Solve integral by finding Fourier series of complex function

[psie]

Homework Statement

Expand the function $u(x)=\frac1{r-e^{ix}},$ where $r>1$, in a Fourier series and use this to compute the integral $\int _0^{2\pi }\frac{dx}{1-2r\cos x+r^2}$.

Relevant Equations

E.g. maybe the formula for the complex Fourier coefficients $c_n=\frac1{2\pi}\int_{-\pi}^\pi u(x)e^{-inx} dx$. Maybe Parseval's formula?

I've mostly worked with real-valued functions, but this seems to be a complex-valued function and the integral for the coefficient doesn't seem that nice, especially when I rewrite the function as $$u(x)=\frac{r-e^{-ix}}{r^2-2r\cos x+1}.$$ I'm stuck on where to even start. Any ideas? I prefer to solve this via Fourier series if this is possible, and not using some other trick.

 

[julian]

Try the function

\begin{align*}
v(x) = A + u(x) e^{ix}
\end{align*}

where $A$ is an appropriately chosen constant. You only need the real part of the function.

 

[psie]

I solved it using Fourier series. It's pretty straightforward actually. The function is $$u(x)=\frac1{r-e^{ix}}=\frac1{r}\frac{1}{1-\frac{e^{ix}}{r}},$$ where $r>1$. You see, this is the sum of a geometric series. Writing that out you will obtain the complex Fourier series and thus the coefficients. Then to solve the integral, you simply apply Parseval's formula. Bingo!

 

[julian]

Yes, that is how you obtain the Fourier coefficients. I was thinking

\begin{align*}
c_0 = \frac{1}{2 \pi} \int_{-\pi}^\pi \dfrac{r - \cos x}{r^2 - 2 r \cos x + 1} dx
\end{align*}

so you can't know the value of the integral you are after ($\int_{-\pi}^\pi \dfrac{1}{r^2 - 2 r \cos x + 1} dx$) from the value of $c_0$ of $u(x)$.

However, you can get the value of the integral you are after from the value of the Fourier coefficient, $c_0$, of $v(x) = 1/2 + u(x) e^{ix}$.

Or, as you say, you can use Parseval's on $u(x)$.