Finding Fourier Series for f(x)=2x+e^x-e^-x (-1< x >1)

[bhajee]

I wondered if someone could help me to find the Fourier series for this function please. I believe it's an odd function.

f(x) = 2x+e^x-e^-x (-1< x > 1)

This is my first post, so I'm going to try this LayTex typing too! Here goes!

f(x)=2x+e^x-e^-^x (-1< x >1)

Thanks

 

[bhajee]

sorry!

(-1< x <1) Doh!

 

[mathman]

You need to evaluate the integral of f(x)sin(pi)nx from -1 to 1. Look it up in an integral table.

 

[bhajee]

Here we are so far (can't get LaTex to work today!)

(e^x - e^-x)/2 = sinh(x)
so
2x + e^x - e^-x = 2x + 2sinh(x)

and for integration by parts
INT u.dv = uv - INT v.du
we have
u=2x+2sinh(x)

du=2+2cosh(x)

dv=sin(j*pi*x)

v=(1/j*pi)*cos(j*pi*x)

how am I going?

 

[mathman]

Try this: xsinax can be integrated by parts easily enough. For the other term, I would use sinax=(e^iax-e^-iax)/2i. Then all you have are the integrals of exponentials.