complex integral and singularity

Question

How can I solve this problem:

I = $\int_{-\infty}^{\infty} \frac{\sin(t)}{t}\;e^{ipt}\;dt$

where p is real and I know the result of this integral is

I = 0, $|p| > 1$

I = $\pi$, $|p| < 1$

Answer 1

If you know $$ \int_{-\infty}^\infty \frac{\sin x}{x}dx =\pi,$$ it is easy for you to solve this integral.

We see that \begin{align} \int_{-\infty}^{\infty} \frac{\sin t}{t}e^{ipt}\,dt&=\int_{-\infty}^{\infty} \frac{\sin t}{t}(\cos pt+i\sin pt)\,dt\\ &=\int_{-\infty}^{\infty} \frac{\sin t\cos pt}{t}\,dt\\ &=\frac{1}{2}\left\{\int_{-\infty}^{\infty} \frac{\sin (1+p)t}{t}\,dt+\int_{-\infty}^{\infty}\frac{\sin (1-p)t}{t}\,dt\right\}. \end{align} Now we recall that $$ \int_{-\infty}^{\infty} \frac{\sin \alpha t}{t}\,dt= \begin{cases} \int_{-\infty}^{\infty} \frac{\sin t}{t}\,dt=\pi , & \alpha >0 \\ -\pi , & \alpha <0 .\end{cases} $$

Then we can see that $$ \int_{-\infty}^{\infty} \frac{\sin t}{t}e^{ipt}\,dt= \begin{cases} \frac{1}{2}(\pi+\pi)=\pi , & -1<p<1 \\ 0 , & |p|>1 .\end{cases} $$

Answer 2

Hint:

Rewrite \begin{align} \frac{\sin t}{t} = \frac{1}{2}\int^1_{-1} e^{ixt}\ dx \end{align} then use Fubini's theorem.

Note: In this method you will need to use the Dirac delta function.