How to do this integral. I need to do this:
$$\int_0^\infty e^{ikt}\sin(pt)dt$$
I already have the solution, I'm just clueless on how to actually calculate it. I've tried several changes,integration by parts and extending it to a $(-\infty,\infty)$ integral so it'd be a simple Fourier transform, but I can't solve it.
I would appreciate solution or hints. Thank you very much.
EDIT:
The result must be of the form:
$$\frac{a}{s}\pm ibs\text{ sign}(k)\delta(-p^2)$$
where $a,b$ are constants, and $s$ is the module of a vector measured with a metric of minkowsky type $\text{diag}(-1,1,1,1)$, where the $0$ component is $k$ and the $3d$ part has module $p$.
All in all, the integral comes from calculating a green function that ends up being:
$$\frac{1}{4\pi(x-x')}\int_0^\infty e^{-ik(t-t')}\sin[k(x-x')]dk$$
And the result should be:
$$\frac{1}{8\pi^2(x-x')^2}-\frac{i}{8\pi}\text{ sign}(t-t')\delta(-(x-x')^2)$$
The constants in each term $\textit{may}$ be wrong, but that's the structure of the result.

Let's illustrate the non-existence.  
I took $k=2,p=1$.
The graph illustrates
$$
\mathrm{Re}\int_0^x e^{i2t}\sin(t)\;dt 
$$
for $x$ from 0 to 20.  It continues with period $2\pi$.

