How can I use Parseval's formula to solve this Fourier transform problem?

[rayman123]

Homework Statement

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
$$<\overbrace{f}^{\wedge},\overbrace{g}^{\wedge}>=2 \pi<f,g>$$
where $$\wedge$$ means the Fourier transform of a function

The Attempt at a Solution

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
$$\mathcal{F}[x^2e^{-x}]=i(\overbrace{f}^{\wedge})^{''})(\xi)=\frac{2}{(1+i \xi)^3}$$

Homework Statement

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

$$\frac{1}{4}<f^{\wedge}, g^{\wedge}>= \frac{\pi}{2}<f,g>$$
how did he get this $$\frac{1}{4}$$ on the left hand side and then $$\frac{\pi}{2}$$ on the right hand side??

 

[HallsofIvy]

He is "norming" the inner product so that the inner product of a "basis vector" with itself is 1.

 

[rayman123]

He is "norming" the inner product so that the inner product of a "basis vector" with itself is 1.

Could you elaborate it a little bit more...I am very new to this...

 

[rayman123]

Of course it is correct, now I see it. The only thing that is done here is


$$\Bigl<\frac{\sin(a\xi}{\xi},\frac{1}{(1+i \xi)^3}\Bigr>=2 \pi \Bigl<\frac{1}{2}\chi_{[-a,a]}, \frac{1}{2}x^2e^{-x}\Bigr>$$
so multiplying everything we simply get
$$*2 \pi \cdot \frac{1}{4}\int_{0}^{a}x^2e^{-x}dx$$ like in the solution