The question is:
A. Calculate the Fourier transform of the function
$$ f(x) = a^2-x^2, |x| < a$$ $$ f(x) = 0, |x| > a$$
where $a \ge 0$
B. Hence, deduce
$$\int_0^{\infty} \frac{\sin x - x\cos x}{x^3}\,\Bbb dx = \frac{\pi}{4}$$
HINT: Start from inverse Fourier transform and note that $f(0) = a^2$.
First time using Math Stack Exchange so I appreciate any patience for me with the MathJax notation! Essentially I'm struggling on part B of the question. What I got for part A was: $$4\left(\frac{\sin(ua)-au\cos(ua)}{u^3}\right)$$
(Don't know if that's right, but took the limits from $-a$ to $a$, since everywhere else it will equal 0.)
When I try deducing the integral, I keep getting it equal to $a^2(\frac{\pi}{4})$.
Any help is appreciated, thank you!
Let $f$ be the function defined as
$$f(x)=\begin{cases}a^2-x^2&,|x|\le a\\\\0&,|x|>a\end{cases}$$
The Fourier Transform of $f$ is then
$$\begin{align} \mathscr{F}\{f\}(u)&=\int_{-\infty}^\infty f(x)e^{iux}\,dx\\\\ &=\int_{-a}^{a} (a^2-x^2)e^{iux}\,dx\\\\ &=\frac{4\sin(au)-4au\cos(au)}{u^3} \end{align}$$
From the Inverse Fourier Transform, we have
$$f(x)=\frac1{2\pi}\int_{-\infty}^\infty \frac{4\sin(au)-4au\cos(au)}{u^3}e^{-iux}\,du\tag1$$
Now, enforcing the substitution $au\mapsto u$, we see that for $|x|\le a$
$$(a^2-x^2)=\frac1{2\pi}\int_{-\infty}^\infty 4a^2\frac{\sin(u)-x\cos(u)}{u}e^{-iux/a}\,du\tag2$$
Setting $x=0$ in $(2)$, we find that
$$\int_{-\infty}^\infty \frac{\sin(u)-u\cos(u)}{u^3}\,du=\frac\pi2\tag3$$
Exploiting the even symmetry of the integrand in $(3)$, we arrive at the coveted result
$$\int_{0}^\infty \frac{\sin(u)-u\cos(u)}{u^3}\,du=\frac\pi4$$