$\displaystyle \int sin(x)exp(-inx)$, where i is the imaginary unit and n is an entire.
Regards.
Actually this doesn't work. Consider briefly that it will return a real answer, which is obviously not correct.
I'd do it by parts. Here's the first step:
$\displaystyle \int sin(x)e^{-inx}dx = \frac{1}{-in}sin(x)e^{-inx} - \int \frac{1}{-in}cos(x)e^{-inx}dx$
Do the second integral by parts and you will have the form:
$\displaystyle \int sin(x)e^{-inx}dx = ~...~ + \frac{1}{n^2} \int sin(x)e^{-inx}dx$
from which you can solve for $\displaystyle \int sin(x)e^{-inx}dx$.
-Dan
Unless I am misreading something I dont think this is purely real.
$\displaystyle \int e^{ix}e^{-inx}dx$
$\displaystyle \int e^{-(n-1)ix}dx=\frac{i}{(n-1)}e^{-(n-1)ix}$
You could also use
$\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
If we don't want to do it by parts, we can write $\displaystyle e^{-inx}$ as $\displaystyle \cos (nx)-i\sin (nx)$. Now we have to compute $\displaystyle \int \sin x \cos(nx) dx$ and $\displaystyle \int \sin x\sin(nx)dx$.
For the first we use $\displaystyle \sin x \cos(nx) =\frac{\sin((n+1))x-\sin((n-1)x)}2 $ and for the second $\displaystyle \sin x\sin (nx)=\frac{\cos((n-1)x)-\cos ((n+1)x)}2$.