$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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\newcommand{\mrm}[1]{\mathrm{#1}}
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\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\int_{0}^{\infty}\expo{-2ax}\,{\sin^{2}\pars{bx} \over x^{2}}\,\dd x =
\int_{0}^{\infty}\expo{-2ax}\,\
\overbrace{{1 - \cos\pars{2bx} \over 2}}^{\ds{\sin^{2}\pars{bx}}}\
\overbrace{\int_{0}^{\infty}t\expo{-xt}\,\dd t}^{\ds{1 \over x^{2}}}\
\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Re\int_{0}^{\infty}t
\int_{0}^{\infty}\bracks{%
\expo{-\pars{t + 2a}x} - \expo{-\pars{t + 2a - 2b\ic}x}}\,\dd x\,\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{\infty}t\bracks{%
{1 \over t + 2a} - {t + 2a \over \pars{t + 2a}^{2} + \pars{2b}^{2}}}\,\dd t
\\[5mm] = &\
{1 \over 2}\int_{0}^{\infty}\bracks{%
1 - {2a \over t + 2a} - {-2a\pars{t + 2a} + \pars{t + 2a}^{2} \over \pars{t + 2a}^{2} + \pars{2b}^{2}}}\,\dd t
\\[5mm] = &
{1 \over 2}\int_{0}^{\infty}\bracks{%
-\,{2a \over t + 2a} + 2a\,{t + 2a \over \pars{t + 2a}^{2} + \pars{2b}^{2}} +
{\pars{2b}^{2} \over \pars{t + 2a}^{2} + \pars{2b}^{2}}}\,\dd t
\\[5mm] & =
\left.{1 \over 2}\,2a\ln\pars{\root{\pars{t + 2a}^{2} + \pars{2b}^{2}} \over t + 2a} +
b\arctan\pars{t + 2a \over 2b}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty}
\\[5mm] = &\
\bbx{\ds{-\,{1 \over 2}\,a\ln\pars{a^{2} + b^{2} \over a^{2}} +
b\arctan\pars{b \over a}}}
\end{align}