Prove that $\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)$.

Question

I have discovered via contour integration that $$\int_0^\infty\,\frac{\exp(t\,u)}{\exp(u)+1}\,\text{d}u={\text{csc}(\pi\,t)}\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\tag{*}$$ for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$. By taking $t\to 0$, I deduce that $$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,y\,\cot(y)\,\text{d}y\,.$$ With a step of integration by parts, I obtain $$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^{\frac{\pi}{2}}\,\ln\big(\sin(y)\big)\,\text{d}y\,.$$ Setting $x:=\sin(y)$, I get $$\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u=-\frac{2}{\pi}\,\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x\,.$$ This shows that $$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u\,.$$ The integral $\displaystyle\int_0^\infty\,\frac{1}{\exp(u)+1}\,\text{d}u$ can be easily obtained since $$\int\,\frac{1}{\exp(u)+1}\,\text{d}u=u-\ln\big(\exp(u)+1\big)+\text{constant}\,.$$ That is, I have $$\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,\text{d}x=-\frac{\pi}{2}\,\ln(2)\,.\tag{#}$$ However, this proof is a very roundabout way to verify the equality above. Is there a more direct way to prove that (#) is true? Any technique is appreciated.

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A nice consequence of (*) is that $$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)+1}\,\text{d}u=\frac{\pi}{2}\,\text{csc}(\pi\,t)-\frac{1}{2\,t}$$ for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$. This provides a proof that $$\eta(2r)=\frac{1}{(2r-1)!}\,\int_0^\infty\,\frac{u^{2r-1}}{\exp(u)+1}\,\text{d}u=\frac{\pi^{2r}}{2}\,\Biggl(\left[t^{2r-1}\right]\Big(\text{csc}(t)\Big)\Biggr)$$ for $r=1,2,3,\ldots$. Here, $\eta$ is the Dirichtlet eta function. In addition, $[t^k]\big(g(t)\big)$ denotes the coefficient of $t^k$ in the Laurent expansion of $g(t)$ about $t=0$. This also justifies the well known results that $$\eta(2r)=\frac{\left(2^{2r-1}-1\right)\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}\text{ and }\zeta(2r)=\frac{2^{2r-1}\,\big|B_{2r}\big|\,\pi^{2r}}{(2r)!}$$ for $r=1,2,3,\ldots$, where $\left(B_j\right)_{j\in\mathbb{Z}_{\geq0}}$ is the sequence of Bernoulli numbers and $\zeta$ is the Riemann zeta function.

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Similarly, $$\begin{align}\int_0^\infty\,\frac{\exp(t\,u)-1}{\exp(u)-1}\,\text{d}u&=\ln(2)+2\,\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-t)\,y)\,\sin(t\,y)}{\sin(y)}\,\text{d}y\\ &\phantom{aaaaa}-\cot(\pi\,t)\,\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}}\,\frac{\sin\big((1-2t)\,y\big)}{\sin(y)}\,\text{d}y\right)\,,\end{align}$$ for all $t\in\mathbb{C}\setminus\mathbb{Z}$ such that $\text{Re}(t)<1$. This gives $$\int_0^\infty\,\frac{\sinh(t\,u)}{\exp(u)-1}\,\text{d}u=\frac{1}{2\,t}-\frac{\pi}{2}\,\cot(\pi\,t)$$ for all $t\in\mathbb{C}\setminus\{0\}$ such that $\big|\text{Re}(t)\big|<1$.

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Another consequence of (*) is that $$\int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y=\frac{\pi}{2}\,\text{sign}(k)$$ for all odd integers $k$. It is an interesting challenge to determine the integral $\displaystyle \int_0^{\frac{\pi}{2}}\,\frac{\sin(k\,y)}{\sin(y)}\,\text{d}y$ for all even integers $k$.

Answer 1

Take $x = \sin t$ then $dx = \cos t \ dt $ so \begin{equation} \frac{\ln (x) }{\sqrt{1 - x^2}} \ dx = \frac{\ln (\sin t) }{\cos t} \cos t \ dt = \ln (\sin t) \ dt \end{equation} SO the integral becomes \begin{equation} A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin t) \ dt \end{equation} or just name $x$ instead of $t$ \begin{equation} A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx \end{equation} Now use the following change of variable \begin{equation} t = \frac{\pi}{2} - x \end{equation} We get \begin{equation} A = - \int\limits_{\frac{\pi}{2}}^{0} \ln (\sin (\frac{\pi}{2} - t)) \ dt = \int\limits_{0}^{\frac{\pi}{2}} \ln (\cos t) \ dt \end{equation} This means \begin{equation} 2A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx + \int\limits_{0}^{\frac{\pi}{2}} \ln (\cos x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} [ \ln (\sin x) \ dx + \ln (\cos x) \ dx ] \end{equation} But $\ln ab = \ln a + \ln b$ so \begin{equation} 2A = \int\limits_{0}^{\frac{\pi}{2}} \ln (\sin x \cos x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} \ln (\frac{1}{2} \sin 2x) \ dx = \int\limits_{0}^{\frac{\pi}{2}} \ln (\frac{1}{2}) + \ln( \sin 2x) \ dx \end{equation} this means that \begin{equation} 2A = \frac{\pi}{2} \ln \frac{1}{2} + B \end{equation} where $B =\int\limits_{0}^{\frac{\pi}{2}} \ln( \sin 2x) \ dx $ \begin{equation} B = \int\limits_{0}^{\frac{\pi}{4}} \ln( \sin 2x) \ dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln( \sin 2x) \ dx \end{equation} Let $ t = 2x - \frac{\pi}{2}$ so \begin{equation} B = \int\limits_{0}^{\frac{\pi}{4}} \ln( \sin 2t) \ dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln( \sin 2(t + \frac{\pi}{2})) \ dx = \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \ln(\sin(t)) \ dt + \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \ln(\cos(t)) \ dt \end{equation} which is \begin{equation} B = \frac{1}{2} A + \frac{1}{2} A = A \end{equation} So \begin{equation} 2A = \frac{\pi}{2}\ln \frac{1}{2} + A \end{equation} So \begin{equation} A = \frac{\pi}{2}\ln \frac{1}{2} \end{equation}

Answer 2

By letting $x=\sin(t)$, and by using the symmetry $\sin(\pi/2-t)=\cos(t)$, we get $$\begin{align}I&=\int_0^1\,\frac{\ln(x)}{\sqrt{1-x^2}}\,dx=\int_0^{\pi/2} \ln (\sin t)\, dt =\frac{1}{2}\left(\int_0^{\pi/2} \ln (\sin t)\, dt+\int_0^{\pi/2} \ln (\cos t)\, dt\right)\\&=\frac{1}{2}\left(\int_0^{\pi/2} \ln(\sin(2t))dt - \int_0^{\pi/2} \ln(2)dt\right)=\frac{I}{2}-\dfrac{\pi}4 \ln(2)\end{align}$$ and the result easily follows.

Answer 3

The antiderivative $$I=\int\,\frac{\ln(x)}{\sqrt{1-x^2}}\,dx$$ can be computed (a CAS gave it).

For $\color{red}{2i I}$, the obtained result is

$$\text{Li}_2\left(e^{-2 i \sin ^{-1}(x)}\right)+2 \log (x) \log \left(\sqrt{1-x^2}+i x\right)+\sin ^{-1}(x)^2-2 i \sin ^{-1}(x) \log \left(1-e^{-2 i \sin ^{-1}(x)}\right)$$ $$\lim_{x\to 1} \, I=-\frac{i \pi }{12} (\pi -6 i \log (2))$$ $$\lim_{x\to 0} \, I=-\frac{i \pi ^2}{12}$$

Answer 4

I would start with

$$ J(a) = \int_0^1 \frac{x^a \; dx}{\sqrt{1-x^2}}, \ a > -1$$ The substitution $x = \sqrt{t}$ gives you $$ J(a) = \frac{1}{2} \int_0^1 t^{(a-1)/2} (1-t)^{-1/2}\; dt = \frac{B((a+1)/2,1/2)}{2} = \frac{\Gamma(a+1/2) \Gamma(1/2)}{\Gamma(a+1)} $$ where $B$ is the Beta function. Then your integral is $$J'(0) = - \frac{-\pi \ln(2)}{2}$$

Answer 5

\begin{align} \int_0^1 x^\alpha\ dx &= \dfrac{x^{\alpha+1}}{\alpha+1}\\ \dfrac{d}{d\alpha}\int_0^1 x^\alpha\ dx &=\dfrac{d}{d\alpha}\dfrac{x^{\alpha+1}}{\alpha+1}\\ \int_0^1 x^\alpha\ln x\ dx &=\dfrac{x^{\alpha+1}((\alpha+1)\ln x-1)}{(\alpha+1)^2}\Big|_0^1=\dfrac{-1}{(\alpha+1)^2} \end{align} \begin{align} I &= \int_{0}^{1} \frac{\ln x}{\sqrt {1-x^2}}dx \\ &= \int_0^1\ln x\sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}x^{2n} dx\\ &= \sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}\dfrac{-1}{(2n+1)^2} \\ &= \color{blue}{-\dfrac{\pi}{2}\ln2} \end{align} using generator function $$\sum_{n=0}^\infty{2n\choose n}\dfrac{1}{4^n}\dfrac{x^{2n+1}}{(2n+1)^2}=\ln\dfrac{1+\sqrt{1-x^2}}{x}-\sqrt{1-x^2}$$

Answer 6

Riemann sums are enough. For any $n\in\mathbb{N}^+$ we have $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n},\tag{1} $$ hence $$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta=\frac{1}{2}\int_{0}^{\pi}\log\sin\theta\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{\log(2n)-n\log 2}{n} = \color{red}{-\frac{\pi}{2}\log 2}.\tag{2}$$

Answer 7

\begin{align} I=\int_0^1 \frac{\ln x}{\sqrt{1-x^2}} dx & \overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x}{\sqrt{x(1-x)}} dx \overset{x\to 1-x}= \frac14\int_0^1 \frac{\ln (1-x)}{\sqrt{x(1-x)}} dx\\ &= \frac18 \int_0^1 \frac{\ln (x(1-x)) }{\sqrt{x(1-x)}} dx \overset{symmetry}=\frac14 \int_0^{\frac12} \frac{\ln (x(1-x)) }{\sqrt{x(1-x)}} dx \\ &\overset{x(1-x)=\frac t4}=\frac12I -\frac{\ln2}4\int_0^1 \frac{dt}{\sqrt{t(1-t)}}= \frac12I- \frac\pi4 \ln2\\ \end{align}

which yields $I=-\frac{\pi}{2}\ln 2$.