10

Here is my question:

How to evaluate this integral: \begin{equation}\int_0^1\frac{\ln(\sqrt{x}+\sqrt{x+1})}{\sqrt{1-x^2}}\ dx=\frac{\pi\ln2}{2} \tag{1} \end{equation}

$\textbf{My Attempt:}$

Let $x=\sin\theta$, and we have

\begin{equation} \int_0^{\frac{\pi}{2}}\ln(\sqrt{\sin\theta}+\sqrt{\sin\theta+1})\ d\theta=\frac{\pi}{2}\ln(\sqrt2+1)-\int_0^{\frac{\pi}{2}}\theta\frac{\cos\theta}{\sqrt{\sin\theta}\sqrt{1+\sin\theta}}d\theta \tag{2} \end{equation}

Then I don't know how to continue. So I try the second method to convert it into a double integral. Notice that:

\begin{aligned} \int_0^1\frac{1}{1-x^2}\int_0^x\frac{1}{\sqrt{y}\sqrt{y+1}}dydx &=\int_{0}^{1}\frac{1}{\sqrt{y}\sqrt{y+1}}\int_y^1\frac{1}{\sqrt{1-x^2}}dxdy \\ &=\int_0^1\frac{\arcsin y}{\sqrt{y}\sqrt{y+1}}dy \end{aligned}

and again I cannot proceed with it.

Could anybody give me some hints? Thanks!

Yunxuan Zhang
  • 1,344
Laura Olatex
  • 1,195

1 Answers1

14

Substitute $x= \cos 2\theta$ \begin{align} &\int_0^1 \frac{\ln(\sqrt{x}+\sqrt{x+1})}{\sqrt{1-x^2}}\ dx\\ = & \int_0^{\pi/4}2 \cosh^{-1}\left(\sqrt2\cos \theta\right) d\theta\\ =&\int_1^{\sqrt2} d_t\bigg[\int_0^{\cos^{-1}\frac1t} 2\cosh^{-1}\left(t\cos \theta\right) d\theta\bigg]\\ =&\int_1^{\sqrt2}\int_0^{\cos^{-1}\frac1t}\frac{2\cos \theta}{\sqrt{t^2\cos^2\theta-1} }d\theta\ dt\\ =&\ \pi\int_1^{\sqrt2}\frac1t dt =\frac\pi2\ln2 \end{align}

Quanto
  • 120,125