Originally Posted by

**olski1** ok so i had to find the fourier series of cos(ax) with period 2 pi where a is not an integer.

So i came to the correct answer iam pretty sure, then they asked me to sub in x =pi into the series to get

i)

$\displaystyle \pi cot(\pi a) -\frac{1}{a } = 2a\sum_{n = 1}^\infty {-1}^{n }/({a}^{2 }- {n}^{2 })\cos (npi) $

$\displaystyle \pi cot(\pi a) -\frac{1}{a } = 2a(\frac{1}{{a}^{ 2} -{1}^{2 } }+ \frac{1}{{a}^{ 2} -{2}^{2}} ........)$

ii)now i have to integrate both sides form a =0 to a=theta to obtain

$\displaystyle \frac{\sin \theta \pi }{\theta \pi }=(1-\frac{{\theta }^{2 } }{{1}^{ 2}.....) } $

but this is where it gets hard for me,

integrating the left hand side of i) i get

$\displaystyle \frac{ln(sin(\pi a)}{\pi } - \frac{lnsin(a\pi )}{\ pi } $

but when calculating from a=0 isnt it undefined?

I dont even know where to start on the right hand side of the equation so any help with that would be greatly appreciated.