Fourier related integration by parts

[medicalphysicsguy]

Hi,

I'm reviewing Fourier transformations and I want to make sure I can do all the prep math by hand. Stuck at the beginning:

\(\displaystyle \int^{\pi}_{-\pi} cos(mx)cos(nx)\,dx \)

equals 0 if \(\displaystyle m \not= n\), \(\displaystyle \pi\) if \(\displaystyle m = n \not= 0\), \(\displaystyle 2\pi\) if \(\displaystyle m = n = 0\)

I am not getting this integrating by parts. I get:

\(\displaystyle u(x) = cos(mx) \) \(\displaystyle v(x) = \frac{1}{n}sin(nx) \)

\(\displaystyle = \frac{1}{n}cos(mx)sin(nx) - \int^\pi_{-\pi} \frac{1}{mn}sin(nx)sin(mx)\,dx \)

And this doesn't look any closer to a solution. Am I messing up my integration by parts or do I just need to keep going?

 

[HallsofIvy]

I started to say you should try to integrate, by parts, again, because sometimes these things will give the same integral again but with an additional part from the \(\displaystyle \left[uv\right]_a^b\) part, and so we can solve for the integral. But when I tried that myself, that additional part is 0 and everything just cancels!

Instead, use the trig identity, \(\displaystyle cos(a)cos(b)= \frac{1}{2}(cos(a+b)+ cos(a- b))\). In this integral, a= mx, b= nx so \(\displaystyle cos(mx)cos(nx)= \frac{1}{2}(cos((m+n)x)- cos((m-n)x)\). The integral can be written \(\displaystyle \frac{1}{2}\int_{-\pi}^\pi cos((m+n)x) dx- \frac{1}{2}\int_{-\pi}^\pi cos((m-n)x) dx\).