Evaluating $\int_0^\infty \frac{\sin^2(xu)du}{u^2}$ How do I evaluate$$\int_0^\infty \frac{\sin^2(xu)du}{u^2}$$
I tried integration by parts, but it kept getting messier.
 A: Using trigonometric formula :
$$I=\int_0^\infty \frac{\sin^2(xu)}{u^2}\,\mathrm{d}u=-\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}_0^\infty\dfrac{\cos\left(2ux\right)-1}{u^2}\,\mathrm{d}u$$
Now integrate by part with $a=\cos\left(2ux\right)-1$ and $b'=\dfrac{1}{u^2}$ :
$${\displaystyle\int}_0^\infty\dfrac{\cos\left(2ux\right)-1}{u^2}\,\mathrm{d}u=-{\displaystyle\int}_0^\infty\dfrac{2x\sin\left(2ux\right)}{u}\,\mathrm{d}u-\left[\dfrac{\cos\left(2ux\right)-1}{u}\right]_0^\infty$$
Now make the change of variables $v=2ux$ (note that the quantity between brackets is equal $0$) :
$$I={\dfrac{1}{2}}\class{steps-node}{\cssId{steps-node-2}{2x}}{\displaystyle\int}_0^\infty\dfrac{\sin\left(v\right)}{v}\,\mathrm{d}v=x{\displaystyle\int}_0^\infty\dfrac{\sin\left(v\right)}{v}\,\mathrm{d}v$$
You recognise Dirichlet integral, thus :
$$I=\frac{\pi x}{2}$$
A: Denote the integral by
$$\mathcal I(x)=\int_0^\infty\frac{\sin^2(xu)}{u^2}\,\mathrm du$$
Then taking the Laplace transform from the $x$ domain into the $s$ domain, you get
$$\mathcal L_s\{\mathcal I(x)\}=\int_0^\infty\frac{\mathrm du}{u^2}\mathcal L_s\{\sin^2(xu)\}=\frac12\int_0^\infty\frac{\mathrm du}{u^2}\mathcal L_s\left\{1-\cos(2xu)\right\}$$
$$\mathcal L_s\{\mathcal I(x)\}=\frac12\int_0^\infty\frac{\mathrm du}{u^2}\left(\frac1s-\frac s{s^2+4u^2}\right)=2\int_0^\infty\frac{\mathrm du}{s(s^2+4u^2)}$$
Evaluating the right hand side yields
$$\mathcal L_s\{\mathcal I(x)\}=\dfrac\pi{2s^2}$$
then taking the inverse transform yields
$$\mathcal I(x)=\dfrac{\pi x}2$$
A: The integrand function is continuous and bounded on $(0,1)$, non-negative and bounded by $\frac{1}{u^2}$ on $[1,+\infty)$, hence it is integrable. With the substitution $u=\frac{t}{|x|}$ the given integral turns into:
$$ |x|\int_{0}^{+\infty}\left(\frac{\sin t}{t}\right)^2\,du \stackrel{IBP}{=}|x|\int_{0}^{+\infty}\frac{\sin(2t)}{t}\,dt\stackrel{t\mapsto\frac{v}{2}}{=}|x|\int_{0}^{+\infty}\frac{\sin v}{v}\,dv = |x|\frac{\pi}{2}.$$
A: Here is a slight different line of reasoning: We begin by noting that
\begin{align*}
\frac{1}{a} &= \int_{0}^{\infty} e^{-ax} \, dx, \qquad (a > 0) \\
\frac{1}{a^2} &= \int_{0}^{\infty} x e^{-ax} \, dx, \qquad (a > 0) \\
\frac{a}{a^2+b^2} &= \int_{0}^{\infty} \cos(bx) e^{-ax} \, dx, \qquad (a > 0) \\
\end{align*}
which can be easily conformed by integration by parts. Then we can write
\begin{align*}
\int_{0}^{\infty} \frac{\sin^2(xu)}{u^2} \, du
&= \int_{0}^{\infty} \sin^2(xu)\left(\int_{0}^{\infty} s e^{-us} \, ds\right) \, du \\
&= \int_{0}^{\infty} s \left( \int_{0}^{\infty} \sin^2(xu) e^{-su} \, du\right) \, ds \\
&= \int_{0}^{\infty} s \left( \int_{0}^{\infty} \frac{1-\cos(2xu)}{2} e^{-su} \, du\right) \, ds \\
&= \int_{0}^{\infty} \frac{s}{2} \left( \frac{1}{s} - \frac{s}{4x^2+s^2}\right) \, ds \\
&= \int_{0}^{\infty} \frac{2x^2}{4x^2+s^2} \, ds.
\end{align*}
The final integral can be evaluated by the substitution $s = 2|x|\tan\theta$, yielding the same answer as others.
A: Hint: consider
$$ f(x)= \int_{0}^{+\infty} \frac{\sin^2(xu)du}{u^2} $$
and calculate $f'(x)$ differentiating under the integral sign (after showing that we are allowed to do so). This will essentially solve the problem (turning it into a simple ODE) provided you are able  to evaluate  :
$$ I_a=\int_{0}^{+\infty} \frac{\sin(ax)}{x} \, dx $$
A: Defining $\quad  f(x)=\int_0^{\infty} \frac{\sin^2(xu)}{u^2}du$, we see that $f$ is even and so we can consider $x>0$. Then you can make the change of variable $ u\rightarrow xu$ which yields $f(x)=x \int_0^{\infty} \frac{\sin^2(u)}{u^2}du$. We now need to evaluate the integral which is just a number and according to Mathematica, it is equal to $\frac{\pi}{2}$
So for a general $x$ you will have $f(x)= \frac{\pi \mid x \mid}{2} $ using the parity of $f$.
