What function has $\int_{-\infty}^\infty f(x) x \exp(-x^2)dx = 0$? What function has
$$\int_{-\infty}^\infty f(x) x \exp(-x^2)dx = 0$$?
The obvious answer is that $f(x)$ is an even function. Then the integral is always zero.
Another answer is $f(x) = H_k(x)$ is a Hermite polynomial with $k\neq 1$, since $H_1(x) = x$ and Hermite polynomials are orthogonal under Gaussian measure.
Are there other special functions or function classes with the above property?
 A: Any function that has no first term in a Hermite expansion
$$
f(x)=\sum_{n=0}^\infty d_n H_n(x)
$$
since
$$
d_n \propto \int_{-\infty}^\infty f(x)H_n(x)e^{-x^2}dx.
$$
By integrating by parts, this can also be written
$$
d_n \propto \int_{-\infty}^\infty \frac{\partial^n f(x)}{\partial x^n}e^{-x^2}dx
$$
so the condition of no first term is equivalent to any function that satisfies
$$
\int_{-\infty}^\infty \frac{\partial f(x)}{\partial x}e^{-x^2}dx=0.
$$
Even functions have this property, and so do all Hermite polynomials above first order.
It is not the even-ness (symmetry) of the function that matters, we can construct a function that is not even but has this property:
$$
f(x) = a\delta(x-c) + b\delta(x+d)
$$
and choose $a,b,c,d$ such that
$$
cae^{-a^2} - dbe^{-b^2}=0.
$$
For instance, $a=1,c=1,d=2$ leads to $b\approx 1.432736294276$.
If you'd prefer a more well behaved function, replace the delta functions with smooth approximations like the Gaussian with very small variances.  For instance
$$
\int_{-\infty}^\infty \left(\exp\left(-\frac{(x-1)^2}{2\cdot 100000}\right)+1.43273629427692367\exp\left(-\frac{(x+2)^2}{2\cdot 100000)}\right)\right)\frac{x\exp(-x^2)}{\sqrt{2\pi\cdot 100000}}dx
$$
is zero to eight decimal places.
