Need help solving what seems like a strange integral

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Homework Statement


I need to find the Fourier series for f(x)=|x|^3.

Homework Equations


f(x)=|x|^3
A_n=\frac{2}{L}\int_{0}^{L}|x|^3cos{\frac{n\pi x}{L}}dx
\int_{0}^{L}x|x|sin{\frac{n\pi x}{L}}dx

The Attempt at a Solution


Since it is an even function I know that it will be a cosine series and so I set out to find the A coefficient like so:
A_n=\frac{2}{L}\int_{0}^{L}|x|^3cos{\frac{n\pi x}{L}}dx

Through integration by parts I end up with a sine term that goes to zero and some coefficients out in front of an integral that looks like:
\int_{0}^{L}x|x|sin{\frac{n\pi x}{L}}dx

I would really appreciate any help with this particular integral or if someone could point me in the direction of a solution where a Fourier series is calculated for a function with an absolute value to the power>2. Thanks in advance.
 
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When dealing with |x| it often helps to use the piece wise expansion of it:
=x, x>0
= -x, x<0.
If you were on the interval from [-L,L] you would have 2 integrals one from [-L,0] and one from [0,L]. However, since you are only interested in the function from [0,L] then $|x|^3$= $x^3$. To solve the integral you have integration by parts like normal.
 
Did you mean \int_0^L |x|^3 cos(\frac{n\pi x)}{L}dx of \int_{L/2}^{Ll/2} |x|^3 cos(\frac{n\pi x)}{L}dx? In the first, which is what you wrote, x is always non-negative so |x|= x. For the second, do the integrals from -L/2 to 0 and from 0 to L/2 separately.
 
HallsofIvy said:
Did you mean \int_0^L |x|^3 cos(\frac{n\pi x)}{L}dx of \int_{L/2}^{Ll/2} |x|^3 cos(\frac{n\pi x)}{L}dx? In the first, which is what you wrote, x is always non-negative so |x|= x. For the second, do the integrals from -L/2 to 0 and from 0 to L/2 separately.
Yeah, that was my bad; the question wants me to find the Fourier series for f(x)=|x|^3 on -L<x<L. So I think I can choose A_{n}=\frac{2}{L}\int_{0}^{L}x^{3}cos{\frac{n\pi x}{L}}dx and then A_{0}=\frac{1}{L}\int_{0}^{L}x^{3}dx?
 
Jufro said:
When dealing with |x| it often helps to use the piece wise expansion of it:
=x, x>0
= -x, x<0.
If you were on the interval from [-L,L] you would have 2 integrals one from [-L,0] and one from [0,L]. However, since you are only interested in the function from [0,L] then $|x|^3$= $x^3$. To solve the integral you have integration by parts like normal.
Thanks for your reply. I was unclear in the question statement; It asks me to find the Fourier series for the function f(x)=|x|^3 on -L<x<L. I guess I can choose A_{n}=\frac{2}{L}\int_{0}^{L}x^{3}cos{\frac{n\pi x}{L}}dx and then A_{0}=\frac{1}{L}\int_{0}^{L}x^{3}dx?
 
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