How to evaluate the integral $\int_{0}^{1} \frac{\ln(x)}{\sqrt{2-x^2}} , dx$?

Question

I'm trying to evaluate the integral $\int_{0}^{1} \frac{\ln(x)}{\sqrt{2-x^2}}dx$ and I'm encountering some difficulties. Here's what I have tried so far:

I used the substitution $x = \sqrt{2} \sin(\theta)$, which transforms the integral into: $\int_{0}^{\frac{\pi}{4}} \ln(\sqrt{2} \sin(\theta)) d\theta.$
This further breaks down into: $\int_{0}^{\frac{\pi}{4}} \left(\ln(\sqrt{2}) + \ln(\sin(\theta)) \right)d\theta.$

This splits into two integrals: $\int_{0}^{\frac{\pi}{4}} \ln(\sqrt{2})d\theta + \int_{0}^{\frac{\pi}{4}} \ln(\sin(\theta))d\theta.$

The first part evaluates to: $\frac{\pi}{8} \ln(2)$. The second part uses a known result: $\int_{0}^{\frac{\pi}{4}} \ln(\sin(\theta))d\theta = -\frac{\pi}{4} \ln(2)$.

Combining these, I get: $\frac{\pi}{8} \ln(2) - \frac{\pi}{4} \ln(2) = -\frac{\pi}{8} \ln(2)$. Numerically, this result evaluates to approximately $-0.273$, but I expect the result to be around $-0.73$ based on numerical methods. Could someone help identify where my evaluation might be going wrong, or confirm the correct method to approach this integral?

Answer 1

The integral $\int_{0}^{\frac{\pi}{4}} \ln(\sin(\theta)) \mathrm{d}\theta$ does not equal $-\frac{\pi}{4} \ln(2)$ but instead $-\frac{G}{2}-\frac{\pi}{4}\ln(2)$ with Catalan's constant G. To get this result you can follow the steps here.

Answer 2

$$ \begin{aligned} \int_0^{\frac{\pi}{4}} \ln (\sin \theta) d \theta = & \int_0^{\frac{\pi}{4}} \ln \left(\frac{e^{\theta i}-e^{-\theta i}}{2 i}\right) d \theta \\ = & \int_0^{\frac{\pi}{4}}\left(\ln \left[e^{\theta i}\left(1-e^{-2 \theta i}\right)\right]-\ln 2-\ln i\right) d \theta\\ = & -\frac{\pi}{4} \ln 2+\Re \int_0^{\frac{\pi}{4}} \ln \left(1-e^{-2 \theta i}\right) d \theta \\ = & -\frac{\pi}{4} \ln 2+\Re\left(\frac{1}{-2 i} \int_0^{\frac{\pi}{4}} \frac{\ln \left(1-e^{-2 \theta i}\right)}{e^{-2 \theta i}} d\left(e^{-2 \theta i}\right)\right] \\ = & -\frac{\pi}{4} \ln 2+\Re\left[\frac{1}{2 i} \operatorname{Li_2} \left(e^{-2 \theta i}\right)\right]_0^{\frac{\pi}{4}} \\ = & -\frac{\pi}{4} \ln 2+\Re\left[\frac{1}{2 i}\left(\operatorname{Li_2} (-i)-\operatorname{Li_2} (1)\right)\right] \\ = & -\frac{\pi}{4} \ln 2+\Re\left[\frac{1}{2 i}\left(-\frac{\pi^2}{48}-i G\right)\right] \\ = & -\frac{\pi}{4} \ln 2-\frac{G}{2} \end{aligned} $$

$$ \begin{aligned} I & =\int_0^{\frac{\pi}{4}} \ln \sqrt{2} d \theta+\int_0^{\frac{\pi}{4}} \ln (\sin \theta) d \theta \\ & =\frac{\pi}{8} \ln 2 -\frac{\pi}{4} \ln 2-\frac{G}{2} \\ & =-\frac{\pi}{8} \ln 2-\frac{1}{2} G \end{aligned} $$ where $G$ is the Catalan’s constant.

Answer 3

With a few substitutions we can unveil two well-known integrals; see here and here.

$$\begin{align*} I &= \int_0^1 \frac{\log x}{\sqrt{2-x^2}} \, dx \\ &= \frac14 \int_0^1 \frac{\log(1-x)}{\sqrt{1-x^2}} \, dx & x\to\sqrt{1-x} \\ &= \frac14 \int_0^\tfrac\pi2 \log(1-\sin x) \, dx & x\to\sin x \\ &= \frac12 \int_0^1 \frac{\log\frac{(1-x)^2}{1+x^2}}{1+x^2} \, dt & x\to2\arctan x \\ &= \int_0^1 \frac{\log(1-x)}{1+x^2} \, dx - \frac12 \int_0^1 \frac{\log\left(1+x^2\right)}{1+x^2} \, dx \\ &= \left(\frac\pi8\log2 - G\right) - \frac12 \left(\frac\pi2\log2 - G\right) \\ &= \boxed{- \frac\pi8 \log2 - \frac G2} \end{align*}$$

Alternatively, after the step $x\to2\arctan x$, putting $x\to\dfrac{1-x}{1+x}$ replaces one of the former log integral with a simpler one,

$$I = \frac12 \int_0^1 \frac{\log\left(2x^2\right) - \log\left(1+x^2\right)}{1+x^2} \, dx \\ \begin{align*} \int_0^1 \frac{\log\left(2x^2\right)}{1+x^2} \, dx &= \frac\pi4 \log2 + 2 \sum_{n\ge0} (-1)^n \int_0^1 x^{2n} \log x \, dx \\ &\!\stackrel{\rm IBP}= \frac\pi4 \log2 - 2 \underbrace{\sum_{n\ge0} \frac{(-1)^n}{(2n+1)^2}}_{=G} \end{align*} \\ \implies I = \frac12 \left(\frac\pi4 \log2 - 2G - \frac\pi2\log2 + G\right) = -\frac\pi8 \log2 - \frac G2$$