I need help with a Fourier Transform.
I know Fourier Transform is defined by:
$$F(\omega)=\int_{-\infty}^{\infty} f(t).e^{-i\omega t}\, dt$$
where $F(\omega)$ is the transform of $f(t)$.
Now, I need to calculate the Fourier Transform of:
$$u(t+\pi) – u(t-\pi)$$
where $u$ is the Heaviside function.
With that, I have to calculate this:
$$\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin{(a\pi)}}{a} \cos{(at)} \,da$$
Any help?
Best Answer
Split the second integral into two pieces,
$$ I(t)=\int_{0}^{\infty}\frac{\sin{\pi a}}{a}\cos(a t)da=\int_{0}^{\infty}\frac{1}{2a}\left(\sin(a(t+\pi))-\sin(a(t-\pi))\right)da $$
Due to the eveness of the integrand we get
$$ 4 I(t)=\int_{-\infty}^{\infty}\frac{1}{a}\sin(a(t+\pi))da-\int_{-\infty}^{\infty}\frac{1}{a}\sin(a(t-\pi))da=\\\ \underbrace{\Im\int_{-\infty}^{\infty}\frac{1}{a}e^{ia(t+\pi)}da}_{I_1}-\underbrace{\Im\int_{-\infty}^{\infty}\frac{1}{a}e^{ia(t-\pi)}da}_{I_2} $$
We can now apply residue theorem. There are two things we have to worry about:
-in which part of the Complex plane our integral converges
-How to avoid the singularity at $0$
We solve the second problem by adding a small semicircle at zero to avoid the divergence.
Now, lets's take $t+\pi>0$ for the moment then we have to close the contour in the upper half plane to calculate $I_1$. The result is
$$ I_1= \pi i $$ If $t+\pi<0$ we have to close in the lhp. we get $$I_1=-\pi i$$ put together both cases yields $$ I_1=\pi i \text{sign}(t+\pi) $$ A similiar reasoning for $I_2$ gives $$ I_2=i\pi\text{sign}(t-\pi) $$ Collecting everything and taking imaginary parts completes our calculation $$ I=\frac{\pi}{4}(\text{sign}(t+\pi)-\text{sign}(t-\pi)) $$
Feel free to ask, if anything is unclear or look at this question of you