Without power series, can we compute $\int_0^{\infty} \frac{\sin (ax)}{x e^x} d x$? Inspired by the question in the post with a beautiful result
$$
\int_0^{\infty} \frac{\sin x}{x e^x} d x=\frac{\pi}{4} ,
$$
I found that it can be generalised further to
$$
\boxed{\int_0^{\infty} \frac{\sin (ax)}{x e^x} d x=\arctan a} ,\quad \textrm{ where }a\ne 0.
$$
which is wonderful too.
To prove it, we can use the power series of
$$
\sin x=\sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n+1}}{(2 n+1) }.
$$
Plugging the series into the integrand yields
$$
\begin{aligned}\int_0^{\infty} \frac{\sin (a x)}{x e^x} d x&=\sum_{n=0}^{\infty} \frac{(-1)^n}{\left(2n+1\right) !} \int_0^{\infty} \frac{(a x)^{2 n+1}}{x e^x} d x\\& =\sum_{n=0}^{\infty} \frac{(-1)^n a^{2 n+1}}{(2 n+1) !} \int_0^{\infty} x^{2 n} e^x d x\\& =\sum_{n=0}^{\infty} \frac{(-1)^n a^{2 n+1}}{(2 n+1) !}\Gamma(2 n+1)\\& =\sum_{n=0}^{\infty} \frac{(-1)^n a^{2 n+1}}{2 n+1}\\&=\arctan a\end{aligned}
$$
My question: Without power series, can we compute $\int_0^{\infty} \frac{\sin (ax)}{x e^x} d x$?
 A: For $f(a)=\int_0^\infty\frac{\sin ax}{xe^x}\,dx$, one may compute $f'(a)=\int_0^\infty e^{-x}\cos ax\,dx=\frac1{1+a^2}$ elementarily.
A: Noticing that
$$
\int_1^{\infty} e^{-t x} d t=-\left[\frac{e^{-t x}}{x}\right]_1^{\infty}=\frac{e^{-x}}{x},
$$
we can evaluate the integral as a double integral. $$
\begin{aligned}
\int_0^{\infty} \frac{\sin (a x)}{x e^x} d x =& \int_0^{\infty} \sin (a x) \int_1^{\infty} e^{-t x} d t d x \\
=& \int_1^{\infty} \int_0^{\infty} \sin (a x) e^{-t x} d x d t \\
=& \int_1^{\infty} \frac{a}{a^2+t^2} d t \quad (\textrm{ via IBP })\\
=& {\left[\arctan \left(\frac{t}{a}\right)\right]_1^{\infty} } \\
=& \frac{\pi}{2}-\arctan \left(\frac{1}{a}\right) \\
=& \arctan a
\end{aligned}
$$
A: The integral can be further generalised by the Feynman’ Technique Integration by considering another integral with parameter $b$. $$
I(b)=\int_0^{\infty} \frac{\sin (a x) e^{-b x}}{x} d x, \quad \textrm{ where }a\ne 0 \textrm{ and }b\geq0.
$$
Differentiating $I(b)$ w.r.t. $b$ and followed by IBP yields $$
\begin{aligned}
I^{\prime}(b) =-\int_0^{\infty} \sin (a x) e^{-b x} d x=-\frac{a}{a^2+b^2}
\end{aligned}
$$
$$
\begin{aligned}
I(b)-I(0) &=-\int_0^b \frac{a}{a^2+y^2} d y \\
&\left.=-\arctan \left(\frac{y}{a}\right)\right]_0^b \\
&=-\arctan \left(\frac{b}{a}\right)
\end{aligned}
$$
Using the famous result: $$
I(0)=\int_0^{\infty} \frac{\sin (a x)}{x} d x \stackrel{ax\mapsto x}{=} sgn (a) \int_0^{\infty} \frac{\sin x}{x} d x=\frac{\pi sgn(a)}{2} \text {. }
$$
We can now conclude that
$$
\boxed{\int_0^{\infty} \frac{\sin (a x)}{x e^{b x}}=I(b)=\frac{\pi sgn(a)}{2}-\arctan \left(\frac{b}{a}\right)=\arctan \left(\frac{a}{b}\right)}
$$
A: We can identify the Laplace transform
$$
\mathcal{L}\left\{\frac{\sin(t)}{t}\right\}(s) = \int_0^\infty \frac{\sin(t)}{t} e^{-st}\, \mathrm{d}t
$$
But since $\mathcal{L}\left\{\frac{f(t)}{t}\right\}(s) = \int_s^{\infty} \mathcal{L}\{f(t)\}(\sigma)\, \mathrm{d}\sigma$ and  since for $s>0$ we have
$$
\mathcal{L}\{\sin(t)\}(s) =\int_0^\infty \sin(t) e^{-st}\,\mathrm{d}t = \Im\left\{ \int_0^\infty e^{-t(s-i)}\right\}\,\mathrm{d}t  = \Im\left\{\frac{1}{s-i} \right\} =\frac{1}{1+s^2}
$$
we get
$$
\int_0^\infty \frac{\sin(t)}{t} e^{-st}\, \mathrm{d}t = \int_{s}^{\infty}\frac{\mathrm{d}\sigma}{1+\sigma^2} = \frac{\pi}{2}-\arctan(s) = \arctan\left(\frac{1}{s}\right), \quad s>0
$$
So for $a>0$ we get
$$
\int_0^{\infty} \frac{\sin (ax)}{x e^x} \,\mathrm{d}x \overset{ax =t}{=}\int_0^\infty \frac{\sin(t)}{t} e^{-\frac{t}{a}}\, \mathrm{d}t \overset{s = 1/a}{=} \arctan(a)
$$
from which the extension to negative $a$'s can be proven from the last equation exploiting the oddness of $\sin$ and $\arctan$.
