TITLE: Solution to this integral?
QUESTION [8 upvotes]: could anyone solve this integral ?
$$\int_0^\infty \frac{e^{-x}\sin(x)\cos(ax)}x~\mathrm dx$$
well i have tried opening up the sin*cos using trigonometric identities but that didn't help so much

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}{\expo{-x}\sin\pars{x}\cos\pars{ax} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}
\int_{0}^{\infty}{\expo{-x}\sin\pars{\bracks{1 + a}x} \over x}\,\dd x +
{1 \over 2}
\int_{0}^{\infty}{\expo{-x}\sin\pars{\bracks{1 - a}x} \over x}\,\dd x
\\[5mm] = &\
{1 \over 2}\pars{1 + a}
\int_{0}^{\infty}\expo{-x}\,
{\sin\pars{\bracks{1 + a}x} \over \pars{1 + a}x}\,\dd x\ +\
\pars{a \leftrightarrow -a}
\\[5mm] = &\
{1 \over 2}\pars{1 + a}
\int_{0}^{\infty}\expo{-x}\,
{1 \over 2}\int_{-1}^{1}\expo{\ic\pars{1 + a}xk}\,\dd k\,\dd x\ +\
\pars{a \leftrightarrow -a}
\\[5mm] = &\
{1 \over 4}\pars{1 + a}
\int_{-1}^{1}\int_{0}^{\infty}\exp\pars{\bracks{- 1 + \ic\pars{1 + a}k}x}
\,\dd x\,\dd k\ +\
\pars{a \leftrightarrow -a}
\\[5mm] = &\
{1 \over 4}\pars{1 + a}
\int_{-1}^{1}{-1 \over -1 + \ic\pars{1 + a}k}\,\dd k\ +\
\pars{a \leftrightarrow -a}
\\[5mm] = &\
{1 \over 4}\pars{1 + a}2
\int_{0}^{1}{\dd k \over \pars{1 + a}^{2}k^{2} + 1}\ +\
\pars{a \leftrightarrow -a} =
{1 \over 2}
\int_{0}^{1 + a}{\dd k \over k^{2} + 1}\ +\
\pars{a \leftrightarrow -a}
\\[5mm] = &\
\bbx{\arctan\pars{1 + a} + \arctan\pars{1 - a} \over 2}
\end{align}