Boundedness of an integral of square function implying zero integral Let $\alpha:\mathbb R\mapsto\mathbb R$ be the smooth function such that $$\int_{-\infty}^{\infty}[\alpha'(x)-x\alpha(x)]^2e^{-\frac{x^2}2}dx<\infty.$$
I wish to prove that $$\int_{-\infty}^{\infty}[\alpha'(x)-x\alpha(x)]e^{-\frac{x^2}2}dx=0.$$
Here is what I got so far,
Using partial integral I have $$\int_{-\infty}^{\infty}\alpha'(x)e^{-\frac{x^2}2}dx=\alpha(x)e^{-\frac{x^2}2}|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}x\alpha(x)e^{-\frac{x^2}2}dx.$$ This implies that $$\int_{-\infty}^{+\infty}[\alpha'(x)-x\alpha(x)]e^{-\frac{x^2}2}dx=\left.\alpha(x)e^{-\frac{x^2}2}\right|_{-\infty}^{+\infty} $$ and I am stack here. Could anyone help me? Any help will be appriciated.
 A: Your proposition is not correct. 
Here is a counter-example.
Set 
$$
\alpha(x):=e^{\frac{x^2}2}\int_0^{x}e^{-\frac{t^2}3+t} {\rm d} t.
$$
Then
$$
\begin{align}
\alpha'(x) & =x\:e^{\frac{x^2}2}\int_0^{x}e^{-\frac{t^2}3+t}{\rm d}x +e^{\frac{x^2}2}e^{-\frac{x^2}3+x}=x\alpha(x)+e^{\frac{x^2}6+x} 
\end{align}
$$
and
$$
\int_{-\infty}^{+\infty}[\alpha'(x)-x\alpha(x)]^2e^{-\frac{x^2}2}{\rm d}x= \int_{-\infty}^{+\infty}e^{-\frac{x^2}6+2x}{\rm d}x<\infty.
$$
But
$$
\begin{align}
\int_{-\infty}^{+\infty}[\alpha'(x)-x\alpha(x)]e^{-\frac{x^2}2}{\rm d}x &=\left.\alpha(x)e^{-\frac{x^2}2}\right|_{-\infty}^{+\infty}\\\\
 &=\left.\int_0^{x}e^{-\frac{t^2}3+t}{\rm d}t\right|_{-\infty}^{+\infty}\\\\
 &=\int_{-\infty}^{+\infty}e^{-\frac{t^2}3+t}{\rm d}t\\\\
 &=\sqrt{3\pi}e^{3/4}\\\\
 &\neq 0.\\\\
\end{align}
$$
Remark. Applying Cauchy-Schwarz inequality to
$$
f(x):=[\alpha'(x)-x\alpha(x)]e^{\Large -\frac c2 x^2}, \quad
g(x):=e^{\Large-\frac c2 x^2}
$$ we have
$$
\left|\int_{-\infty}^{+\infty}[\alpha'(x)-x\alpha(x)]e^{\Large-c x^2}{\rm d}x\right|\leq \int_{-\infty}^{+\infty}[\alpha'(x)-x\alpha(x)]^2e^{-\frac{x^2}2}{\rm d}x<+\infty
$$ when $c\leq\dfrac14$.
A: the first formula $$\int_{-\infty }^{\infty } e^{-\frac{x^2}{2}} \left(x f(x)-f'(x)\right)^2 \, dx<\infty$$ it is necesary condition for the converge of the integral using Mellin transfro it very easy to see that
$$\int_{-\infty }^{\infty } e^{-\frac{x^2}{2}} \left(x f(x)-f'(x)\right) \, dx=0$$
XE f(x)=Sin(x) or E^-x gives Zero
and whith any rational c 
$$\int_{-\infty }^{\infty } e^{-c x^2} \left(2 c x f(x)-f'(x)\right) \, dx=0$$
or
$$\int_{-\infty }^{\infty } e^{-c (b+x)^2} \left(f(x) (2 b c+2 c x)-f'(x)\right) \, dx=0$$
