Integrating $\int_{-\infty}^0e^x\sin(x)dx$ I ask if anyone could solve the following:
$$\int_{-\infty}^0e^x\sin(x)dx=?$$
I can visually see that it will converge and that it should be less than $1$:
$$\int_{-\infty}^0e^x\sin(x)dx<\int_{-\infty}^0e^xdx=1$$
But I am unsure what its exact value is.
Trying to find the definite integral by integrating by parts 4 times only results in $e^x\sin(x)$, which got me nowhere.
How should I evaluate this?

$$
\int_{-\infty}^0e^x\sin(x)dx = 
\mbox{Im}\int_{-\infty}^0 e^{x(1+i)}dx
=\mbox{Im}\left.\frac{1}{1+i}e^{x(1+i)}\right|_{-\infty}^0
=\mbox{Im}\frac{1}{1+i}=\mbox{Im}\frac{1-i}{2}=-\frac{1}{2}.
$$
