Compute $\int_{0}^{\infty}\frac{e^{-ax-\frac{b}{x}}}{x^{3/2}}dx$

Question

Problem: Compute the integral

$$\int_{0}^{\infty}\frac{e^{-ax-\frac{b}{x}}}{x^{3/2}}dx.$$

What I have done is u-substitution. I let $u=\frac{1}{x^{1/2}}$ and I end up with: $$2\int_{0}^{\infty}e^{-\frac{a}{u^2}-bu^2}du.$$

But I'm stuck here. I appreciate any hints given. Thank you.

Answer 1

As explained in this answer it holds that

$$ \int_{-\infty}^\infty f\left(x - \frac{\color{Purple}{\alpha}}{x}\right)\, dx= \int_{-\infty}^\infty f(x)\,dx \qquad \alpha >0 $$

In your case, notice that $$ 2\int_{0}^{\infty}e^{-\frac{a}{u^2}-bu^2}\, du \overset{\color{Blue}{u\sqrt{b} =x}}{=} \frac{\color{green}{2}}{\sqrt{b}}\int_0^\infty e^{-\left(x^2 + \frac{ab}{x^2}\right)} \, dx = \frac{1}{\sqrt{b}}\int_{\color{green}{-\infty}}^\infty e^{-\left(x - \frac{\color{Purple}{\sqrt{ab}}}{x}\right)^2 - 2 \sqrt{ab}} \ dx $$ where we used the fact that the integral is even in the last step to change the limits of integration. We can now and apply the formula from the linked answer and get $$ I = \frac{1}{\sqrt{b}}\int_{-\infty}^\infty e^{-x^2 - 2 \sqrt{ab}} \ dx = \frac{1}{\sqrt{b}}e^{- 2 \sqrt{ab}}\int_{-\infty}^\infty e^{-x^2 } \ dx = \frac{1}{\sqrt{b}}e^{- 2 \sqrt{ab}}\sqrt{\pi} = \boxed{e^{- 2 \sqrt{ab}}\sqrt{\frac{\pi}{b}}} $$ for $a\ge 0, b > 0$ (otherwise you can't guarantee convergence) and where on the penultimate equality we used the evaluation of the Gaussian integral.

Answer 2

This is similar to Robert Lee's answer, but a bit more self-contained: $$ \begin{align} \int_0^\infty\frac{e^{-ax-\frac{b}{x}}}{x^{3/2}}\,\mathrm{d}x &=2\left(\frac ab\right)^{1/4}\int_0^\infty\frac{e^{-\sqrt{ab}\left(x^2+\frac1{x^2}\right)}}{x^2}\,\mathrm{d}x\tag1\\ &=2\left(\frac ab\right)^{1/4}e^{-2\sqrt{ab}}\int_0^\infty\frac{e^{-\sqrt{ab}\left(x-\frac1x\right)^2}}{x^2}\,\mathrm{d}x\tag2\\ &=2\left(\frac ab\right)^{1/4}e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac1x\right)^2}\,\mathrm{d}x\tag3\\ &=\left(\frac ab\right)^{1/4}e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac1x\right)^2}\,\mathrm{d}\!\left(x-\tfrac1x\right)\tag4\\ &=\left(\frac ab\right)^{1/4}e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\,u^2}\,\mathrm{d}u\tag5\\ &=\sqrt{\frac\pi{b}}\,e^{-2\sqrt{ab}}\tag6\\ \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto\sqrt{\frac ba}\,x^2$
$(2)$: complete the square
$(3)$: substitute $x\mapsto\frac1x$
$(4)$: average $(2)$ and $(3)$
$(5)$: substitute $u=x-\frac1x$
$(6)$: $\int_{-\infty}^\infty e^{-\lambda x^2}\,\mathrm{d}x=\sqrt{\frac\pi\lambda}$