Complex Fourier Series of f: Evaluating Parseval's Relation

[AkilMAI]

Define f(t)=e^{-t} ont he interval [-\pi,\pi),and extend f to 2\pi-periodic.Find the complex
Fourier series of f.Then, apply Parseval's relation to f to evaluate
\sum^{\infty}_0 \frac{1}{1+k^{2}}

For the first part when I calculate c_k \frac{1}{2\pi}\ \int^\pi_{-\pi} e^{-ikt-t}dt...I get the following \frac{-e^{-(ik+1)\pi} + e^{(ik+1)\pi}}{2\pi(ik+1)}...is there any way to simplify it?Also for the second part,how can I apply Parseval's relation to evaluate the sum?
Thanks in advance

 

[vela]

You can use

\sinh x = \frac{e^x-e^{-x}}{2}

Not sure if it would really help much though.

 

[AkilMAI]

maybe I did something wrong with the integral

 

[vela]

You have

\frac{-e^{-(ik+1)\pi} + e^{(ik+1)\pi}}{2\pi(ik+1)} = \frac{\sinh \pi(1+ik)}{\pi(1+ik)}

Use the identity \sinh(a+b) = \sinh a \cosh b + \cosh a \sinh b.

 

[AkilMAI]

thanks vela... plugging that result in the Parseval’s relation provides me with another result...the questions is how can I use it to evaluate the sum from above?

 

[vela]

What does Parseval's relation tell you?

 

[AkilMAI]

that the integral of the square of a function is equal to the sum of the square of its transform...?

 

[vela]

So what do you get when you apply it to this particular problem?

 

[AkilMAI]

\frac{ \frac{\pi^{2}}{4}*cosh(iK)^{2}}{k^{2} +1}=\frac{1}{2\pi}*(e^{-2\pi} - e^{2\pi})

 

[vela]

Check your algebra. You should have a factor of \sinh \pi in there, and you can simplify \cosh ik. Also, where's the sum?

 

[AkilMAI]

sorry I do have a pi in there ,latex typo also is the forum,the loading time is very high...yes the sum is in the l.h.s...