dwsmith said:
So on the recent graduate problem of the week, I saw that $\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$, but so does, $\int_0^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$.
How can they both be the same?
Let us use integration by parts to compute $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx$. At the end, we will need to use the fact that $\displaystyle\int_0^{\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$
Let $u=\sin^2x$ and $\,dv=\dfrac{\,dx}{x^2}$. Then $\,du=2\sin x\cos x\,dx=\sin(2x)\,dx$ and $v=-\dfrac{1}{x}$. Therefore,
\[\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx = \left[-\frac{\sin^2 x}{x}\right]_0^{\infty}+\int_0^{\infty}\frac{\sin(2x)}{x}\,dx=\int_0^{\infty}\frac{\sin(2x)}{x}\,dx.\]
(We note that $|\sin x|\leq 1\implies |\sin^2 x|\leq 1$ and thus $\displaystyle\lim_{x\to\infty} \frac{\sin^2 x}{x}\sim \lim_{x\to\infty} \frac{1}{x}=0$; We also note that $\displaystyle\lim_{x\to 0}\frac{\sin^2 x}{x}=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\sin x=0$. Hence, that's why the $\displaystyle\left[-\frac{\sin^2 x}{x}\right]_0^{\infty}$ term goes to zero.)
Now let $t=2x\implies\,dt=2\,dx$. Therefore,
\[\int_0^{\infty}\frac{\sin(2x)}{x}\,dx\xrightarrow{t=2x}{} \int_0^{\infty}\frac{\sin t}{t/2}\frac{\,dt}{2}=\int_0^{\infty}\frac{\sin t}{t}=\frac{\pi}{2}.\]
And thus, we also have that $\displaystyle\int_0^{\infty}\frac{\sin^2 x}{x^2}\,dx =\frac{\pi}{2}$.
I hope this makes sense!