This is a question from one of the past papers of my university which I am unable to do. I am not being able to do question 2 from below.
Let $f(x)= a^2-x^2 \,\,\,\,\, |x|<a \\\,\,\,\,\,\,\,\,\,\,\, =0 \,\,\,\,\,|x|>a>0$.
Calculate the fourier transform of this function, and hence evaluate: $$1. \int_0^{\infty} \frac{\sin{2x}-2x\cos{2x}}{x^3} \,dx$$ $$2. \int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}\, dx$$
The fourier transform is easy to transform, I have checked this many times and I am sure this is correct. Only only needs to multiply by $\cos$ as it is even.
$\mathcal{F(f)}=2\int_0^{a}(a^2-x^2)\cos{\xi x}\,dx=2\frac{2 \xi^2 a^2 sin{a \xi} + 2 a \xi \cos{a \xi} - 2a \sin{a \xi}}{\xi ^3}$
The first question is obvious. substitute $a=2$, then use the fourier integral representation $\frac{2}{2 \pi}\int_0^{\infty}\hat{f}(\xi) \cos{\xi x}\,d{\xi}=\frac{f(o^{+})+f(o^{-})}{2}$, at $x=0$, so that the cos becomes 1 after subtracting the first term which is the dirichlet integral, you get question 1.
How do I do question 2? Do I have to do integration by parts on each of the function? I have no idea how to handle the square.
First of all, your expression for the FT is incorrect. I get
$$\int_{-a}^a dx \, (a^2-x^2) e^{i k x} = 4 \frac{\sin{a k} - a k \cos{a k}}{k^3}$$
By Parseval-Plancherel, we may write
$$\int_{-\infty}^{\infty} dk \: \left ( 4 \frac{\sin{a k} - a k \cos{a k}}{k^3} \right )^2 = 2 \pi \int_{-a}^a dx \, \left ( a^2-x^2 \right )^2 $$
Take it from there...
To elaborate, Parseval-Plancherel states that, for a function $f$ and its FT $\hat{f}$, we have
$$\int_{-\infty}^{\infty} dx \, |f(x)|^2 = \frac{1}{2 \pi} \int_{-\infty}^{\infty} |\hat{f}(k)|^2 $$
In any case, I get
$$\int_0^{\infty} dk \frac{(\sin{2k}-2k\cos{2k})^2}{k^6} = \frac{32 \pi}{15}$$