Looking for a closed form for the integral

$$\int_0^{\infty } e^{-\left(\frac{a-\log (x)}{b}\right)^2} \left(\frac{1}{2} \text{erf}\left(\frac{a-\log (x)}{b}\right)+\frac{1}{2}\right) \, \mathrm{d}x,$$ where erf is the error function $erf (z)=\frac{2}{\sqrt{\pi }}\int _0^ze^{-t^2}\mathrm{d}t.$

I've tried all manner of tricks, to no avail.

  • $\begingroup$ what kind of tricks? $\endgroup$ – tired Feb 17 '15 at 15:42
  • 2
    $\begingroup$ Subbing $x=e^y$ will at least do no harm. $\endgroup$ – Ron Gordon Feb 17 '15 at 15:47
  • $\begingroup$ For $a=0$, one term is a Gaussian integral, and the other one is $0$, due to the parity of the integrand. $\endgroup$ – Lucian Feb 17 '15 at 16:10
  • $\begingroup$ Substitution $x=e^y$ which removes the log didn't work. $\endgroup$ – Nero Feb 17 '15 at 18:09
  • $\begingroup$ I'm not sure what you mean, "it didn't work." Do you mean that you get an integral that you can't evaluate? $\endgroup$ – Ron Gordon Feb 17 '15 at 19:15

Your integral admits a closed form.

Theorem. Let $a$ and $b$ be any real numbers. Then

$$\int_0^{\infty }\!\! e^{\large-\left(\frac{a-\log x}{b}\right)^2}\!\! \left(\text{erf}\left(\frac{a-\log x}{b}\right)+1\right) \mathrm{d}x=\sqrt{\pi}\:|b|\:e^{{\large {a+\frac{b^2}{4}}}}\left(1-\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right)\! \right) \tag1$$

where $\displaystyle \text{erf}(z)=\frac{2}{\sqrt{\pi }}\int _0^ze^{\large-t^2}\mathrm{d}t.$

Proof. Let's denote the left hand side of $(1)$ by $I(a,b)$.

By the change of variable $$\displaystyle U:=\frac{a-\log x}{b}, \quad dx=-b\:e^a \:e^{-b U}dU,$$ we have $$ \begin{align} I(a,b)&=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \left(\text{erf}\left(U\right)+1\right) \mathrm{d}U\\\\ &=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U+|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U \tag2 \end{align} $$ Clearly, by the gaussian integral: $$ \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U =e^{\large b^2/4}\int_{-\infty }^{+\infty } e^{\large-(U+b/2)^2} \mathrm{d}U =\sqrt{\pi}\:e^{ b^2/4} .\tag3 $$ Set $$ f(b):=\int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U \tag4 $$ and observe that $$ f(0)=\frac{\sqrt{\pi}}{4}\left[(\text{erf}(U))^2\right]_{-\infty }^{+\infty }=0. \tag5$$ Differentiating $(4)$ with respect to $b$ and performing an integration by parts gives $$ \begin{align} f'(b)&=-\int_{-\infty }^{+\infty } U\:e^{\large-U^2-bU} \text{erf}(U)\:\mathrm{d}U\\\\ f'(b)&=\left[\frac 12 e^{\large-U^2}\left(e^{\large-bU} \text{erf}(U)\right)\right]_{-\infty }^{+\infty }-\frac 12\int_{-\infty }^{+\infty } e^{\large-U^2} \left(-b\:e^{\large-bU} \text{erf}(U)+\frac{2}{\sqrt{\pi }}e^{\large-bU}e^{\large-U^2}\right)\mathrm{d}U\\\\ f'(b)&=\frac{b}{2}f(b)-\frac{1}{\sqrt{\pi }}\int_{-\infty }^{+\infty } e^{\large-2U^2-bU}\mathrm{d}U \end{align} $$ or $$ f'(b)=\frac{b}{2}f(b)-\frac{\sqrt{2}}{2}\:e^{\large b^2/8}. \tag6 $$ We classically solve the ODE $(6)$, using $(5)$, to obtain $$ f(b)=-\sqrt{\pi }\:e^{\large b^2/4}\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right) \tag7 $$ then plugging $(7)$, $(4)$ and $(3)$ into $(2)$ gives $(1)$ as desired.

  • 2
    $\begingroup$ Thanks a million. This is it. $\endgroup$ – Nero Feb 18 '15 at 0:28

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