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!
 A: 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
$$ I = \frac{\pi}{6}(1 - e^{-2x}) - \frac{\pi}{6}(1 - e^{-x}) = \frac{\pi}{6}(e^{-x} - e^{-2x}). $$
A: Partial answer.
$$I=\int_0^\infty \frac{t\sin{(tx)}}{(1+t^2)(4+t^2)}\,dt=\Im\left(\int_0^\infty \frac{t\,e^{itx}}{(1+t^2)(4+t^2)}\,dt \right)$$
$$\frac{t}{(1+t^2)(4+t^2)}=\frac{1}{6 (t+i)}-\frac{1}{6 (t-2 i)}-\frac{1}{6 (t+2 i)}+\frac{1}{6 (t-i)}$$ So, you basically face four integrals
$$I_k=\int_0^\infty\frac {e^{itx}}{t+ki}$$ With simple change of variables,
$$J_k=\int\frac {e^{itx}}{t+ki}=e^{k x} \,\text{Ei}[(i t-k) ]$$ where appears the exponential integral function and
$I_k+I_{-k}$ does simplify.
A: Suppose $x\ge0$. Let
$$ I(x)=\int_0^\infty \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt.$$
Then
\begin{eqnarray}
I'(x)&=&\int_0^\infty \frac{t^2\cos(tx)}{(1+t^2)(4+t^2)}dt\\
&=&\int_0^\infty \frac{\cos(tx)}{4+t^2}dt-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt\\
&=&\frac\pi4e^{-2x}-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt
\end{eqnarray}
and hence
$$ I''(x)=-\frac{\pi}2e^{-2x}+I(x) $$
or
$$ I''(x)-I(x)=-\frac{\pi}2e^{-2x}. \tag1$$
Now (1) has a particular solution $I_p(x)=-\frac{\pi}{6}e^{-2x}$ and hence (1) has the general solution
$$ I(x)=c_1e^x+c_2e^{-x}-\frac{\pi}{6}e^{-2x}. $$
Since $I(0)=0, I'(0)=\frac{\pi}{6}$, it is easy to get $c_1=0,c_2=\frac{\pi}{6}$. So
$$ I(x)=\frac{\pi}{6}(e^{-x}-e^{-2x}). $$
The calculation of the integral
$$ \int_0^\infty \frac{\cos(tx)}{4+t^2}dt $$
can be found, for example, from this.
A: 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
$$
\int_{0}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{1}{2} \int_{-\infty}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{6} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
.
$$
A: 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!
A: hint: With $f(x)=e^{-ax}$ the definition of fourier transform shows
$$\hat{f}(w)=\sqrt{\dfrac{2}{\pi}}\dfrac{a}{a^2+w^2}$$
then
$$f(x)=\int_{-\infty}^{\infty}\hat{f}(w)e^{2\pi ixw}dw=a\sqrt{\dfrac{2}{\pi}}\int_{0}^{\infty}\dfrac{2i \sin(2\pi xw)}{a^2+w^2}dw=e^{-ax}$$
so from
$$\dfrac{1}{(1+w^2)(4+w^2)}=\dfrac13\dfrac{1}{1+w^2}-\dfrac13\dfrac{1}{4+w^2}$$
you find an integral and finally you find the answer with differentiating respect to $x$. the formula holds for $Re~x>0$.
