Calculate $\int^{+\infty}_{x=0} e^{-x}\sin x dx$ This one has been tricky due to trig functions persisting in the integral, but I believe they can be ignored:
Using integration by parts I calculated as follows:
$$I = \int^{+\infty}_{x=0} e^{-x}\sin x  dx$$
$$ u = e^{-x}; du = -e^{-x}; dv = \sin x; v = -\cos x $$
$$I = -e^{-x}\cos x - \int_{x=0}^{\infty}e^{-x}\cos x$$
Integration by parts a second time:
$$ u = e^{-x}; du = -e^{-x}; dv = \cos x; v = \sin x $$
$$I = -e^{-x}\cos x - (e^{-x}\sin x + \int^{+\infty}_{x=0} e^{-x}\sin x)$$
$$I = -e^{-x}\cos x - e^{-x}\sin x - I $$
$$2I = -e^{-x}\cos x - e^{-x}\sin x$$
$$I = \frac{-e^{-x} (\cos x + \sin x)}{2} \biggr |_{x=0}^{\infty}$$
My understanding at this point is that, although the trig functions don't have a limit as x approaches infinity, they do have a ceiling on their combined possible value that is negligible due to $-e^{-x}$ dominating:
$$\lim\limits_{x \to \infty} -\frac{1}{e^x} = 0$$
So I believe the final answer should be 0, but it's not. Where, then, am I miscalculating? 
EDIT:
As pointed out, I gave as the answer only the upper bound.
$$ I = \frac{-e^{-x}}{2} \frac{\cos x + \sin x}{2} \biggr |_{x=0}^{\infty}$$
$$ I = 0 - \frac{-e^0 (\cos 0 + \sin 0)}{2}$$
$$ I = 0 - \frac{-1 (1 + 0)}{2} = \frac{1}{2}$$
 A: The upper bound as $x \to \infty$ is $0$, due to the reasons you state. However, the lower bound of the integration results at $x = 0$, that you subtract, becomes
$$-e^{-0}\left(\frac{\cos(0) + \sin(0)}{2}\right) = \frac{-1}{2} \tag{1}\label{eq1A}$$
As such, it seems your final answer should be $\frac{1}{2}$.
Note that anomaly's question comment indicates you added an extra factor of $\frac{1}{2}$ in going to your last line, which I originally had in my answer but which I've now corrected.
A: An exercise:
$\Im \displaystyle{\int_{0}^{\infty}}e^{x(i-1)}dx=$
$\Im \frac{1}{i-1}e^{x(i-1)}\big ]_0^{\infty}=$
$\Im \frac{i+1}{(-2)}e^{-x}(\cos x +i \sin x)\big ]_0^{\infty}=$
$(-1/2)e^{-x}(\cos x+\sin x)\big ]_0^{\infty}=$
$-(-1/2)(1)=1/2$.
A: To make your reasoning more formal, if $f(x):=e^{-x}$ has $x\to\infty$ limit $0$ while $g(x)$ is bounded for sufficiently large $x>0$, as happens when $g$ is the sine or cosine function, say $|g|\le A$ (in this case with $A=1$), $fg\to0$ too because, if you fix a large enough $M>0$ such that $|f(x)|<\frac{\delta}{A}$ for all $x>M$, such $x$ also satisfy $|f(x)g(x)|<\delta$.
A: I'm pretty sure you only made a rookie mistake of confusing the limit as the function approaches infinity with the limit of the integral as its upper bound approaches infinity, but I'll include a little explanation here anyways.
To understand why the integral as a whole wouldn't be zero, consider the first interval $(0, \pi)$ and the successive interval $(\pi, 2\pi)$ for your integrand, $f(x) = e^{-x}\sin(x)$. Now, since $$|\sin(x)| = |\sin(x+\pi)|,$$ and since $$|e^{-x}| > |e^{-(x+\pi)}|,$$ we know that $$|e^{-x}\sin(x)| > |e^{-(x+\pi)}\sin(x+\pi)| \longrightarrow |e^{-x}\sin(x)|-|e^{-(x+\pi)}\sin(x+\pi)|>0.$$
Additionally, since $\sin(x)$ and $\sin(x+\pi)$ are always of opposite signs, and since $e^{-x}$ is always positive, we know that $e^{-x}\sin(x)$ and $e^{-(x+\pi)}\sin(x+\pi)$ are always of opposite signs. So if $e^{-x}\sin(x)$ is greater than 0, which it is on $(0,\pi)$, then $e^{-(x+\pi)}\sin(x+\pi)$ is less than zero, giving us, for all $x \in (0,pi)$:
$$|e^{-x}\sin(x)|-|e^{-(x+\pi)}\sin(x+\pi)|>0 \longrightarrow e^{-x}\sin(x)+e^{-(x+\pi)}\sin(x+\pi)>0.$$
Summing up all these values, or integrating from 0 to pi, would not change this equality, so: 
$$\int_0^{\pi}e^{-x}\sin(x)+e^{-(x+\pi)}\sin(x+\pi) dx > 0$$
$$\int_0^{\pi}e^{-x}\sin(x)+e^{-(x+\pi)}\sin(x+\pi) dx = \int_0^{\pi}e^{-x}\sin(x)dx+ \int_{\pi}^{2\pi}e^{-(x)}\sin(x) dx = \int_0^{2\pi}e^{-x}\sin(x)dx.$$
$$\int_0^{2\pi}e^{-x}\sin(x)dx > 0.$$
We can apply this logic for every other interval $(2\pi n, 2\pi(n+1))$to show that the integral over each of these intervals is positive, meaning that the integral over the entire positive x axis, the sum of these integrals, must also be positive, but certainly small.
For the actual calculation of integrals like these, I like to use $e^{ix} = \cos(x) + i\sin(x)$ just to make it a little easier. For this problem in particular, consider the integral $$\int_0^{\infty} e^{-x} e^{ix}dx = \int_0^{\infty} e^{-x} (\cos(x) + i\sin(x))dx = i\int_0^{\infty} e^{-x}\sin(x)dx + \int_0^{\infty} e^{-x} \cos(x)dx.$$
Since the actual values that come out of both $\int_0^{\infty} e^{-x}\sin(x)dx$ and $\int_0^{\infty} e^{-x} \cos(x)dx$ will be purely real, we know that when $\int_0^{\infty} e^{-x} e^{ix}dx$ evaluates to "$a+bi$", $$a=\int_0^{\infty} e^{-x} \cos(x)dx$$ $$b=\int_0^{\infty} e^{-x}\sin(x)dx.$$
Evaluating the integral would work in the same way you evaluate any exponential function with real coefficients:
$$\int_0^{\infty} e^{-x} e^{ix}dx = \int_0^{\infty} e^{-x+ix}dx = [\frac{e^{-x+ix}}{(i-1)}]_0^{\infty} = [\frac{(1+i)e^{-x+ix}}{-2}]_0^{\infty} = \frac{-1}{2}[(1+i)(\cos(x)+i\sin(x))e^{-x}]_0^{\infty}$$ $$= \frac{-1}{2}[(\cos(x)-\sin(x))e^{-x}]_0^{\infty} +\frac{-i}{2}[(\cos(x)+\sin(x))e^{-x}]_0^{\infty}$$ $$=\frac{1}{2}+i\frac{1}{2}.$$
So $$\frac{1}{2}=\int_0^{\infty} e^{-x} \sin(x)dx$$ (and as a little bonus: $\frac{1}{2}=\int_0^{\infty} e^{-x}\cos(x)dx.)$
