# Fourier series(past exam question)

1. Oct 17, 2009

### oddiseas

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

This is a past exam question that involves solving a differential equation. I am using a fourier series. in the solution they simply state that it is obvious that this equals pi^2/4 for n=1, 0 for n is odd>1, and -4n/(n^2-1) for even n.

A{n}=2/pi∫(xsinx)sin nx dx

I always integrate by parts, which for this equation is way too long, especially for an exam. So i am wondering how can i evaluate the value of the integral without integrating? i have tried using the addition formulas, but that seems like just as much work.

Another point of confusion is knowing when a function is even or odd. isnt the function xsinx an even function since f(-X)=-xsin(-X)=-(-sinx)=xsinx and thus an odd function * an even gives an odd, and i thought that the integral would then be zero. But in the solution they specify to use a sine series to fulfill the initial conditions that u(x,t)=u(pi,t)=0.

2. Relevant equations

3. The attempt at a solution

2. Oct 17, 2009

### ╔(σ_σ)╝

Well i used the addition formula and i got

-0.5( cos(x(n+1) -cos(x(n-1))

Those seem very easily integrated by parts

INT( xcos(x(n+1))dx = (1/(n+1))*xsin(x(n+1) - (1/(n+1))*INT(sin(x(n+1))dx

where INT is the integral symbol.