Help evaluate $\int_0^\infty x\operatorname{erfc}(a + b\ln (x)) \,dx$.

Question

I am trying to evaluate $$ I = \int_0^\infty x\operatorname{erfc}(a + b\ln (x)) \,dx $$ where $a \ge 0$ and $b> 0$. $$ I = \frac{2}{\sqrt{\pi}}\int_0^\infty \int_{a + b\ln (x)}^{\infty} \mathrm{e}^{-u^2}\,du \,dx $$

Some options I have explored towards finding the solution.

Use Series Expansion of the exponential function: Since the upper limit of the inner integral is infinite I think using the Maclaurin series of exponential function will not work.

Change the order of integration: I think the region of integration is $x < a + b \ln(x) < \infty$ & $0 < x < \infty$, but I am not sure what it's corresponding region is. I suspect it is $0 < x < a + b \ln(x)$ & $0 < a + b \ln(x) < \infty$. If that is correct then is the below integral the correct formulation ?

$$I = \frac{2}{\sqrt{\pi}}\int_{x = e^{-a/b}}^{x = \infty} \int_{u = 0}^{u = a + b \ln(x)} \mathrm{e}^{-u^2}\,du \,dx$$

Thanks in advance for your help.

Answer 1

Let $x=e^t$. Then

$$ I = \frac{2}{\sqrt{\pi}}\int_{-\infty}^{+\infty}\int_{0}^{+\infty}\exp\left(2t-(u+a+bt)^2\right)\,du\,dt $$ is just the integral of $\exp(-q(u,t))$ with $q$ being a quadratic form in $u,t$, hence it can be computed by completing the square or just switching the order of integration:

$$ I = \frac{2e^{1/b^2}}{b}\int_{0}^{+\infty}\exp\left(-2\,\frac{a+u}{b}\right)\,du = \color{red}{\exp\left(\frac{1-2ab}{b^2}\right)}.$$