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!

Let $f(t) = \frac{1}{(1+t^2)(4+t^2)}.$ Its Fourier transform is
$$
\hat{f}(x)
= \int_{-\infty}^{\infty} \frac{e^{-itx}}{(1+t^2)(4+t^2)}dt
= \int_{-\infty}^{\infty} \frac{1}{3}\left( \frac{1}{1+t^2} - \frac{1}{4+t^2} \right) e^{-itx} dt \\
= \frac{1}{12} \int_{-\infty}^{\infty} \left(2 \frac{2\cdot 1}{1+t^2} - \frac{2\cdot 2}{4+t^2} \right) e^{-itx} dt 
= \frac{1}{12} \left( 2 \cdot 2\pi\,e^{-|x|} - 2\pi\, e^{-2|x|} \right) \\
= \frac{\pi}{6} \left( 2e^{-|x|} - e^{-2|x|} \right)
,
$$
where rule 208 and rule 105 from the third column of these tables have been used.
Taking the derivative we get
$$
\hat{f}'(x)
= \int_{-\infty}^{\infty} \frac{(-it)\,e^{-itx}}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{6} \left( -2e^{-|x|}\operatorname{sign}(x) + 2 e^{-2|x|}\operatorname{sign}(x) \right) \\
= -\frac{\pi}{3} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
$$
so
$$
\int_{-\infty}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= -\operatorname{Re} \int_{-\infty}^{\infty} \frac{(-i)t\,e^{-itx}}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{3} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
$$
and
$$
\int_{0}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{1}{2} \int_{-\infty}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{6} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
.
$$
