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!

  • 3
    $\begingroup$ What do you think about the substitution $ax=t^2-ab$ ? $\endgroup$ – Bumblebee Aug 26 '17 at 19:11
  • $\begingroup$ Yeah, comes natural to the mind $\endgroup$ – Francesco Alem. Aug 26 '17 at 19:28

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

  • $\begingroup$ You mean why does it from change from $0$ and $\infty$ to $\sqrt{ab}$ and $\infty$? $\endgroup$ – tattwamasi amrutam Aug 26 '17 at 19:33
  • $\begingroup$ Never mind, I got it $\endgroup$ – Crescendo Sep 3 '17 at 17:24

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