Integral with trigonometric

[Tolaso J Kos]

Let $a>0$.Evaluate the integral:

$$\int_0^\infty \frac{\sin^2 x}{x^2 \left(x^2+a^2 \right)}\, {\rm d}x$$

[Grigorios Kostakos]

For \(a>0\):
\begin{align*}
I&=\int_0^{+\infty} {\frac{\sin^2 x}{x^2 \,(x^2+a^2 )}\, {\rm d}x}\\
&=\frac{1}{a^2}\int_0^{+\infty}{\frac{\sin^2 x}{x^2}\, {\rm d}x}-\frac{1}{a^2}\int_0^{+\infty} {\frac{\sin^2 x}{x^2+a^2}\, {\rm d}x}\\
&=\frac{1}{a^2}\int_0^{+\infty}{\frac{\sin^2 x}{x^2}\, {\rm d}x}-\frac{1}{2a^2}\int_0^{+\infty} {\frac{1-\cos(2x)}{x^2+a^2}\, {\rm d}x}\\
&=\frac{1}{a^2}\int_0^{+\infty} {\frac{\sin^2 x}{x^2}\, {\rm d}x}-\frac{1}{2a^2}\int_0^{+\infty} {\frac{1}{x^2+a^2}\, {\rm d}x}+\frac{1}{2a^2}\int_0^{+\infty} {\frac{\cos(2x)}{x^2+a^2}\, {\rm d}x}\\
&=\frac{1}{a^2}\,I_1-\frac{\pi}{4a^3}+\frac{1}{2a^2}\,I_2\,.
\end{align*}
\begin{align*}
I_1&=\int_0^{+\infty} {\frac{\sin^2 x}{x^2}\, {\rm d}x}\\
&=\cancelto{0}{\bigg[-\frac{\sin^2 x}{x}\bigg]_0^{\infty}}+\int_0^{+\infty} {\frac{2\cos{x}\,\sin{x}}{x}\, {\rm d}x}\\
&=\int_0^{+\infty} {\frac{\sin (2x)}{x}\, {\rm d}x}\\
&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,2x}\\
{\frac{1}{2}\,dt\,=\,dx} \\
\end{subarray}} \,\int_0^{+\infty}{\frac{\sin {t}}{t}\, {\rm d}t}\\
&\stackrel{(*)}{=}\frac{\pi}{2}\,.
\end{align*}
For the evaluation of \(I_2\) we consider the complex function \(f(z)=\dfrac{{\rm{e}}^{2i\,z}}{z^2+a^2}\), which is meromorphic in \(\mathbb{C}\) with simple poles at \(z=\pm ia\). We integrate \(f(z)\) along the positive oriented simple curve \(\gamma\) which is the sum of upper semicircle \(C_r\,, \; r>a\,,\) and its diameter \(d_r\,.\)

upper_semicircle.png (9.37 KiB) Viewed 2451 times

[/centre]

Only the simple pole \(z=ia\) lies inside \(\gamma\), so, by residue theorem we have that \begin{align*}
\displaystyle\int_{C_r}{f(z)\, {\rm d}z}+\int_{d_r}{f(z)\, {\rm d}z}&=\int_{\gamma}{f(z)\, {\rm d}z}=2\pi i\,{\rm{Res}}\big(f(z);z=ia\big)&\Rightarrow\\
\displaystyle\int_{-r}^{r}{\dfrac{{\rm{e}}^{2it}}{t^2+a^2}\, {\rm d}t}&=2\pi i\,\Big(-\frac{i}{2a\,{\rm{e}}^{2a}}\Big)-\displaystyle\int_{C_r}{f(z)\, {\rm d}z}&\Rightarrow\\
\displaystyle\int_{-r}^{r}{\dfrac{{\rm{e}}^{2it}}{t^2+a^2}\, {\rm d}t}&=\frac{\pi}{a\,{\rm{e}}^{2a}}-\displaystyle\int_{C_r}{f(z)\, {\rm d}z}\quad (1)
\end{align*} For the evaluation of \(\displaystyle\int_{C_r}{f(z)\, {\rm d}z}\), by estimation lemma, we have that \begin{align*}
\displaystyle\bigg|\int_{C_r}{\dfrac{{\rm{e}}^{2i\,z}}{z^2+a^2}\,dz}\bigg|&\leqslant\int_{C_r}{\bigg|\dfrac{{\rm{e}}^{2i\,z}}{z^2+a^2}\bigg|\,dz}\\
&=\int_{C_r}{\dfrac{\big|{\rm{e}}^{-2\Im(z)}\big|}{\big|z^2+a^2\big|}\,dz}\\
&\stackrel{-2\Im(z)<0}{\leqslant}\int_{C_r}{\dfrac{1}{\big|z^2+a^2\big|}\,dz}\\
& \leqslant\frac{\pi\,r}{r^2+a^2}\xrightarrow{r\to +\infty}0\,.
\end{align*} So \((1)\) becomes \begin{align*}
\displaystyle\int_{-\infty}^{+\infty}{\dfrac{{\rm{e}}^{2it}}{t^2+a^2}\, {\rm d}t}&=\lim_{t\to+\infty}\int_{-r}^{r}{\dfrac{{\rm{e}}^{2it}}{t^2+a^2}\, {\rm d}t}=\frac{\pi}{a\,{\rm{e}}^{2a}}&\Rightarrow\\
\displaystyle\int_{-\infty}^{+\infty}{\dfrac{\cos(2t)}{t^2+a^2}\, {\rm d}t}&=\Re\Big(\frac{\pi}{a\,{\rm{e}}^{2a}}\Big)=\frac{\pi}{a\,{\rm{e}}^{2a}}&\Rightarrow\\
I_2&=\frac{1}{2}\int_{-\infty}^{+\infty}{\dfrac{\cos(2t)}{t^2+a^2}\, {\rm d}t}=\frac{\pi}{2a\,{\rm{e}}^{2a}}\,.
\end{align*} Finally \begin{align*}
I&=\frac{1}{a^2}\,\frac{\pi}{2}-\frac{\pi}{4a^3}+\frac{1}{2a^2}\,\frac{\pi}{2a\,{\rm{e}}^{2a}}\\
&=\frac{\pi}{2a^2}-\frac{\pi}{4a^3}+\frac{\pi}{4a^3\,{\rm{e}}^{2a}}\\
&=\frac{\pi}{4a^3}\,(2a-1+{\rm{e}}^{-2a})\,.
\end{align*}


\((*)\; \int_0^{+\infty} {\frac{\sin {t}}{t}\, {\rm d}t}=\frac{\pi}{2}\) it is a well known integral.

[Tolaso J Kos]

Nicely done Grigoris. Here is a solution to the integral \( \displaystyle \int_0^\infty \frac{\sin x}{x}\, {\rm d}x \) using Laplace Transfomation.

$$\begin{aligned} \int_{0}^{\infty}\frac{\sin x}{x}\, {\rm d}x
&=\int_{0}^{\infty}\sin x \left( \int_{0}^{\infty}e^{-xt}\, {\rm
d}t\right)\; {\rm d}x \\ &=\int_{0}^{\infty}\left ( \int_{0}^{\infty}\sin x e^{-xt}\, {\rm d}x \right )\, {\rm d}t \\ &=\int_{0}^{\infty}\frac{{\rm d}t}{t^2+1} = \frac{\pi}{2} \end{aligned}$$

Of course contour integration also works but I leave it to someone else. If needed, I can write it up another day.