Without power series, can we compute $\int_0^{\infty} \frac{\sin (ax)}{x e^x} d x$?

Question

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 $\int_0^{\infty} \frac{\sin (ax)}{x e^x} d x$?

Answer 1

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.

Answer 2

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. $$ \begin{aligned} \int_0^{\infty} \frac{\sin (a x)}{x e^x} d x =& \int_0^{\infty} \sin (a x) \int_1^{\infty} e^{-t x} d t d x \\ =& \int_1^{\infty} \int_0^{\infty} \sin (a x) e^{-t x} d x d t \\ =& \int_1^{\infty} \frac{a}{a^2+t^2} d t \quad (\textrm{ via IBP })\\ =& {\left[\arctan \left(\frac{t}{a}\right)\right]_1^{\infty} } \\ =& \frac{\pi}{2}-\arctan \left(\frac{1}{a}\right) \\ =& \arctan a \end{aligned} $$

Answer 3

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 $$ \boxed{\int_0^{\infty} \frac{\sin (a x)}{x e^{b x}}=I(b)=\frac{\pi sgn(a)}{2}-\arctan \left(\frac{b}{a}\right)=\arctan \left(\frac{a}{b}\right)} $$

Answer 4

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 $\arctan$.