Prove the relation related to this Fourier Transformate

Question

I have this Fourier trasformate $$\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx$$ and I have to prove this relation

$$\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx =2 \int_{0}^{1}(1-x^2)\cos(px) dx$$ Hint: prove that $\hat{f}(p)$ is real and even.

When I solve an exercise about Fourier antitrasformate, I usually analyse what happens for p>0 and p<0 and this is the proceeding that I'm following for this one.

If p>0, $\displaystyle \hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-ipx}dx$

If p<0, $\displaystyle \hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{ipx}dx$ (now with "p" I'm indicating the absolute value of p)

So I'd say that $\displaystyle\hat{f}(p)=\int_{-1}^{+1}(1-x^2)e^{-i|p|x}dx$

I also know that $\displaystyle\cos (px)=\frac{e^{ipx}-e^{-ipx}}{2}$

But now I don't know to to go on... Many thanks for your help!

Answer 1

I think it goes like that: $$\int_{-1}^1 (1-x^2)e^{-ipx}dx=\int_{0}^1 (1-x^2)e^{-ipx}dx+\int_{-1}^0 (1-x^2)e^{-ipx}dx$$ now in the second integral make the coordinate transformation $x\rightarrow -x$ and you'll have : $$\int_{0}^1 (1-x^2)e^{-ipx}dx-\int_{1}^0 (1-x^2)e^{ipx}dx$$ ok now knowing that$$\int_a^bfdx=-\int_b^afdx$$ you can see that the integral above just becomes: $$\hat{f(p)}=\int_{0}^1 (1-x^2)(e^{-ipx}+e^{ipx})dx=2\int_0^1(1-x^2)\cos(px) dx$$

Answer 2

Use the symmetry of the problem

$ \int_{-1}^{1} (1−x^2)e^{−ipx}dx = -\int_{0}^{-1} (1−x^2)e^{−ipx}dx + \int_{0}^{1} (1−x^2)e^{−ipx}dx$

change variables on the first integral $x\rightarrow -x$

$\int_{0}^{1} (1−x^2)e^{ipx}dx + \int_{0}^{1} (1−x^2)e^{−ipx}dx$

combine

$\int_{0}^{1} (1−x^2) (e^{ipx} + e^{−ipx}) dx = 2\int_{0}^{1} (1−x^2) cos(px) dx$

try to always use symmetry to your advantage, it typically is the only way to help solve an analytical problem.