Integral $\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt$ I am preparing for a master's degree entrance exam. One of the questions from the past exam asks to show the following formula. Could someone provide some hint?
$$\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt = \frac{\pi}{6} (e^{-x} - e^{-2x})$$ for $x \geq 0.$
Thank you!

1st Solution. Following @Claude Leibovici's suggestion, let us utilize complex-analytic technique. First, symmetrize the integral to write
$$ I
= \frac{1}{2}\int_{-\infty}^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}\,\mathrm{d}t
= \frac{1}{2}\operatorname{Im}\biggl(\int_{-\infty}^{\infty}\frac{te^{itx}}{(1+t^2)(4+t^2)}\,\mathrm{d}t\biggr). $$
Now let $R > 2$, and let $\mathcal{C}_R = L_R \cup \Gamma_R $ be the contour where

*

*$L_R = [-R, R]$ is the line segment traced from left to right, and

*$\Gamma_R = \{ Re^{i\theta} : 0 \leq \theta \leq \pi \}$ is the upper semicircular arc traversed counter-clockwise.

Then by the residue theorem, for $x \geq 0$,
\begin{align*}
\int_{\mathcal{C}_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z
&= 2\pi i \left( \mathop{\underset{z=i}{\mathrm{Res}}} \frac{ze^{izx}}{(1+z^2)(4+z^2)} + \mathop{\underset{z=2i}{\mathrm{Res}}} \frac{ze^{izx}}{(1+z^2)(4+z^2)}\right) \\
&= 2\pi i \left( \left. \frac{ze^{izx}}{(2z)(4+z^2)} \right|_{z=i} + \left. \frac{ze^{izx}}{(1+z^2)(2z)} \right|_{z=2i} \right) \\
&= \frac{\pi i}{3} \left( e^{-x} - e^{-2x} \right).
\end{align*}
On the other hand,
$$ \left| \int_{\Gamma_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \right|
\leq \int_{\Gamma_R} \frac{R}{(R^2 - 1)(R^2 - 4)} \, \left|\mathrm{d}z\right|
= \frac{\pi R^2}{(R^2 - 1)(R^2 - 4)} $$
and so, the contour integral along $\Gamma_R$ vanishes as $R\to\infty$. Therefore
\begin{align*}
I
&= \frac{1}{2}\operatorname{Im}\biggl(\lim_{R\to\infty} \int_{L_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \biggr) \\
&= \frac{1}{2}\operatorname{Im}\biggl(\lim_{R\to\infty} \int_{\mathcal{C}_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \biggr) \\
&= \frac{\pi}{6} \left( e^{-x} - e^{-2x} \right).
\end{align*}

2nd Solution. Using the decomposition $ \frac{t}{(1+t^2)(4+t^2)} = \frac{4}{3t(4+t^2)} - \frac{1}{3t(1+t^2)} $, we may write
$$ I = \frac{4}{3} \int_{0}^{\infty} \frac{\sin(xt)}{t(4+t^2)} \, \mathrm{d}t - \frac{1}{3} \int_{0}^{\infty} \frac{\sin(xt)}{t(1+t^2)} \, \mathrm{d}t. $$
In order to compute this, let $a>0$ and consider
$$ J(x) := \int_{0}^{\infty} \frac{\sin(xt)}{t(a^2+t^2)} \, \mathrm{d}t. $$
Then by direct computation, we find that $J$ solves
$$ J''(x) - a^2 J(x) = - \int_{0}^{\infty} \frac{\sin(xt)}{t} \, \mathrm{d}t = -\frac{\pi}{2}. $$
Among the general solution $ \frac{\pi}{2a^2} + c_1 e^{ax} + c_2 e^{-ax} $ of this ODE, the unique solution that satisfies $J(0) = 0$ and remains bounded as $x\to\infty$ is
$$ J(x) = \frac{\pi}{2a^2}(1 - e^{-ax}). $$
Therefore
$$ I = \frac{\pi}{6}(1 - e^{-2x}) - \frac{\pi}{6}(1 - e^{-x}) = \frac{\pi}{6}(e^{-x} - e^{-2x}). $$
