} = \frac{i}{2} \log{\left( \frac{i k + 1}{i k -1}\right)} = \arctan{\frac{1}{k}} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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)}
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\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}
A: 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
$$F(k)=\arctan(\frac 1k)$$
A: One more option:$$\begin{align}\int_0^\infty\int_0^\infty e^{-(k+y)x}\sin x\mathrm{d}x\mathrm{d}y&=\Im\int_0^\infty\int_0^\infty e^{-(k+y-i)x}\mathrm{d}x\mathrm{d}y\\&=\int_0^\infty\tfrac{1}{(k+y)^2+1}\mathrm{d}y\\&=[\arctan(k+y)]_0^\infty\\&=\tfrac{\pi}{2}-\arctan k.\end{align}$$
