# Why is this definite integral antisymmetric in $s\mapsto s^{-1}$?

I recently happened into the following integral identity, valid for positive $s>0$:

$$\int_0^1 \log\left[x^s+(1-x)^{s}\right]\frac{dx}{x}=-\frac{\pi^2}{12}\left(s-\frac{1}{s}\right).$$

The obvious question is how to show this (feel free to do so!). But what stirs my curiosity is that the right-hand expression implies the integral to be antisymmetric under $s\mapsto s^{-1}$, which I would not have expected. Is there a simple explanation for this property?

• Nice...............+1@semiclassical – Bhaskara-III Dec 25 '15 at 6:19

HINT: Your integral can be brought to this form $$\int_{-1}^1 \frac{\log \left((1-x)^s+(1+x)^s\right)-s \log (2)}{x+1} \, dx$$ and then you can split the interval and calculate each integral separately. Use on the positive side that $\frac{1-x}{1+x}\mapsto x$ and $\frac{1+x}{1-x}\mapsto x$ on the negative side. Let me know if that works.

One can do this as follows :

$I =\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { ({ x }^{ s }+{ (1-x) }^{ s }) } }{ x } dx }$

Write it like as follows :

$\displaystyle I = \int _{ 0 }^{ 1 }{ \frac { s\ln { (1-x) } }{ x } dx } +\int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ (\frac { x }{ 1-x } ) }^{ s } } ) }{ x } dx }$

$\Rightarrow I = J+K$

Where $\displaystyle J = \int _{ 0 }^{ 1 }{ \frac { s\ln { (1-x) } }{ x } dx }$

and $\displaystyle K = \int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ (\frac { x }{ 1-x } ) }^{ s } } ) }{ x } dx }$

For evaluating $J$ use the taylor expansion of $\ln(1-x)$ :

$\displaystyle J = -s\int _{ 0 }^{ 1 }{ \sum _{ r=1 }^{ \infty }{ \frac { { x }^{ r-1 } }{ r } } dx }$

Interchanging summation and integral we have :

$\displaystyle J = (-s)\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } \int _{ 0 }^{ 1 }{ { x }^{ r-1 }dx } }$

$\displaystyle J = (-s)\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { r }^{ 2 } } } = -s\zeta(2)=\dfrac { -s{ \pi }^{ 2 } }{ 6 }$

For evaluating $K$ we will substitute $y = \dfrac{x}{1-x}$

$\displaystyle K = \int _{ 0 }^{ \infty }{ \frac { \ln { (1+{ y }^{ s }) } }{ y(1+y) } dy }$

Split it into two parts :

$\displaystyle K =\int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ y }^{ s }) } }{ y(1+y) } dy } +\int _{ 1 }^{ \infty }{ \frac { \ln { (1+{ y }^{ s }) } }{ y(1+y) } dy }$

In the second part put $t=\dfrac{1}{s}$

$\displaystyle K = \int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ y }^{ s }) } }{ y(1+y) } dy } +\int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ t }^{ s }) } -s\ln { (t) } }{ (1+t) } dy }$

$\displaystyle \Rightarrow K = \int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ y }^{ s }) } }{ y } -\frac { s\ln { (y) } }{ 1+y } dy }$

We can write $K = L-M$

where $\displaystyle L=\int _{ 0 }^{ 1 }{ \frac { \ln { (1+{ y }^{ s }) } }{ y } dy }$

$\displaystyle M = \int _{ 0 }^{ 1 }{ \frac { s\ln { (y) } }{ 1+y } dy }$

For evaluting $L$ we have to put ${y}^{s}=t$ to get :

$\displaystyle L = \frac { 1 }{ s } \int _{ 0 }^{ 1 }{ \frac { \ln { (1+t) } }{ t } dt }$

Using taylor series and interchanging summation and integral we have :

$\displaystyle L = \frac { 1 }{ s } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } \int _{ 0 }^{ 1 }{ { x }^{ r-1 }dr } }$

$\Rightarrow \displaystyle L = \frac { 1 }{ s } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ { r }^{ 2 } } }$

$\displaystyle L = \frac { 1 }{ s } (1-{ 2 }^{ 1-2 })\zeta (2)=\frac { { \pi }^{ 2 } }{ 12s }$

Now $\displaystyle f(a) = \int _{ 0 }^{ 1 }{ { y }^{ a }dy } =\frac { 1 }{ a+1 }$

Differentiating both sides with respect to $a$ we have :

$\Rightarrow \displaystyle \int _{ 0 }^{ 1 }{ { y }^{ a }\ln { (y) } dy } =\frac { -1 }{ { (a+1) }^{ 2 } }$

For evaluting $M$ we will use the series of $\dfrac{1}{1+x}$ and interchanging summation and integral we have :

$\displaystyle M = s\sum _{ r=0 }^{ \infty }{ { (-1) }^{ r }\int _{ 0 }^{ 1 }{ { y }^{ r }\ln { (y) } dy } }$

Using the property I have proved above we have :

$\displaystyle M = s\sum _{ r=0 }^{ \infty }{ \frac { { (-1) }^{ r+1 } }{ { (r+1) }^{ 2 } } } =s\sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r } }{ { r }^{ 2 } } } =\frac { -s{ \pi }^{ 2 } }{ 12 }$

Finally we have $K = \dfrac { { \pi }^{ 2 } }{ 12 } (s+\dfrac { 1 }{ s } )$

Using all the results we have :

$\displaystyle I = \frac { { \pi }^{ 2 } }{ 12 } (s+\frac { 1 }{ s } )-\frac { s{ \pi }^{ 2 } }{ 6 } = \frac { { \pi }^{ 2 } }{ 12 } (\frac { 1 }{ s } -s)$

Hence Proved.

Comment: I can't comment quantitatively that why it is antisymmetric but qualitatively it can be seen as :

For $s>1$ :

${x}^{s}<x , {(1-x)}^{s}<(1-x)$

$\Rightarrow {x}^{s}+{(1-x)}^{s} < 1$

$\Rightarrow \ln({x}^{s}+{(1-x)}^{s}) < 0$

Hence the resultant integral is negative : Going with similar reasoning we can say that :

for $s<1$ the integral is positive.