Proving that $\int_0^1 \frac{\ln(x)+\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx=0$

I am looking, if a simple way exists , to show that K=0 , using symmetry !, a clever change of variables or integrations by parts !, without taking separately integrals , where $\boxed{K=\int_0^1 \frac{\ln(x)+\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx=0}$

If we take separately $I=\int_0^1 \frac{\ln(x)}{\sqrt {1-x^2}}$ and $J=\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}}$, the integrals $I$ and $J$ define the same series to a sign (two series of opposite sums) $ I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$ and $J=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$

Explanation By Fourier series $\ln\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)=\sum_{k=0}^\infty\frac{(2k)!}{4^k(2k+1)(k!)^2}\sin((2k+1)x)$ then $J=\int_0^{\frac{\pi}{2}} \ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)dt=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$ $ I=\int_{0}^{1}\frac{\log(t)}{\sqrt{1-t^{2}}}dt$ We know that $\frac{1}{\sqrt{1-t^{2}}}=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}}t^{2n} $ and $ \int_{0}^{1}\log(t)t^{2n}dt=-\frac{1}{(2n+1)^{2}}$ then $I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}} $

similarly @Zacky question prove that $\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$ without calculating separately integrals.

We can write K as $K=\displaystyle \int_0^{\dfrac{\pi}{2}} \Big(\ln(\sin t )+\ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)\Big)dt=0 $

Remark : Wolframalpha can calculate K But do not know how to calculate $\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx$. I believe that Wolframe uses a simple way to see that $K$ is zero . same observation for the Integral of Zaky

Solution 1:

I know this is not exactly what it is asked for. On the path of Zacky,

\begin{align*} J&=\int_0^1\frac{\text{arcsinh}\left(\sqrt{x}\right)}{\sqrt{1-x^2}}dx\\ &\overset{\text{IBP}}=\Big[\arcsin (x)\text{arcsinh}\left(\sqrt{x}\right)\Big]_0^1-\frac{1}{2}\underbrace{\int_0^1 \frac{\arcsin x}{\sqrt{x}\sqrt{1+x}}dx}_{_{z=\sqrt{\frac{1-x}{1+x}}}}\\ &=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\arcsin\left(\frac{1-z^2}{1+z^2}\right)}{(1+z^2)\sqrt{1-z^2}}dz\\ &=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\left(\frac{\pi}{2}-2\arctan z\right)}{(1+z^2)\sqrt{1-z^2}}dz\\ &=\frac{\pi}{2}\left(\ln\left(1+\sqrt{2}\right)-\sqrt{2}\underbrace{\int_0^1 \frac{z}{(1+z^2)\sqrt{1-z^2}}dz}_{=\text{K}}\right)+2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz\\ K&=\frac{1}{2\sqrt{2}}\left[\ln\left(\frac{\sqrt{2}-\sqrt{1-z^2}}{\sqrt{2}+\sqrt{1-z^2}}\right)\right]_0^1=\frac{1}{\sqrt{2}}\ln\left(1+\sqrt{2}\right)\\\ J&=2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz \end{align*} Define on $[0,1]$, \begin{align*}F(a)&=\int_0^1 \frac{z\arctan (az)}{(1+z^2)\sqrt{1-z^2}}dz\\ F^\prime(a)&=\int_0^1 \frac{z^2}{(1+z^2)(1+a^2z^2)\sqrt{1-z^2}}dz\\ &=-\frac{1}{2}\left[\frac{\sqrt{2}\arctan\left(\frac{z\sqrt{2}}{\sqrt{1-z^2}}\right)-\frac{2}{\sqrt{1+a^2}}\arctan\left(\frac{z\sqrt{1+a^2}}{\sqrt{1-z^2}}\right)}{1-a^2}\right]_{z=0}^{z=1}\\ &=\frac{\pi\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{2(1-a^2)\sqrt{2}} \end{align*} Since $F(0)=0$ then, \begin{align*}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz&=F(1)-F(0)\\ &=\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{1-a^2}da\\ &=\frac{\pi}{2\sqrt{2}}\left[\text{arctanh}\left(\frac{\sqrt{2}a}{\sqrt{1+a^2}}\right)-\text{arctanh}(a)\right]_0^1\\ &=\frac{\pi}{2\sqrt{2}}\lim_{a\rightarrow 1}\ln\left(\frac{\sqrt{1+a^2}+a\sqrt{2}}{1+a}\right)\\ &=\frac{\pi\ln 2}{4\sqrt{2}} \end{align*} Therefore, \begin{align*}\boxed{J=\frac{\pi\ln 2}{2}}\end{align*}