Finding the Fourier series I am trying to write the Fourier series for the following function
$$f(x) = \left\{\begin{aligned}
&0 && -\pi<x<0\\
&e^{-x} && 0<x<\pi
\end{aligned}
\right.$$
I used all the same formulas I always use, but the coefficients I got look a bit strange to me
\begin{aligned}
a_0 = \frac{1}{\pi}\int_0^\pi e^{-x}dx = \frac{1-e^{-\pi}}{\pi}
\end{aligned}
\begin{aligned}
a_n = \frac{1}{\pi}\int_0^\pi e^{-x} \cos(nx)dx = \frac{1}{\pi}\int_0^\pi e^{-x}\frac{e^{inx}+e^{-inx}}{2} = \frac{1-e^{-\pi}(-1)^n}{\pi(1+n^2)}
\end{aligned}
\begin{aligned}
b_n = \frac{1}{\pi}\int_0^\pi e^{-x} \sin(nx)dx = \frac{1}{\pi}\int_0^\pi e^{-x}\frac{e^{inx}-e^{-inx}}{2i} = \frac{n-ne^{-\pi}(-1)^n}{\pi(1+n^2)}
\end{aligned}
...and my final solution for the series is
\begin{aligned}
F(x) = \frac{1-e^{-\pi}}{2\pi} + \sum_{n=1}^\infty \frac{1-e^{-\pi}(-1)^n}{\pi(1+n^2)} \cos(nx) + \sum_{n=1}^\infty \frac{n-ne^{-\pi}(-1)^n}{\pi(1+n^2)} \sin(nx)
\end{aligned}
I tried plotting $F(x)$ and it looked nothing like my function. My other method of checking the solution is to evaluate $F(x)$ at e.g. $x=0$ and see if I get the right sum, but I had no luck doing that either.
I'd be very grateful if anybody could tell me if there's something wrong with my math.
 A: Since I cannot comment, here's the plot of your Fourier transform:

Your solution is correct. If you're interested this code was used to calculate the function(in Matlab):
 function [ y ] = Untitled2( x)
suma = (1-exp(-pi))./(2 * pi);
for n = 1:1000
   suma = suma + cos(n.*x).*(1- exp(-pi).*((-1).^n))./(pi.*(1 + n^2)); 
   suma = suma + sin(n.*x).*(n- n.*exp(-pi).*((-1).^n))./(pi.*(1 + n^2));
end
y = suma;
end

A: The approach suggested by @user1949350 is correct and will work most of the time but I may have found something. It MUST be changed so that it incorporates the idea of convergence within a particular tolerance level. So I am just gonna take his code and extend it to include that part too. 
function [ y ] = abc(x,tol)
    suma = (1-exp(-pi))/(2 * pi) * ones(1,length(x));
    sumb = Inf(1,length(x));
    n = 1;

    while(max(abs(sumb-suma)) > tol)
       disp(n);
       sumb = suma;
       suma = suma + cos(n.*x).*(1- exp(-pi).*((-1).^n))./(pi.*(1 + n^2)); 
       suma = suma + sin(n.*x).*(n- n.*exp(-pi).*((-1).^n))./(pi.*(1 + n^2));
       n = n+1;
    end
    y = suma;
end

When I set tol = $10^{-4}$, it took about 3000 iterations to converge to that tolerance. So, 1000 terms is a pretty good approximation for this function, but it doesn't lie within $10^{-4}$ tolerance.
The plot of the Fourier series looks pretty much the same:
 
Cheers!
