How do I calculate the integral $\int_{-L}^L{e^{i\frac{(m-n)\pi}{L}x}dx}$?

Question

Let $0 < L \in \mathbb{R}$ and $m,n \in \mathbb{N}$.
How can I calculate this integral? Here it is once more, a bit larger: $$\int_{-L}^L{e^{i\frac{(m-n)\pi}{L}x}dx}$$

Answer 1

If $m=n$ your integral is

$$ \int_{-L}^{L}\mathbb{d}x=2L $$

If $m\neq n$ your integral is

$$ \int_{-L}^{L}e^{\mathbb{i}\pi x(m-n)/L}\mathbb{d}x=\frac{L}{\mathbb{i}\pi(m-n)}(e^{\mathbb{i}\pi(m-n)}-e^{\mathbb{i}\pi(-m+n)}) $$

Recalling the Euler formula $e^{\mathbb{i}\theta}=\cos(\theta)+\mathbb{i}\sin(\theta)$

$$ (e^{\mathbb{i}\pi(m-n)}-e^{\mathbb{i}\pi(-m+n)})=\cos(\pi(m-n))+\mathbb{i}\sin(\pi(m-n))-\left[\cos(\pi(n-m))+\mathbb{i}\sin(\pi(n-m))\right] $$

The $\sin$ terms are zero above because $n,m\in\mathbb{N}$ and $\cos$ is even so that $\cos(\pi(m-n))-\cos(\pi(n-m))=0$. Therefore the integral is zero when $n\neq m$. We can summarize the result nicely by using the delta-kronecker symbol.

$$ \int_{-L}^{L}e^{\mathbb{i}\pi x(m-n)/L}\mathbb{d}x=2L\delta_{m,n} $$

$\delta_{m,n}=1$ when $m=n$ and is equal to zero otherwise.