Finding an inverse integral transform

Question

I have the following integral transform: $$ f(y) = \frac{1}{\sqrt{4\pi y}} \int_0^\infty x \exp\left(-\frac{x^2}{4y}\right) g(x) dx$$ where $g(x)$ is an even polynomial in $x$. Does somebody know the inverse integral transform, expressing $g(x)$ in terms of $f(y)$? Thanks in advance.

Answer 1

Since $g(x)$ is a polynomial in $x$, you face $$I_n = \frac{1}{\sqrt{4\pi y}} \int_0^\infty x^n\exp\left(-\frac{x^2}{4y}\right)\, dx$$ Let $x=2 \sqrt{ty}$ to make $$I_n=\frac{2^{n-1} y^{n/2}}{\sqrt{\pi }}\int_0^\infty e^{-t} t^{\frac{n-1}{2}}\,dt$$ Now, using the gamma function $$I_n=\frac{2^{n-1} \Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\pi }}\,y^{n/2}$$

Suppose that $g(x)=a+b x+c x^2+d x^3$ which would give $$f(y)=\frac{a }{\sqrt{\pi }}\sqrt{y}+b y+\frac{4 c}{\sqrt{\pi }} y^{3/2}+6 d y^2$$ If $y$ is small, you could consider that this is a series and use series reversion