How to find the integral of implicitly defined function? Let $a$ and $b$ be real numbers such that $ 0<a<b$. The decreasing continuous function
$y:[0,1] \to [0,1]$ is implicitly defined by the equation $y^a-y^b=x^a-x^b.$
Prove
$$\int_0^1 \frac {\ln (y)} x \, dx=- \frac {\pi^2} {3ab}.
$$
 A: $$F(x,y)=y^a-y^b-x^a+x^b=0$$
by line parameterization $x=m\,y$
$$F(m\,y,y)=m^ay^a-m^by^b-y^a+y^b=0$$
$$\Rightarrow (m^a-1)y^a-(m^b-1)y^b=0$$
$$\Rightarrow \frac{m^a-1}{m^b-1}=y^{b-a}\Rightarrow y=\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}$$
and 
$$x=m\,y=m\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}$$
You can see that the integral limits for $m$ are $0$ and $+\infty$.
We can transform the integral such that
$$x=m\,y\Rightarrow dx=dm\,y+m\,dy$$
$$\int \frac {\ln (y)} x \, dx=\int \frac {\ln (y)} {m\,y} \, (dm\,y+m\,dy)=\int_0^{\infty} \frac{\ln\bigg(\frac{m^a-1}{m^b-1}\bigg)^{\frac 1{b-a}}}{m}dm+\int_0^1 \frac{\ln(y)}{y}dy $$
If you try to calculate the integral for any value of $0\lt a\lt b$ you can see that the integrals are not convergent. Therefore I guess that it is a wrong statement?
A: OK, at long last, I have a solution. Thanks to @Occupy Gezi and my colleague Robert Varley for getting me on the right track. As @Occupy Gezi noted, some care is required to work with convergent integrals.
Consider the curve $x^a-x^b=y^a-y^b$ (with $y(0)=1$ and $y(1)=0$).  We want to exploit the symmetry of the curve about the line $y=x$. Let $x=y=\tau$ be the point on the curve where $x=y$, and let's write
$$\int_0^1 \ln y \frac{dx}x = \int_0^\tau \ln y \frac{dx}x + \int_\tau^1 \ln y \frac{dx}x\,.$$
We make a change of coordinates $x=yu$ to do the first integral: Since $\dfrac{dx}x = \dfrac{dy}y+\dfrac{du}u$, we get (noting that $u$ goes from $0$ to $1$ as $x$ goes from $0$ to $\tau$)
\begin{align*}
\int_0^\tau \ln y \frac{dx}x &= -\int_\tau^1 \ln y \frac{dy}y + \int_0^1 \ln y \frac{du}u \\ &= -\frac12(\ln y)^2\Big]_\tau^1 + \int_0^1 \ln y \frac{du}u = \frac12(\ln\tau)^2 + \int_0^1 \ln y \frac{du}u\,.
\end{align*}
Next, note that as $(x,y)\to (1,0)$ along the curve,
$(\ln x)(\ln y)\to 0$ because, using $(\ln x)\ln(1-x^{b-a}) = (\ln y)\ln(1-y^{b-a})$, we have
$$(\ln x)(\ln y) \sim \frac{(\ln y)^2\ln(1-y^{b-a})}{\ln (1-x^{b-a})} \sim \frac{(\ln y)^2 y^{b-a}}{a\ln y} = \frac1a y^{b-a}\ln y\to 0 \text{ as } y\to 0.$$
We now can make the "inverse" change of coordinates $y=xv$ to do the second integral. This time we must do an integration by parts first.
\begin{align*}
\int_\tau^1 \ln y \frac{dx}x &= (\ln x)(\ln y)\Big]_{(x,y)=(\tau,\tau)}^{(x,y)=(1,0)} + \int_0^\tau \ln x \frac{dy}y \\ & = -(\ln\tau)^2 + \int_0^\tau \ln x \frac{dy}y \\ 
&= -(\ln\tau)^2 - \int_\tau^1 \ln x \frac{dx}x + \int_0^1 \ln x \frac{dz}z \\
&= -\frac12(\ln\tau)^2 + \int_0^1 \ln x \frac{dz}z\,.
\end{align*}
Thus, exploiting the inherent symmetry, we have
$$\int_0^1 \ln y\frac{dx}x = \int_0^1 \ln y \frac{du}u + \int_0^1 \ln x \frac{dz}z = 2\int_0^1 \ln x \frac{dz}z\,.$$
Now observe that
\begin{multline*}
x^a-x^b=y^a-y^b \implies x^a(1-x^{b-a}) = x^az^a(1-x^{b-a}z^{b-a}) \\ \implies x^{b-a} = \frac{1-z^a}{1-z^b}\,,
\end{multline*}
and so, doing easy substitutions,
\begin{align*}
\int_0^1 \ln x \frac{dz}z &= \frac1{b-a}\left(\int_0^1 \ln(1-z^a)\frac{dz}z - \int_0^1 \ln(1-z^b)\frac{dz}z\right) \\ &=\frac1{b-a}\left(\frac1a\int_0^1 \ln(1-w)\frac{dw}w - \frac1b\int_0^1 \ln(1-w)\frac{dw}w\right) \\ &= \frac1{ab}\int_0^1 \ln(1-w)\frac{dw}w\,.
\end{align*}
By expansion in power series, one recognizes that this dilogarithm integral gives us, at long last,
$$\int_0^1 \ln y\frac{dx}x = \frac2{ab}\int_0^1 \ln(1-w)\frac{dw}w = \frac2{ab}\left(-\frac{\pi^2}6\right) = -\frac{\pi^2}{3ab}\,.$$
(Whew!!)
