compute a integral value given another integral value given for $\displaystyle t>0;\int_0^\infty\frac{\sin x}{x}e^{-tx}dx=\frac{\pi}{2}-\arctan t$, compute the value of $\displaystyle\int_0^\infty\frac{(1-e^{-x})\sin x}{x^2}dx$
idk where to start, i thinked in integration by part but
$$u=\frac{1-e^{-x}}{x^2}\Rightarrow du=\frac{(x+2)e^{-x}-2}{x^3}dx\\
dv=\sin xdx\Rightarrow v=-\cos x\\
\int_0^\infty\frac{(1-e^{-x})\sin x}{x^2}dx=\left.\frac{(e^{-x}-1)\cos x}{x^2}\right|_0^\infty+\int_0^\infty\frac{[(x+2)e^{-x}-2]\cos x}{x^3}dx$$
but i don't think that help here, so i don't know what to do or how to arive in a form where i can use first result.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\pars{1 - \expo{-x}}\sin\pars{x} \over x^{2}}\,\dd x & =
\int_{0}^{\infty}{\sin\pars{x} \over x}\
\overbrace{\pars{\int_{0}^{1}\expo{-xt}\,\dd t}}
^{\ds{1 - \expo{-x} \over x}}\
\,\dd x =
\int_{0}^{1}\color{red}{\int_{0}^{\infty}{\sin\pars{x} \over x}\expo{-xt}
\dd x}\,\dd t
\\[5mm] & =
{\pi \over 2} - \int_{0}^{1}\arctan\pars{t}\,\dd t =
{\pi \over 2} - \bracks{{\pi \over 4} - \int_{0}^{1}{t \over t^{2} + 1}\,\dd t}
\\[5mm] & =
\bbx{{\pi \over 4} + {1 \over 2}\,\ln\pars{2}}
\end{align}
A: Hint
Your intuition was correct, you need to integrate by part, taking:
$$u=(1-e^{-x}) \sin(x)$$
$$dv=\frac{1}{x^2}dx$$
then:
$$v=\frac{-1}{x}$$
and:
$$ du=\left(e^{-x} \sin(x)+(1-e^{-x}) \cos(x) \right) dx$$
