compute a integral value given another integral value

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 2

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: $$ du=\left(e^{-x} \sin(x)+(1-e^{-x}) \cos(x) \right) dx$$