integration - Closed form for this integral with Beta function $int_0^infty x mathrm B (x,x)~dx$

Is it possible to find the closed form of

$$\int_0^\infty x \mathrm B (x,x)~dx=2.44333\dots$$

This integral converges because for small $x$ Beta function behaves like $2/x$.

Using the integral definition of the Beta function, we can transform it to the following forms:

$$\int_0^\infty x \mathrm B (x,x)~dx=\int_0^1 \frac{dt}{t(1-t) \ln^2 (t(1-t))}=^{t(1-t)=z}$$

$$=2 \int_0^{1/4} \frac{dz}{z \sqrt{1-4z} \ln^2 z}=^{4z=\sin^2 y}= \int_0^{\pi/2} \frac{dy}{\sin y \ln^2 (\sin (y)/2)}$$

Another substitution will get rid of the logarithm:

$$=2 \int_0^{1/4} \frac{dz}{z \sqrt{1-4z} \ln^2 z}=^{z=\exp (-u)}=2 \int_{ \ln 4}^{\infty} \frac{du}{u^2\sqrt{1-4 \exp (-u)}}=^{v=1/u}$$

$$=2 \int_0^{ 1/ \ln 4} \frac{dv}{\sqrt{1-4 \exp (-1/v)}}$$

However, I have not been able to proceed further to some special function or even a series.

It might surrender to residues, but I'm useless with them.

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Edit

Trying integration by parts. Take $u=1/\sqrt{1-4z}$ and $dv=dz/(z \ln^2 z)$

$$\int_0^{1/4} \frac{dz}{z \sqrt{1-4z} \ln^2 z}=-\frac{1}{\ln z \sqrt{1-4z}} |_0^{1/4}+2 \int_0^{1/4} \frac{dz}{(1-4z)^{3/2} \ln z}$$

Not working out, the first part blows up at the limits.

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A series solution in terms of elementary functions would be acceptable too! I've found one, but with incomplete Gamma function, which is just an integral in disguise (see my answer).

Answer

There is a kind of series 'solution' which I will put up as an answer in hope that someone will offer a better one.

Let's work with this form of the integral:

$$2 \int_0^{ 1/ \ln 4} \frac{dv}{\sqrt{1-4 \exp (-1/v)}}$$

We are simply going to use the following series for $|p|<1$:

$$\frac{1}{\sqrt{1-p}}=\frac{1}{\sqrt{\pi}} \sum_{k=0}^\infty \frac{\Gamma \left(k+\frac12 \right)}{k!}p^k$$

$$\frac{1}{\sqrt{1-4 e^{-1/v}}}=\frac{1}{\sqrt{\pi}} \sum_{k=0}^\infty \frac{\Gamma \left(k+\frac12 \right)}{k!}4^k e^{-k/v}$$

The integration by terms seems to work:

$$\int_0^{ 1/ \ln 4} e^{-k/v}dv=\frac{1}{4^k \ln 4}-k \Gamma(0,k \ln 4)$$

Thus, the integral is equal to:

$$2 \int_0^{ 1/ \ln 4} \frac{dv}{\sqrt{1-4 e^{-1/v}}}=\frac{1}{\sqrt{\pi} \ln 2} \sum_{k=0}^\infty \frac{\Gamma \left(k+\frac12 \right)}{k!} \left(1-k ~4^k \ln 4~ \Gamma(0,k \ln 4) \right)$$

There is some kind of trouble for $k=0$, but the limit exists:

$$\lim_{q \to 0} q~ \Gamma(0,q)=0$$

Finally we can write:

$$\int_0^\infty x \mathrm B (x,x)~dx=\frac{1}{\ln 2} +\frac{1}{ \sqrt{\pi}~\ln 2} \sum_{k=1}^\infty \frac{\Gamma \left(k+\frac12 \right)}{k!} \left(1-k ~4^k \ln 4~ \Gamma(0,k \ln 4) \right)$$

$$\int_0^\infty x \mathrm B (x,x)~dx= \frac{1}{\ln 2} + \sum_{k=1}^\infty \binom{2k}{k} \left(\frac{1}{4^k \ln 2}-2k ~ \Gamma(0,k \ln 4) \right)$$