<s>[INST] 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.
 [/INST] 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.
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