# Difficult integrals (important für Fourier series)

1. Nov 2, 2008

### keenPenguin

Hello,

I couldn't figure out the following problem so far.

1. The problem statement, all variables and given/known data

$$m,n \in \mathbb{N}$$. Show the following relations:

2. Relevant equations

a) is no problem, but b) and c) seem tricky. I tried partial integration (two times) on b), which gave me a term which contained the original integral. So I sorted the term and got that:

$$\int_{0}^{2\pi}\sin\left(mx\right)\cos\left(nx\right)dx=\frac{1}{1+\frac{n^{2}}{m^{2}}}\left(\left[-\frac{\cos\left(mx\right)}{m}\cos\left(nx\right)\right]_{0}^{2\pi}+\frac{n}{m}\left[\frac{\sin\left(mx\right)}{m}\sin\left(nx\right)\right]_{0}^{2\pi}\right)$$

But I can't see why this should be zero.

Any help appreciated!

kP

2. Nov 2, 2008

### Redbelly98

Staff Emeritus
Welcome to PF.

I think the trig identities for angle addition and subtraction would be useful here. I mean the ones that involve
sin(a+b), sin(a-b), cos(a+b), and cos(a-b)

EDIT:
I looked more carefully at what you did get. Okay, if m and n are integers, what can you say about cos(mx) and cos(nx) when they are evaluated at 0 vs. when they are evaluated at 2pi?

Last edited: Nov 2, 2008
3. Nov 2, 2008

### keenPenguin

Thank you Redbelly,

you are right, I don't know why I missed that, n and m being integers, b) is quite obviously 0.

Concerning c): Yes, I also thought about those trig identities, but still couldn't figure it out. It's

cos(mx+nx)+sin(mx)sin(nx)=cos(mx)cos(nx), but trying to integrate the LHS, I get into even more trouble.

4. Nov 2, 2008

### Redbelly98

Staff Emeritus
Try

cos(mx+nx) + cos(mx-nx) = ?

5. Nov 2, 2008

### keenPenguin

That's it, thanks a lot. Works fine! :-)