<s>[INST] 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!
 [/INST] First $$I(x):=\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt$$ then applying a Laplace transform we get $$\mathscr{L}_{x\to s}\{I(x)\}=\int_0^{\infty} e^{-sx}\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dtdx$$ here by the $x\to s$ I just denote which vairables I change to which, to avoid confusion. Now since all integrals converge we change order of Integration getting. $$\int_0^{\infty} \int_0^{\infty} e^{-sx}\frac{t\sin(tx)}{(1+t^2)(4+t^2)}dxdt=\int_0^{\infty} \frac{t}{(1+t^2)(4+t^2)}\mathscr{L}_{x \to s} \{ \sin(tx) \}dt$$ Now using the fact that $\mathscr{L}_{x\to s} \{ \sin(tx)\}=\frac{t}{t^2+s^2}$ which is obtainable by using $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$, we get $$\int_0^{\infty} \frac{t}{(1+t^2)(4+t^2)}\mathscr{L}_{x \to s} \{ \sin(tx) \}dt=\int_0^{\infty} \frac{t^2}{(1+t^2)(4+t^2)(s^2+t^2)}dt$$ Now using partial fraction decomposition $$\frac{t^2}{(1+t^2)(4+t^2)(s^2+t^2)}=-\frac{1}{3(s^2-1)(1+t^2)}+\frac{4}{3(s^2-4)(4+t^2)}-\frac{s^2}{(s^2-4)(s^2-1)(s^2+t^2)}$$ Plugging this back in we get $$\int_0^{\infty} -\frac{1}{3(s^2-1)(1+t^2)}+\frac{4}{3(s^2-4)(4+t^2)}-\frac{s^2}{(s^2-4)(s^2-1)(s^2+t^2)}dt=$$$$=-\frac{1}{3(s^2-1)}\int_0^{\infty} \frac{1}{1+t^2}dt+\frac{4}{3(s^2-4)}\int_0^{\infty} \frac{1}{4+t^2}dt-\frac{s^2}{(s^2-4)(s^2-1)}\int_0^{\infty} \frac{1}{s^2+t^2}dt$$ Which using the fact that for $z>0$ we have $\int_0^{\infty} \frac{1}{z^2+t^2}dt=\frac{\pi}{2z}$ is equal to $$\pi(-\frac{1}{6(s^2-1)}+\frac{1}{3(s^2-4)}-\frac{s}{2(s^2-4)(s^2-1)})$$ Now this is equal to $\mathscr{L}_{x\to s} \{I(x)\}$ thus the inverse Laplace transform of this is I(x). Which can be easy be found using partial fraction decompositon to be $$\boxed{\frac{\pi}{6}(e^{-x}-e^{-2x})}$$
Very neat solution!
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