Integrate $\int_{-\infty}^\infty xe^{-3x^{2}}\cos xdx$

Question

I need to solve this: $$\int_{-\infty}^\infty xe^{-3x^{2}}\cos(x)dx$$

Or maybe I don't need to compute the integral because the limits are infinite?

Answer 1

If we let $f(x)=xe^{-3x^{2}}\cos(x)$ , we notice that $$-f(x)=f(-x)$$ Hence, it is an odd function and then it can be easily seen that the 'positive' area from $0$ to $\infty$ get cancelled by 'negative' area from $0$ to $-\infty$.

So the integral is zero.

Here's the graph:

We also see the integral converges since the exponential terms 'dominates' the other two terms and makes the expression zero as we go to $\pm \infty$...This can also be done rigorously by bounding the function.

Answer 2

I think we would need to prove first that the integral $\exists$. $f(x)=x$ is also an odd function, $\int\limits_{-a}^{a}f=0$ $\forall a \in \mathbb{R}$, but $\nexists \int\limits_{-\infty}^{\infty}f$.
We have that $\forall x\geq0$: $$xe^{-3x^2} \geq xe^{-3x^2}\cos(x) \geq -xe^{-3x^2}$$ So: $$\int_{0}^{\infty}xe^{-3x^2}\geq \int_{0}^{\infty}\cos(x)xe^{-3x^2} \geq \int_{0}^{\infty}-xe^{-3x^2}$$ The integrals on the left and right side can be computed easily, it's $\frac{1}{6}$ and $-\frac{1}{6}$. So we have: $$\frac{1}{6}\geq\int_{0}^{\infty}\cos(x)xe^{-3x^2}\geq -\frac{1}{6}$$ We can do the same for the $\int\limits_{-\infty}^{0}$ case.
Now we can conclude that both of the integrals $\exists$, so the original integral $\exists$ as well. Now, we can do the same thing as you can see in the other answers to conclude that the integral of the function on $\mathbb{R}$ is zero since it's odd.