Fourier Series Coefficients for f(x)=xcos(x) on [-∏,∏]

[tigertan]

Hey there,

First time user of this forum.

I have a question regarding an integral I've been stuck on for the past few days. I would really appreciate any eye opener into this problem!

How do I find the Fourier coefficients for f(x)=xcos(x), x [-∏,∏]

So when calculating the coefficient b_n I figure that the equation I'll be working with is (1/∏)∫xcos(x)sin(kx)dx (FROM -∏ to ∏).

I keep getting down to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏) . I some how always get a 0 for b₁. What am I doing incorrectly??

 

[I like Serena]

Welcome to PF, tigertan!

I'm not getting 0 for b₁.
So what did you do?

 

[tigertan]

Am I right in breaking it up to 1/2∏∫xsin((k+1)x)dx (FROM -∏ to ∏) + 1/2∏∫xsin((k-1)x)dx (FROM -∏ to ∏)?

I then end up with 1/2∏[-(k+1)cos((k+1)x) + sin((k+1)x)](FROM -∏ to ∏) + 1/2∏[-(k-1)cos((k-1)x) + sin((k-1)x)](FROM -∏ to ∏)

This equals to 0 when I put b₁

 

[I like Serena]

You lost a factor x in your integration.
It seems you did not properly apply integration by parts.

You should have an expression containing: x cos(k+1)x.
When you substitute the bounds -pi and +pi, these add up.