An Integral involving nature logarithm

Question

I wish to find the closed form expression of the following integral $$\int\limits_{0}^1\frac{\log^{p-1}(x)}{x}\log\left(\frac{2}{1+\sqrt{1+x}}\right)dx=?\quad(p\in\mathbb{N}).$$ By using the identity \begin{align}\label{b1} \sum\limits_{n=1}^\infty \frac{x^n}{4^n n} \binom{2n}{n} =2\log\left(\frac{2}{1+\sqrt{1-x}}\right), \end{align} we can find that $$\int\limits_{0}^1\frac{\log^{p-1}(x)}{x}\log\left(\frac{2}{1+\sqrt{1+x}}\right)dx=\frac{(-1)^{p-1}(p-1)!}{2}\sum\limits_{n=1}^\infty \frac{(-1)^n}{4^n n^{p+1}}\binom{2n}{n}.$$

Answer 1

The latter series is quite easy only when $p+1 = 0$, which doesn't belong to your case, or when $p+1 = 1$, which might be your case depending on how you define the set $\mathbb{N}$, which usually doesn't take the zero into account.

Assuming then $p\in\mathbb{W}$, where $\mathbb{W}$ denotes the Whole set, that is, $$\mathbb{W} = \{\mathbb{N} + \{0\}\}$$

we find that the latter series converges to

$$\ln(4) - 2\ln(1 + \sqrt{2})$$

Considering that I'm talking about the series only, the coefficient in front of it becomes trivial for $p = 0$.

All the other cases, where $p+1 = k$, $k>1$ make the series to converge as well but you won't fine any close numerical forms, in terms of elementary functions or values, but only via HyperGeometric Functions (Google for more).

The only closed form is then the one for $p = 0$, taking your range of definition for $p$.