# 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.

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.

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).

• WolframAlpha only manages to calculate your original integral for finite upper bounds, suggesting a closed form may not be avaliable. – Simply Beautiful Art Aug 2 '16 at 22:44
• @SimpleArt, I wouldn't rely on Wolfram Alpha about things like that. I tried Mathematica as well, but again - CAS are limited, unlike us, superior human beings – Yuriy S Aug 3 '16 at 5:49

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:

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

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

• I've accepted this answer for the lack of a better one. If someone will offer a better answer, I will gladly accept it instead – Yuriy S Dec 16 '16 at 12:12