Two integrals exhibit the same closed form

Motivated by this interesting question.

$$\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag1$$


$$\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag2$$

Making an attempt

$$u={x\over 1-x}\implies -(1-x)^2du=dx\tag a$$

$$u=\sqrt{x\over 1-x}\implies 2x(1-x)^2du=dx\tag b$$

I try on $(a)$ and $(b)$ it doesn't seem to be working. So I try another

$u=\sqrt{1-x}\implies -2\sqrt{1-x}du=dx$, apply to $(1)$ then we have

$$2\int_{0}^{1}\sqrt{1-u^2}\ln\left({1-u^2\over u^2}\right)\mathrm du=I_1-I_2\tag3$$

$$I_1-I_2=2\int_{0}^{1}\sqrt{1-u^2}\ln(1-u^2)\mathrm du-2\int_{0}^{1}\sqrt{1-u^2}\ln u^2\mathrm du\tag4$$

For $I_1$ make $v=\sqrt{1-u^2}\implies -{\sqrt{1-u^2}\over u}dv=du$, then

$$I_1=4\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln v \mathrm dv\tag5$$

Recall $${1\over \sqrt{1+x}}=\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}(-x)^k\tag6$$

then $(5)$ becomes

$$I_1=4\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln v \mathrm dv\tag7$$

$$I_1=-\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\tag8$$

Apply same substitution, $I_2$ becomes

$$I_2=2\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln(1-v^2)\mathrm dv\tag9$$

$$I_2=2\sum_{0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln(1-v^2)\mathrm dv\tag c$$

Surely this is going wrong because there is not sign of $\pi $ anywhere in $I_1$ nor $I_2$!

How do we show that $(1)=(2)$ and prove one of them?

  • 1
    $\begingroup$ This is another question that can be tackled through Euler's Beta function and differentiation under the integral sign. $\endgroup$ – Jack D'Aurizio Apr 18 '17 at 15:07
  • $\begingroup$ $\lim_{x\rightarrow 1^{-}} \sqrt{\dfrac{x}{1-x}}=\lim_{x\rightarrow 1^-} \dfrac{x}{1-x}=+\infty$ $\endgroup$ – FDP Apr 18 '17 at 15:20
  • $\begingroup$ If $u=\sqrt{\dfrac{x}{1-x}}$ then $dx=\dfrac{2u}{{{\left( 1+{{u}^{2}}\right) }^{2}}}du$. $\endgroup$ – FDP Apr 18 '17 at 16:02

The integrals are equal Note that $$ \int_0^1\sqrt\frac{x}{1-x}\ln\frac{x}{1-x}\,dx-\int_0^1x\sqrt\frac{x}{1-x}\ln\frac{x}{1-x}\,dx =\int_0^1\sqrt{x(1-x)}\ln\frac{x}{1-x}\,dx. $$ The last integral becomes zero, since $f(x)=\sqrt{x(1-x)}\ln\frac{x}{1-x}$ satisfies $f(x)+f(1-x)=0$.

The value of the integrals

To calculate the integrals, introduce $$ I(a)=\int_0^1x^a(1-x)^{-a}\,dx. $$ The integral in (1) is then just $I'(1/2)$. Now, however, $I(a)$ is a beta-integral that via the symmetry of the Gamma function reduces to $$ I(a)=B(1+a,1-a)=\frac{\Gamma(1+a)\Gamma(1-a)}{\Gamma(2)}=a\Gamma(a)\Gamma(1-a)=\frac{a\pi}{\sin(a\pi)}, $$ and differentiating yields $$ I'(1/2)=\frac{\pi}{\sin(\pi/2)}-\frac{\pi^2/2}{\sin^2(\pi/2)}{\cos(\pi/2)}=\pi. $$


On the path of Latte,

$\displaystyle J=\int_0^1 \sqrt{\dfrac{x}{1-x}}\ln\left(\dfrac{x}{1-x}\right)dx$

Perform the change of variable $y=\sqrt{\dfrac{x}{1-x}}$ and then, use integration by parts,

$\begin{align} J&=4\int_0^{\infty} \dfrac{x^2\ln x}{(1+x^2)^2}dx\\ &=\left[-\dfrac{2x\ln x}{1+x^2}\right]_0^{\infty}+2\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx+2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\ &=2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\ &=2\Big[\arctan x\Big]_0^{\infty}\\ &=2\times \dfrac{\pi}{2}\\ &=\boxed{\pi} \end{align}$

$\begin{align} \int_0^{\infty}\dfrac{\ln x}{1+x^2}dx=\int_0^{1}\dfrac{\ln x}{1+x^2}dx+\int_1^{\infty}\dfrac{\ln x}{1+x^2}dx \end{align}$

In the latter integral perform the change of variable $y=\dfrac{1}{x}$,

$\begin{align} \int_0^{\infty}\dfrac{\ln x}{1+x^2}dx&=\int_0^{1}\dfrac{\ln x}{1+x^2}dx-\int_0^{1}\dfrac{\ln x}{1+x^2}dx\\ &=0 \end{align}$


Let $$I=\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx,J=\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx. $$ Then $$ I-J= \int_{0}^{1}\sqrt{x(1-x)}\cdot\ln\left({x\over 1-x}\right) dx.$$ Using $x\to1-x$, it is easy to get $$ I-J=-(J-I) $$ and hence $I-J=0$ or $I=J$. The rest follows @mickep's answer.


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