# Showing $\int_0^\infty\frac {r} {x^2}e^{-x^2-\frac {r^2} {x^2}}dx=\int_0^\infty e^{-x^2-\frac {r^2} {x^2}}dx$

As part of a parameter integral problem, I need to show that given the function $$\operatorname{F}\left(r\right) = \int_{0}^{\infty}\exp\left(-x^{2} - \frac{r^{2}}{x^{2}}\right){\rm d}x\,,\qquad \operatorname{F}\hspace{0.1mm}'\!\left(r\right) = -2\operatorname{F}\left(r\right)\,,$$ which amounts to what the title demands. However I cannot find a way to do so, either by integration by parts, substitution or anything else, even though it should be simple.

What am I missing $$?$$.

• Substitute $u=\frac{r}{x}$ – leoli1 Jan 21 at 21:41
• Incidentally, the larger problem this is a part of, in which you show $F^\prime=-2F$ and hence $F=F(0)e^{-2r}$, can be circumvented by evaluating $F$ directly, using Glasser's master theorem and the fact that the integrand is even. – J.G. Jan 21 at 21:43

Use a substitution $$x\mapsto r/x$$. It may be easier to follow if you write $$y=r/x$$ so the right-hand side transforms to $$\int_0^\infty e^{-r^2/y^2-y^2}dy$$.