TITLE: Proving two integrals are equal to each other
QUESTION [5 upvotes]: Question: How do you show that$$\int\limits_{0}^{\infty}\frac {e^{-ax}}{\sqrt{b+x}}\,\mathrm dx=\int\limits_{\sqrt{ab}}^{\infty}\frac 2{\sqrt{a}}e^{ab-t^2}\,\mathrm dt$$

This problem emerged when I was trying to prove an identity by working backwards, and I'm not sure how to tackle it. I need to find a way to get the limits of the right-hand side to zero to infinity but am unsure how to do that.
Should I make the substitution$$u^2=ab$$If so, how do I get rid of the $\sqrt{a}$ term in the denominator?
Note: Here, $a,b>0$ and are real numbers!

REPLY [6 votes]: Let $$t^2=ab+ax$$ Then $2t\,dt=a\,dx$. The integral becomes $$\int\limits_{\sqrt{ab}}^{\infty}\frac{e^{ab-t^2}}{\frac{t}{\sqrt{a}}}\frac{2t}{a}\,dt=\int\limits_{\sqrt{ab}}^{\infty}\frac{2}{\sqrt{a}}e^{ab-t^2}\,dt$$