$\int_{0}^{\infty}{\sin^2bx\over x^2}\cdot{\mathrm dx\over e^{2ax}}=b\tan^{-1}\left({b\over a}\right)-{a\over 2}\ln{\left(a^2+b^2\over a^2\right)}?$ Consider $(1)$

$$\int_{0}^{\infty}{\sin^2bx\over x^2}\cdot{\mathrm dx\over e^{2ax}}=I\tag1$$
  $(a,b)>0$

How does one show that $$I=\color{blue}{b\tan^{-1}\left({b\over a}\right)-{a\over 2}\ln{\left(a^2+b^2\over a^2\right)}}?$$
An attempt:
Using $\sin^2(x)={1-\cos(2x)\over 2}$
$(1)$ becomes
$${1\over 2}\int_{0}^{\infty}{e^{-2ax}\over x^2}-{e^{-2ax}\cos(2bx)\over x^2}{\mathrm dx}=I\tag2$$
$$I_1-I_2=I\tag3$$
$I_1$ is diverges, so else can we tackle $(1)?$
 A: Consider $\displaystyle F(a)=\int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx\\$
Double differentiating w.r.t 'a' we have,
$\displaystyle F''(a)=4\int\limits_0^\infty e^{-2ax}\sin^2 bx \; dx$
And a little integration by parts gives the last integral which simplifies as,
$\displaystyle F''(a)=\dfrac{1}{a}-\dfrac{a}{a^2+b^2}\\$
Now integrating , $\displaystyle F'(a)=\ln a-\dfrac{1}{2}\ln|a^2+b^2|+C_1$
Integrating for the last time we have,
$\displaystyle \int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx=b\arctan\left(\dfrac{a}{b}\right)-\dfrac{a}{2}\ln\left(\dfrac{a^2+b^2}{a^2}\right)+C_1a+C_2$
To eliminate the constants use $F'(1)=-\dfrac{1}{2}\ln|1+b^2|$ which will give $C_1=0$ and putting $b=0$ in the last equation will give $C_2=0$ and the result follows,
$\displaystyle \color{blue}{\int\limits_0^\infty e^{-2ax}\dfrac{\sin^2 bx}{x^2}\; dx}=\color{red}{b\arctan\left(\dfrac{a}{b}\right)-\dfrac{a}{2}\ln\left(\dfrac{a^2+b^2}{a^2}\right)}$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}\expo{-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}
