# Math Help - Fourier transform problems

1. ## Fourier transform problems

Compute the integral
$\int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi$

by using Parseval's formula for Fourier transform
$<\overbrace{f}^{\wedge},\overbrace{g}^{\wedge}>= 2 \pi$
where $\wedge$ means the Fourier transform of a function

Using Parseval's formula we can rewrite the integral as

$\int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi = \Bigl<\overbrace{\frac{\sin(a\xi)}{\xi}}^{\mathcal {F}},\overbrace{\frac{1}{(1+i\xi)^3}}^{\mathcal{F} }\Bigr>$

with their inversions as follows

$\Bigl(\frac{\sin(a \xi)}{\xi}\Bigr)^{\wedge \wedge }=\frac{1}{2} \chi_{[-a,a]}(x)$ by the table 2 in Folland p.223

and the other one
$\Bigl(\frac{2}{(i \xi+1)^3}\Bigr)^{\wedge \wedge}=\frac{1}{2}x^2e^{-x}$

which I calculated myslef as follows

we let $g(x)= e^{-x}$ and then the Fourier transform of g(x) is $\mathcal{F}[g(x)]= \frac{1}{1+i \xi}$

then using formula in Folland again table 2 $\mathcal{F}[xf(x)]=i(\overbrace{f}^{\wedge})^{'})(\xi)$

and applying it to our function $x^2e^{-x}$
we get
$\mathcal{F}[x^2e^{-x}]=i(\overbrace{f}^{\wedge})^{''})(\xi)=\frac{2}{(1+ i \xi)^3}$

going back to the integral and plugging in everything in the Parseval's formula gives me smth like this

$\int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi = \Bigl<\overbrace{\frac{\sin(a\xi)}{\xi}}^{\mathcal {F}},\overbrace{\frac{1}{(1+i\xi)^3}}^{\mathcal{F} }\Bigr>=2 \pi \Bigl<\frac{1}{2} \chi_{[-a,a]}, \frac{1}{2}x^2e^{-x}\Bigr>$

in the solution given by our teacher we find

$\frac{1}{4}= \frac{\pi}{2}$
how did he get this $\frac{1}{4}$ on the left hand side and then $\frac{\pi}{2}$ on the right hand side??

2. ## Re: Fourier transform problems

Originally Posted by rayman
Compute the integral
$\int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi$

by using Parseval's formula for Fourier transform
$<\overbrace{f}^{\wedge},\overbrace{g}^{\wedge}>= 2 \pi$
where $\wedge$ means the Fourier transform of a function

<snip>

$\frac{1}{4}= \frac{\pi}{2}$
how did he get this $\frac{1}{4}$ on the left hand side and then $\frac{\pi}{2}$ on the right hand side??
Divide the first of these by 4 to get the second...

CB