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?

In combination with yhhuang's answer, I think I have a similar explanation:
$$\int e^x\cdot\sin(x)dx=\int e^x\cdot\text{Im}(\cos(x)+i\sin(x))dx$$
$$=\int e^x\cdot\text{Im}(e^{xi})dx$$
$$=\text{Im}\int e^{x(1+i)}dx$$
I found this easier to understand.
