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?

It can easily be done by applying by parts twice:
Let
$$ I=\int e^x \cdot \sin x \quad dx = e^x\sin x-\int e^x\cos x\,dx$$
for the second one apply by parts,
$$\int e^x\cos x\,dx=e^x\cos x+I$$
so, $$I = \frac{1}{2} e^x (\sin x -\cos x)$$
