Analytical solution of integral $x^{1+n}\exp(-x) \sin(x)$ $$\int\limits_0^\infty x^{1+n} \exp(-x) \sin(xy) \mathrm{d}x $$
for real $n, x, y$ and $n\ge-3$ has the following analytic solution according to mathematica:
$$(1 + y^2)^{\left(-1 - \frac{n}{2}\right)} \Gamma(2 + n) \sin((2 + n) \arctan(y))$$
I tried to do the integration by hand, but don't know how to do it when $1+n$ is not a positive integer (if it is one can do partial integration and gets a factorial instead of the gamma function).
Is there any way to do this?
Or does one need to solve for integer $n$ and can then generalize to non-integer $n$ by inserting the gamma function for factorials?
I thought about doing this in the Complex plane using the residue theorem, but there are no poles, so this doesn't seem to work.

The integral is the imaginary part of
$\displaystyle \int\limits_0^\infty x^{1+n} \exp(-(1-i y)x)\mathrm{d}x$,
which suggests the change of variables $z=(1-i y)x$. You can check that the change in the contour doesn't matter, and you end up with
$\displaystyle\frac{\Gamma(2+n)}{(1-iy)^{2+n}}$,
the imaginary part of which is your answer.
