how to compute this integral for fourier series

Question

I am trying to find the Fourier sine and cosine series of $\frac{1}{(1+x^2)}$ from $0$ to $2$, and do not know where to even begin to evaluate this integral: $\int \frac{sin(nx)}{(1+x^2)} dx$ (and similarly for the cosine series too). I found $a_0$ since integrating the function gives the arctan function. Can someone please help with the sine and cosine extensions?

Answer 1

Initially, I want to solve this problem by using residue theorem. But I find that the integrand is odd function so that this method is invalid personally.

Then, I guess this integral can not be presented as a closed form. I solve this problem as follow

We can factorize the integrand as: $$ \dfrac{\sin(nx)}{1+x^2}=\dfrac{\sin(nx)}{(x-i)(x+i)}=\dfrac{1}{2i}(\dfrac{\sin(nx)}{x-i}-\dfrac{\sin(nx)}{x+i}) $$

So the integral is

$$ I=\dfrac{1}{2i}(\int_0^2\dfrac{\sin(nx)}{x-i}dx-\int_0^2\dfrac{\sin(nx)}{x+i}dx) $$ Note that $$ \int_0^2\dfrac{\sin(nx)}{x-i}dx \stackrel{x-i=t}{=} \int_{-i}^{2-i}\dfrac{\sin(n(t+i))}{t}dt=\int_{-i}^{2-i}\dfrac{\sin(nt)\cos(ni)+\cos(nt)\sin(ni)}{t}dt\\ =\cosh(n)\int_{-i}^{2-i}\dfrac{\sin(nt)}{t}dt+i\sinh(n)\int_{-i}^{2-i}\dfrac{\cos(nt)}{t}dt\\ =\cosh(n)(Si((2-i)n)+Si(in))+i\sinh(n)(Ci((2-i)n)-Ci(-in)) $$ where $$ Si(x)=\int_0^x\frac{\sin(t)}{t}dt\qquad Ci(x)=\gamma+\ln(x)+\int_0^x\frac{\cos(t)-1}{t}dt $$ and $\gamma$ is Euler constant.

Similarly, we have $$ \int_0^2\dfrac{\sin(nx)}{x+i}dx=\cosh(n)(Si((2+i)n)-Si(in))+i\sinh(n)(-Ci((2+i)n)+Ci(in)) $$

In summary, we have: $$ I=\dfrac{1}{2}[(Ci(in)-Ci(n(i+2))-Ci(-in)+Ci(n(2-i)))\sinh(n)-(Si(n(i+2))+Si(n(2-i)))i\cosh(n)] $$ I can only calculate to here.