Natural log integral

Step one) Using the familiar series $\frac{1}{1-x}=1+x+x^2+\dots \ \ \ \ 0<x<1$ and integrating it for writing $\ln(1-x)$ as $-\sum_{n=1}^{\infty} \frac{x^n}{n}$ Step two) Rearrange the integral as $I=\int_{0}^{1} \ln(x) \left(-\sum_{n=1}^{\infty} \frac{x^n}{n}\right) \ \text{d}x =-\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1} \ln(x) x^n \ \text{d}x=-\sum_{n=1}^{\infty} \frac{1}{n} I'$ Step three) Applying integration by parts for the integral $I'=\int_{0}^{1} \ln(x) x^n \ \text{d}x$ with $u=x^n$ and $\text{d}v=\ln(x) \text{d}x$ which gives $I'=\left(x^n (x \ln(x)-x) \right)_{0}^{1} - n \left(\int_{0}^{1} (x \ln(x)-x) x^{n-1} \text{d}x \right)$ $I'=-1-n\left(\int_{0}^{1} (x^n \ln(x)-x^n) \text{d}x \right)$ $I'=-1-n\left(I'-\int_{0}^{1} x^n \text{d}x \right)=-1-n\left(I'-\frac{1}{n+1} \right)$ $I'=-\frac{1}{(n+1)^2}$ Step four) Evaluating the resulting sum $I=-\sum_{n=1}^{\infty} \frac{1}{n} I' =\sum_{n=1}^{\infty} \frac{1}{n(n+1)^2}$ $I=\sum_{n=1}^{\infty} \frac{1}{n(n+1)}- \sum_{n=1}^{\infty}\frac{1}{(n+1)^2}=1-\left(\frac{\pi^2}{6}-1 \right)=2-\frac{\pi^2}{6}$

A bit of an overpowered solution, but it's more intuitive and easier to spot immediately with the right knowledge:

Consider the beta function $\displaystyle \mathrm{B}(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \ dt = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ . Taking the partial derivatives wrt $x$ and $y$ , we have

$\frac{\partial}{\partial x} \frac{\partial}{\partial y} \mathrm{B}(x,y) = \int_0^1 \frac{\partial}{\partial x} \frac{\partial}{\partial y} t^{x-1}(1-t)^{y-1} \ dt = \int_0^1 t^{x-1}(1-t)^{y-1}\ln t \ln(1-t) \ dt$

as well as

$\frac{\partial}{\partial x} \frac{\partial}{\partial y} \mathrm{B}(x,y) = \frac{\partial}{\partial x} \frac{\partial}{\partial y} \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = ([\psi(x)-\psi(x+y)][\psi(y)-\psi(x+y)]-\psi^{(1)}(x+y))\mathrm{B}(x,y)$

where $\psi$ is the digamma function. Thus, evaluated at $x=1,y=1$ , we have

$\int_0^1 \ln t \ln(1-t) \ dt = ([\psi(1)-\psi(2)]^2-\psi^{(1)}(2))\mathrm{B}(1,1) = [\psi(1)-\psi(2)]^2-\psi^{(1)}(2)$

We note that $\displaystyle \psi(z) = -\gamma+H_{z-1} = -\gamma+\int_0^1 \frac{1-t^{z-1}}{1-t} \ dt$ . Hence,

$\psi^{(1)}(2) = \int_0^1 \left. \frac{d}{dz} \frac{1-t^{z-1}}{1-t} \right|_{z=2} \ dt = \int_0^1 \frac{t\ln t}{t-1} \ dt$

Substituting $u=1-t$ , we get

$\int_0^1 \frac{t\ln t}{t-1} \ dt = \int_0^1 \frac{-(1-u)\ln(1-u)}{u} \ du = \sum_{n=1}^\infty \int_0^1 \frac{u^{n-1}-u^n}{n} \ du = \sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n(n+1)} = \frac{\pi^2}{6}-1$

Therefore, we have

$[\psi(1)-\psi(2)]^2-\psi^{(1)}(2) = [-\gamma - (-\gamma + H_1)]^1 - (\frac{\pi^2}{6}-1) = \boxed{2-\dfrac{\pi^2}{6}}$