3
$\begingroup$

Two integrals exhibit the same closed form

Motivated by this interesting question.

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

and

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

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?

$\endgroup$
  • 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
5
$\begingroup$

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 $$ I'(1/2)=\frac{\pi}{\sin(\pi/2)}-\frac{\pi^2/2}{\sin^2(\pi/2)}{\cos(\pi/2)}=\pi. $$

$\endgroup$
4
$\begingroup$

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 $y=\dfrac{1}{x}$,

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.