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.


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:

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

  • $\begingroup$ Can you also prove that $~\displaystyle\lim_{n\to\infty}~n-\int_{\tfrac1n}^n\text{erf }(a+\ln x)~dx ~=~ 2~\exp\Big(\tfrac14-a\Big)~$ ? $\endgroup$ – Lucian May 19 '16 at 18:27
  • 1
    $\begingroup$ @Lucian: isn't it just a consequence of the same technique? Set $x=e^t$, write $\text{erf}$ as an integral, switch integrals, profit. $\endgroup$ – Jack D'Aurizio May 19 '16 at 18:37
  • $\begingroup$ Jack, I believe that the exponent in the first integral should lead with $2t$, not $t$. $x\,dx\to e^{2t}\,dt$. -Mark $\endgroup$ – Mark Viola May 19 '16 at 20:40
  • $\begingroup$ @Dr.MV: you're right, but the technique's still good. I will fix this soon. $\endgroup$ – Jack D'Aurizio May 19 '16 at 20:42
  • $\begingroup$ And now a +1 ... $\endgroup$ – Mark Viola May 19 '16 at 20:45

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