Proving that $\int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx=\frac{2G}{\pi}$

Question

In my research about distribution theory in the topic of probability and statistic, I came across the following integral: $$\int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx$$ I run it using WolframAlpha and as shown here I have got $\dfrac{2C}{\pi}$, where $C$ is Catalan's constant. The latter let me to believe that $$\int_0^t \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx$$ have a nice closed form which i didn't Get it using integration by part and series asymptotic of both error function and complementary error function, Any way to get that closed form?

Note: $\mathrm{erfc}$ is the complementary error function.

Answer 1

I'm not sure what definition of the error function you have, but I will use: $$\operatorname{erf}(x)=\frac{2x}{\sqrt \pi}\int_0^1 e^{-x^2z^2}dz,\quad \operatorname{erfc}(x)=\frac{2x}{\sqrt \pi}\int_1^\infty e^{-x^2y^2}dy$$ Now notice that via the substitution $x\to \frac{1}{x}$ your integral is: $$I=\int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx=\int_0^\infty \frac{\operatorname{erf}(x)\operatorname{erfc}(x)}{x}dx$$ Using the integral representation of the error function that I mentioned we get: $$\require{cancel}I=\left(\frac{2}{\sqrt{\pi}}\right)^2 \int_0^1 \int_1^\infty \int_0^\infty \frac{xe^{-x^2z^2}\cdot \cancel xe^{-x^2y^2}dz}{\cancel x}dx dydz$$ $$\overset{\large x^2=t}=\frac{2}{\pi}\int_0^1 \int_1^\infty \int_0^\infty e^{-t(z^2+y^2)}dtdydx=\frac{2}{\pi}\int_0^1 \int_1^\infty \frac{1}{z^2+y^2}dydz$$ $$=\frac{2}{\pi}\int_0^1 \frac{\arctan z}{z}dz=\frac{2}{\pi}\operatorname{Ti}_2(1)=\frac{2G}{\pi}$$

You might try the same approach for your more general integral, although at first sight it won't look that nice.

Answer 2

\begin{eqnarray*} I=\int_{0}^{\infty} \text{erf} \left( \frac{1}{x} \right) \text{erfc} \left( \frac{1}{x} \right) \frac{dx}{x} = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{1/x} e^{-t^2} dt \int_{1/x}^{\infty} e^{-u^2} du \frac{dx}{x} \end{eqnarray*} substitute $X=1/x$ \begin{eqnarray*} I = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{X} e^{-t^2} dt \int_{X}^{\infty} e^{-u^2} du \frac{dX}{X} \end{eqnarray*} Now change the order of the integrations a couple of times \begin{eqnarray*} I &=& \frac{4}{\pi} \int_{0}^{\infty} \int_{t}^{\infty} e^{-t^2-u^2} \int_{t}^{u} \frac{dX}{X} du dt \\ &=& \frac{4}{\pi} \int_{0}^{\infty} \int_{t}^{\infty} e^{-t^2-u^2} \ln(u/t) du dt \\ \end{eqnarray*} Now rescale $u=ty$ \begin{eqnarray*} I &=& \frac{4}{\pi} \int_{0}^{\infty} \int_{1}^{\infty} e^{-t^2(1+y^2)} dt \ln(y) t dy dt \\ &=& \frac{2}{\pi} \int_{1}^{\infty} dt \frac{\ln(y)}{1+y^2} dy \\ \end{eqnarray*} This integral is a well known representation of the Catalan constant ... see https://en.wikipedia.org/wiki/Catalan%27s_constant#Integral_identities