1. ## Fourier Series Coefficients

There is a part of this question I don't understand:

Eventually they ended up with:

$f(x)=\sum^{\infty}_{k=1} \frac{-2k \pi sinh 1 cos(k \pi)}{1+k^2 \pi^2} sin(k \pi x)$

So the part I don't understand right at the begining where they have:

$b_k= \int^1_{-1} sinh (x) sin (k \pi x) dx$

Should it not be $b_k= \frac{1}{2} \int^1_{-1} sinh (x) sin (k \pi x) dx$?

Because in my texbook $b_k$ is given by:

$b_k= \frac{1}{b-a} \int^b_a f(x)sin(kx) dx$

I'm very confused, can anyone help?

2. If you can determine an orthonormal basis on [-1,1], you can avoid the messy integration of that infinite sum.

3. My point is that in this question they found the coefficient b_k like this:

$b_k= \int^1_{-1} sinh (x) sin (k \pi x) dx$

But I think it should be:

$b_k= \frac{1}{b-a} \int^b_a f(x)sin(kx) dx = \frac{1}{2} \int^1_{-1} sinh (x) sin (k \pi x) dx$

Why did they ignore the $\frac{1}{b-a}$ term?

4. The convention for what coefficients go outside the integral and and which go under the exponential vary from person to person. The important thing is that you get the correct series at the end.