Why does $\int_{-\pi}^{\pi}\sin(nx)e^{inx}$=$-\pi i$? I tried couple of times to compute $\int_{-\pi}^{\pi}\sin(nx)e^{inx}$. According to W.A it should be $-\pi i$, I'm losing my mind trying to understand where I got wrong.
Assuming the whole process that $e^{-in \pi}=\cos(n \pi)-i \sin(n\pi)=(-1)^{n}$, and $[\cos(nx)e^{-inx}]_{-\pi}^{\pi}$=0.
Here is what I did:
$\int_{-\pi}^{\pi}\sin(nx)e^{inx}=[\frac{(-\cos(nx))}{n}e^{-inx}]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}\frac{ine^{-inx}\cos(nx)}{n}=-\frac{1}{n}[\cos(nx)e^{-inx}]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}ie^{-inx}\cos(nx)e^{inx}=0-i[\frac{e^{-inx}}{-in}]_{-\pi}^{\pi}\cos(nx)=0.$
I'd really love to understand what is wrong with that.
Thank you very much.
 A: You should use
$$\sin(n x)=\frac{e^{inx}-e^{-inx}}{2i}.$$
Then
$$\int_{-\pi}^\pi \frac{e^{inx}-e^{-inx}}{2i} e^{inx}dx=\frac{1}{2i}\int_{-\pi}^\pi(e^{2inx}-1)dx$$
but
$$\frac{1}{2i}\int_{-\pi}^\pi e^{2inx}dx=0$$
and you are left with $i\pi$.
A: For what it's worth, and just for kicks, let's try to use the "integrate by parts twice and solve for the integral trick":
$$\eqalign{
\int \underbrace{\vphantom{(}e^{Bx}}_u \underbrace{\sin (Ax)\, dx}_{dv}
&= \underbrace{-A^{-1}\cos(Ax)}_v \underbrace{e^{Bx}\vphantom{(}}_u
-\int \underbrace{-A^{-1}\vphantom{(}}_v \underbrace{B\cos(Ax)e^{Bx}\,dx}_{du}\cr
&= -A^{-1}\cos(Ax)e^{Bx}+A^{-1}B\Bigl[ \int \underbrace{e^{Bx}\vphantom{(}}_s 
\underbrace{ \cos(Ax)\,dx}_{dw}\Bigr]  \cr           
&= -A^{-1}\cos(Ax)   e^{Bx} +A^{-1}B\Bigl[ 
\underbrace{  A^{-1}\sin(Ax)}_w \underbrace{e^{Bx}\vphantom{(}}_s   - \int  \underbrace{A^{-1} \sin(Ax)}_w\underbrace{Be^{Bx}\vphantom{(}}_{ds}   \Bigr]  \cr

&= -A^{-1}\cos(Ax)e^{Bx}+A^{-2}B   \sin(Ax)e^{Bx}   -  A^{-2}B^2\int \sin(Ax)e^{Bx}  . 
   \cr

}
$$
Whence
$$\eqalign{\int \sin (nx)e^{inx}\, dx
&= -n^{-1}\cos(nx)e^{inx}+n^{-2}(in)  \sin(nx)e^{inx}   -  n^{-2}(in)^2\int \sin(nx)e^{inx}   
  \cr
&= {-e^{inx}  ( \cos(nx)- i    \sin(nx)) \over n } -i^2 \int \sin(nx)e^{inx}   \cr
&= {-e^{inx}  e^{-inx} \over n } -i^2 \int \sin(nx)e^{inx}   \cr
&= {-1 \over n } + \int \sin(nx)e^{inx}.   
}

$$
 So, the trick used in the real case (solving for the original integral) breaks down  here. Note the above is true because there are constants attached to to general antiderivatives... 
