Evaluating $\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-|x|} e^{-i k x} d x , \quad k \in \mathbb{Z}$

Question

I will like to find the complex coefficients of the Fourier series for $f(x)=e^{-|x|},x\in [-\pi,\pi]$: $$ c_{k}(f):=\left\langle f, e_{k}\right\rangle= \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-|x|} e^{-i k x} d x = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-|x|-ikx} d x , \quad k \in \mathbb{Z} $$

I tried $u$-substitution: $u(x)=-|x|-ikx$. I got $c_k(f)=0,\forall k$. This is clearly not right.

I was thinking whether I could use the fact $\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{ikx} d x = \delta_{k,0},k \in\mathbb{Z}$? I was lucky to be able to use this in some other problems but I can se a way to use it here. Any help is welcome.

Kind regards,

Answer 1

The real part of the integrand is even, and the imaginary part is odd, giving a real integral.

Then

$$\int_0^{\pi}e^{-x-ikx}dx=\left.\frac{e^{-x-ikx}}{-1-ik}\right|_0^\pi=\left.\frac{e^{-\pi-ik\pi}-1}{-1-ik}\right|_0^\pi=-\frac{(-1)^ke^{-\pi}-1}{1+k^2}(1-ik)$$

and finally

$$I=\frac{1-(-1)^ke^{-\pi}}{\pi(1+k^2)}.$$