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

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

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.