Specific approach to integral Problem: Evaluate $$\int_{0}^{\infty} \dfrac{\sin^3{x}}{x \cdot e^x} dx=\dfrac{A\pi}{B}-\dfrac{\tan^{-1} (C)}{D},$$
Attempt through Differentiation under the Integral Sign:
Using $\sin^3x=\frac{3\sin x-\sin 3x}{4}$, rewrite as
$$ \frac{3}{4}\int_0^\infty \frac{e^{-x}\sin x}{x}\,dx-\frac{1}{4}\int_0^\infty \frac{e^{-x}\sin 3x}{x}\,dx $$
$$ I(b)=\int_0^\infty \frac{e^{-x}\sin bx}{x}\,dx\qquad,\qquad\mbox{where}\,\,b>0 $$
$$ \begin{align} I'(b)&=\int_0^\infty e^{-x}\cos bx\,dx\\ \end{align} $$
How to proceed, please?
 A: We may use:
$$ \int_{0}^{+\infty}\frac{f(x)}{x\,e^{x}}\,dx = \int_{1}^{+\infty}(\mathcal{L}\,f)(t)\,dt \tag{1}$$
so we just need to find the Laplace transform of $\sin^3(x)$. 
Pretty easy task through the triplication formulas, since $\mathcal{L}(\sin x)=\frac{1}{1+t^2}$ implies:
$$ \mathcal{L}(\sin^3 x) = \frac{3}{4}\left(\frac{1}{t^2+1}-\frac{1}{t^2+9}\right)=\frac{6}{(1+t^2)(9+t^2)}\tag{2}$$
then:
$$ \int_{0}^{+\infty}\frac{\sin^3 x}{x\,e^{x}}\,dx = \frac{3\pi}{16}-\frac{3}{4}\int_{1}^{+\infty}\frac{dt}{t^2+9}=\color{red}{\frac{3\pi}{16}-\frac{\arctan 3}{4}}.\tag{3}$$
A: Hint. You may just write, for $b \in \mathbb{R}$,
$$
\int_0^{\infty}e^{-x}\cos bx\: dx=\Re\int_0^{\infty}e^{-(1+ib)x} dx=\Re\left. \frac{e^{-(1+ib)x}}{-(1+ib)}\right|_0^{\infty}=\frac{1}{1+b^2}.
$$
Alternatively, integrating by parts twice:
$$
\begin{align}
I(b)=\int_0^{\infty}e^{-x}\cos bx\: dx&=\left.-e^{-x}\cos bx\right|_0^{\infty}-b\int_0^{\infty}e^{-x}\sin bx\: dx\\\\
&=1-b\times\left(\left.-e^{-x}\sin bx\right|_0^{\infty}+b\int_0^{\infty}e^{-x}\cos bx\: dx\right)\\\\
&=1-b^2I(b)
\end{align}
$$ giving $\displaystyle (1+b^2)I(b)=1$ and
$$
I(b)=\frac{1}{1+b^2}.
$$
