I recently happened to look 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!). However, what stirs my curiosity is that the right-hand expression implies that the integral is antisymmetric under $s\mapsto s^{-1}$, which I would not have expected.
Is there a simple explanation for this property?
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.
The antisymmetry stems from the integrand with inherent symmetric itself, revealed by averaging following $x\to 1-x$ $$I(s)=\int_0^1 \frac{\ln\left[x^s+(1-x)^{s}\right]}x dx =\frac12 \int_0^1 \frac{\ln\left[x^s+(1-x)^{s}\right]}{x(1-x)} dx $$ Further symmetry below is seen from $\left(\frac {1-x}x \right)^s= \frac {1-y}y$ $$x^s+(1-x)^{s}=\frac1{[y^{1/s}+(1-y)^{1/s}]^s}, \>\>\>\>\>\>\>\>\frac{s \ dx}{x(1-x)}= \frac{dy}{y(1-y)} $$ As a result $$I(s)=-\frac12 \int_0^1 \frac{\ln\left[y^{1/s}+(1-y)^{1/s}\right]}{y(1-y)} dy=-I\left(1/s\right)$$