A logarithm integral

Question

Calculate the integral \begin{align} \int_{0}^{1} \frac{ \ln(\sqrt{x} - \sqrt{1-x}) }{ \sqrt{x} } \ dx \end{align} and show the value is negative.

Answer 1

$$I=\int_{0}^{1}\frac{\ln (\sqrt{x}-\sqrt{1-x})}{\sqrt{x}}dx\\=2\sqrt{x}\ln(\sqrt{x}-\sqrt{1-x})|_{0}^1-\int_{0}^1\frac{\sqrt{x}+\sqrt{1-x}}{\sqrt{1-x}}dx$$ The first part is $0$ and the second integral is $$-1-1/2\int_{0}^{1}\frac{x+1-x}{\sqrt{x(1-x)}}dx=-1-1/2\int_{0}^{\pi/2}\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}d\theta=-1-\pi/2$$

Answer 2

This integral cannot be real. The issue in Samrat's proof is that $\log(\sqrt{x} - \sqrt{1-x})$ has sort of jump at $x = 1/2$.

Let us use the standard branch cut so that $\log(-x) = \log x + \pi i$ for $x > 0$. Then we can write

\begin{align*} \int_{0}^{1} \frac{\log(\sqrt{x} - \sqrt{1-x})}{\sqrt{x}} \, dx &= \int_{0}^{1/2} \frac{\log(\sqrt{1-x} - \sqrt{x}) + \pi i}{\sqrt{x}} \, dx \\ &\quad + \int_{1/2}^{1} \frac{\log(\sqrt{x} - \sqrt{1-x})}{\sqrt{x}} \, dx \\ &= \sqrt{2}\pi i + I_{1} + I_{2}, \end{align*}

where

$$ I_{1} = \int_{0}^{1/2} \frac{\log(\sqrt{1-x} - \sqrt{x})}{\sqrt{x}} \, dx \quad \text{and} \quad I_{2} = \int_{0}^{1/2} \frac{\log(\sqrt{1-x} - \sqrt{x})}{\sqrt{1-x}} \, dx. $$

For $I_{1}$, integration by parts shows that

\begin{align*} I_{1} &= \left[ (2\sqrt{x} - \sqrt{2}) \log(\sqrt{1-x} - \sqrt{x}) \right]_{0}^{1/2} \\ &\quad + \frac{1}{2} \int_{0}^{1/2} \frac{2\sqrt{x} - \sqrt{2}}{\sqrt{1-x} - \sqrt{x}}\left( \frac{1}{\sqrt{1-x}} + \frac{1}{\sqrt{x}} \right) \, dx. \end{align*}

(Here, the bizarre choice for the antiderivative is introduced to cancel out the logarithmic singularity.) Likewise,

\begin{align*} I_{2} &= \left[ (\sqrt{2} - 2\sqrt{1-x}) \log(\sqrt{1-x} - \sqrt{x}) \right]_{0}^{1/2} \\ &\quad + \frac{1}{2} \int_{0}^{1/2} \frac{\sqrt{2} - 2\sqrt{1-x}}{\sqrt{1-x} - \sqrt{x}}\left( \frac{1}{\sqrt{1-x}} + \frac{1}{\sqrt{x}} \right) \, dx. \end{align*}

Adding them together,

$$I_{1} + I_{2} = - \int_{0}^{1/2} \left( \frac{1}{\sqrt{1-x}} + \frac{1}{\sqrt{x}} \right) \, dx = -\int_{0}^{1} \frac{dx}{\sqrt{x}} = -2. $$

Therefore the final answer is

$$ \int_{0}^{1} \frac{\log(\sqrt{x} - \sqrt{1-x})}{\sqrt{x}} \, dx = -2 + \sqrt{2}\pi i. $$