$I=-i\oint \frac{dz}{z} \, z^{(n_1-n_2)}$

Question

How to calculate the following integral: $$\int_0^{2\pi}d\phi \, e^{i(n_2-n_1)\phi}$$ Making the substitution $z=e^{i\phi}$, I obtained $$-i\oint_{|z|=1} \frac{dz}{z} \, z^{(n_2-n_1)}$$ I can imagine that you have to use the residues but I don't know how to do.

Answer 1

Hint. The integrand function has a pole at $0$. The residue at $0$ is $2\pi i$ if $n_1=n_2$, otherwise it is zero.

P.S. What is integrating path?

Answer 2

$$I=-i\oint_{|z|=1} \frac{dz}{z} \, z^{(n_2-n_1)}=-i\oint_{|z|=1}z^{(n_2-n_1-1)}dz=-i.2\pi i.Res(z^{(n_2-n_1-1)},0)$$ $$ Res(z^{(n_2-n_1-1)},0)= \begin{cases} 0, n_1\neq n_2\\ 1, n_1=n_2 \end{cases}. $$

Answer 3

No residue theory required, actually:

• $n_1 = n_1\implies$ constant integrand $= 1\implies$ integral $= 2\pi i$.
• $n_1\ne n_1\implies\displaystyle\int_0^{2\pi}e^{i(n_2-n_1)\phi}\,d\phi = \frac{e^{i(n_2-n_1)\phi}}{i(n_2-n_1)}\Big|_{\phi = 0}^{\phi = 2\pi} = 0.$