Why does $\int_{-\infty}^{\infty}e^{-x^2}\sin x\,dx=0?$ I can't get my head around something...
Why does $\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\sin x\,dx=0$ but $\displaystyle\int_{-\infty}^{\infty} \sin x\,dx$ or $\displaystyle\int_{-\infty}^{\infty} \frac{\sin x}{x^{2n}}\,dx$ doesn't converge?
I thought maybe the first equality can be justified by saying the integrand is odd, but since this is also the case for the others, I don't understand why they aren't $0$. Does this have something to do with the exponential function "dominating" the sine?
 A: The function $e^{-x^2}\sin x$ is an odd function. So if we have convergence, the answer of $0$ is automatic. And convergence is clear, because of the rapid decay of $e^{-x^2}$. 
A: Why does any improper integral converge?  Its really based on how quickly the integrand "goes to zero."
The exponential is negative in the first integral, which means the value of the integrand will decrease very rapidly as $x$ increases or decreases.
Sine just bounces up and down between -1 and 1, so it doesn't ever converge to a given value.
In the third integral, as $x$ gets bigger, the oscillations will get bigger, which certainly doesn't converge.
EDIT:
For the new third integral--it is kinda curious that it doesn't converge.  (That is, you can't tell just by looking without practice.)  However, evidently, the $x^{-2n}$ term doesn't decay fast enough to make the integral converge.
EDIT V2:
see @sos440's comment as to why the third integral doesn't converge.
A: If $f$ is measurable  bounded by $B$, then $|\int_{-\infty}^{\infty} e^{-x^2} f(x)dx| \leq B \int_{-\infty}^{\infty} e^{-x^2} dx \leq B(\int_{-1}^{1} e^{-x^2} + 2 \int_1^\infty e^{-x} dx) \leq 2B(1+e)$.
