# Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform

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.

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Hint: Use Plancherel's formula. –  user60725 Apr 30 '13 at 12:39
@BarackObama: I didn't know the plancherel's formula, so I looked it up on the internet, which correct me in I am wrong is $\int_{\mathbb{R}}||f(x)||^2 \, dx = \int_{\mathbb{R}}||\hat{f(\xi)}||^2 \, d {\xi}$. But in my problem, the integrand is not the square of the entire fourier transform. There is an extra term, which I won't be able to integrate after squaring. How do I do it? –  ramanujan_dirac Apr 30 '13 at 13:11

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

$$\int_0^{\infty} dk \frac{(\sin{2k}-2k\cos{2k})^2}{k^6} = \frac{32 \pi}{15}$$

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Hi. Should I have taken the sine fourier transform? My sine and cos are interchanged. But the function is even, so the sine part would be zero right? –  ramanujan_dirac Apr 30 '13 at 13:30
@ramanujan_dirac: no, you were justified in what you did - it just didn't make things easier for you. I just took the regular FT, and things worked out. Likely, looking at your result, you got a sign wrong. –  Ron Gordon Apr 30 '13 at 13:31
At first glance, I am flabbergasted. The first term $\int_o^{\infty}a^2 \cos{\xi x} dx=a^2\frac{\sin{a \xi}}{\xi}=\frac{a^2 \xi ^2 \sin{a \xi} }{\xi ^3}$, which is clearly missing from your answer, and clearly present in mine. The second part done by parts, should't cancel it out. Thanks a lot for the trouble though. :) –  ramanujan_dirac Apr 30 '13 at 13:36
@ramanujan_dirac: here's what WA has to say: wolframalpha.com/input/… –  Ron Gordon Apr 30 '13 at 13:38
Thanks! I missed a term in the second integral. I am really clumsy in my calculations. Sorry for the trouble. –  ramanujan_dirac Apr 30 '13 at 13:52