I have a problem to solve and I have something that i don't know how to do.
The half-space Gaussian integral is given :
$$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$
I have to calculate
$$\int_{0}^\infty \exp \left(-ax^2 - \frac{b}{x^2} \right)dx$$
a and b are real and positive
Is someone have an idea ? :)
Thanks for your answers
Mathieu
Note that in THIS ANSWER, I presented a solution to a more general version of the integral of interest herein.
Let $I(a,b)$ be the integral given by
for $a>0$ and $b>0$.
$$I(a,b)=\int_0^\infty e^{-\left(ax^2+\frac{b}{x^2}\right)}\,dx \tag 1$$
Enforcing the substitution $x\to \sqrt[4]{b/a}x$ into $(1)$ reveals
$$\begin{align} I(a,b)&=\sqrt[4]{\frac{b}{a}}\int_0^\infty e^{-\sqrt{ab}\left(x^2+\frac{1}{x^2}\right)}\,dx \tag 2 \end{align}$$
Next, noting that $x^2+\frac{1}{x^2}=\left(x-\frac1x\right)^2+2$, we can write $(2)$ as
$$\begin{align} I(a,b)&=\sqrt[4]{\frac{b}{a}}e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,dx\tag 3 \end{align}$$
Enforcing the substitution $x\to 1/x$ in $(3)$ yields
$$I(a,b)=\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,\frac{1}{x^2}\,dx\tag4$$
Adding $(3)$ and $(4)$, we obtain
$$\begin{align} I(a,b)&=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,d\left(x-\frac{1}{x}\right)\\\\ &=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\,x^2}\,dx\\\\ &=\frac1{2\sqrt a} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-x^2}\,dx\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\sqrt{\pi}}{2\sqrt a}e^{-2\sqrt{ab}}} \end{align}$$
$$\exp \left(-ax^2 - \frac{b}{x^2} \right)$$ $$ = \exp \left( -ax^2\right)*\exp \left( -\frac{b}{x^2} \right)$$ $$=\exp \left( -ax^2\right)*\exp \left( -\frac{b*x^2}{x^4} \right)$$ $$=\exp \left( -ax^2\right)*\exp \left( -bx^2\right)*\exp \left( -\frac{1}{x^4} \right)$$
Now something with partial integration?
