$$\int_0^1 \frac{\ln x}{x-1}dx$$
I thought of solving the problem by using $|z|=1$ and indenting the singularity around $z=1$.
Or can we use log trick (as used in $-\infty$ to $+\infty$) problems in order to solve this? in yes then $\log x$ can be taken as the branch cut. Please help.
Hint (no complex contour). Note that $$\int_0^1\frac{\ln x}{x-1}dx=\lim_{s\to 1^-}\int_0^s\frac{-\ln (1-t)}{t}dt$$ and for $t\in [0,1)$, $$\frac{-\ln (1-t)}{t}=\sum_{k\geq 1}\frac{t^{k-1}}{k}.$$
Instead of using the series for $\log(1-x)$ as Robert Z does, we can also use the series for $\frac1{1-x}$ and integrate by parts ($u=\log(x)$ and $\mathrm{d}v=x^k\,\mathrm{d}x$): $$ \begin{align} \int_0^1\frac{\log(x)}{x-1}\,\mathrm{d}x &=-\sum_{k=0}^\infty\int_0^1\log(x)\,x^k\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac1{k+1}\int_0^1x^k\,\mathrm{d}x \end{align} $$
Or perhaps a dilogarithm approach. Under the change of variable $x \mapsto 1-x$, the integral becomes $$\int^1_0 \frac{\ln x}{x - 1} \, dx = - \int^1_0 \frac{\ln (1 - x)}{x} \, dx.$$ And from the definition of the dilogarithm function, namely $$\text{Li}_2 (x) = - \int^x_0 \frac{\ln (1 - u)}{u} \, du,$$ we have $$\int^1_0 \frac{\ln x}{x - 1} \, dx = \left [\text{Li}_2 (x) \right ]^1_0 = \text{Li}_2(1) - \text{Li}_2(0).$$ Since $\text{Li}_2 (1) = \zeta (2) = \dfrac{\pi}{6}$ and $\text{Li}_2(0) = 0$, this gives $$\int^1_0 \frac{\ln x}{x - 1} \, dx = \frac{\pi^2}{6},$$ as expected.
Failing that, you could always crash this small pea under foot with a sledgehammer using an approach based on the derivative of the beta function as follows. From the definition for the beta function we have $$\text{B}(x,y) = \int^1_0 t^{x - 1} (1 - t)^{y - 1} \, dt.$$ Observing that $$\partial_x \text{B}(x,y) = \int^1_0 t^{x - 1} \ln t (1 - t)^{y - 1} \, dt,$$ we see that $$\int^1_0 \frac{\ln t}{1 - t} \, dt = \lim_{y \to 0^+} \lim_{x \to 1} \partial_x \text{B}(x,y).$$ Thus $$\int^1_0 \frac{\ln x}{x - 1} \, dx = -\lim_{y \to 0^+} \lim_{x \to 1} \partial_x \text{B}(x,y).$$
Since $$\text{B} (x,y) = \frac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)},$$ where $\Gamma (z)$ is the gamma function, on taking the logarithm before differentiating with respect to $x$ we have $$\partial_x \text{B}(x,y) = \text{B} (x, y) \left [\frac{\Gamma'(x)}{\Gamma (x)} + \frac{\Gamma'(x + y)}{\Gamma (x + y)} \right ] = \text{B}(x,y) [\psi (x) - \psi (x + y)].$$ Here the result $\psi (z) = \Gamma'(z)/\Gamma (z)$ where $\psi (z)$ is the digamma function has been used. It follows that \begin{align*} \int^1_0 \frac{\ln x}{x - 1} \, dx &= - \lim_{y \to 0^+} \lim_{x \to 1} \text{B}(x,y) [\psi (x) - \psi (x + y)]\\ &= -\lim_{y \to 0^+} \text{B}(1,y) [\psi(1) - \psi(1 + y)]. \end{align*}
The $y$ limit takes a little more effort to evaluate and pays to consider it carefully. Since $$\text{B}(1,y) = \frac{\Gamma (1) \Gamma (y)}{\Gamma (1 + y)} = \frac{\Gamma (y)}{\Gamma (1 + y)},$$ we have \begin{align*} \int^1_0 \frac{\ln x}{x - 1} \, dx &= -\lim_{y \to 0^+} \frac{\Gamma (y)}{\Gamma (1 + y)} [\psi(1) - \psi (1 + y)]\\ &= -\lim_{y \to 0^+} \frac{1}{\Gamma (1 + y)} \cdot \lim_{y \to 0^+} \Gamma (y) [\psi(1) - \psi (1 + y)]\\ &= -\lim_{y \to 0^+} \Gamma (y) [\psi(1) - \psi (1 + y)]\\ &= -\lim_{y \to 0^+} \frac{\psi(1) - \psi (1 + y)}{1/\Gamma (y)}. \end{align*} The limit is now indeterminate and of the form $\frac{0}{0}$ and can be found by applying l'Hôpital's rule. Doing so yields \begin{align*} \int^1_0 \frac{\ln x}{x - 1} \, dx &= -\lim_{y \to 0^+} \frac{-\psi^{(1)}(1 + y)}{-\Gamma'(y)/\Gamma^2 (y)}. \end{align*} Here $\psi^{(m)} (x)$ is the polygamma function of order $m$ and the following result for the derivative of the digamma function has been used $$\psi^{(m)}(z) = \frac{d^m}{dz^m} \psi (z).$$
As $\psi (x) = \Gamma' (x)/\Gamma (x)$, we can write \begin{align*} \int^1_0 \frac{\ln x}{x - 1} \, dx &= -\lim_{y \to 0^+} \frac{\Gamma (y) \psi^{(1)} (1 + y)}{\psi (y)}\\ &= -\lim_{y \to 0^+} \psi^{(1)} (1 + y) \cdot \lim_{y \to 0^+} \frac{\Gamma (y)}{\psi (y)} \\ &= - \psi^{(1)}(1) \cdot \lim_{y \to 0^+} \frac{y \, \Gamma (y)}{y \, \psi(y)}. \end{align*} Now since $$y \Gamma (y) = \Gamma (1 + y),$$ and from the functional relation equation for the digamma function, namely $$\psi (1 + y) = \psi (y) + \frac{1}{y},$$ we can write \begin{align*} \int^1_0 \frac{\ln x}{x - 1} \, dx &= - \psi^{(1)} (1) \cdot \lim_{y \to 0^+} \frac{\Gamma (1 + y)}{y \, \psi (1 + y) - 1} = \psi^{(1)}(1). \end{align*}
The value for $\psi^{(1)}(1)$ can be found from the series representation for the polygamma function. Since $$\psi^{(m)} (z) = (-1)^{m + 1} m! \sum^\infty_{n = 0} \frac{1}{(z + n)^{m + 1}},$$ setting $m = 1$ and $z = 1$ gives $$\psi^{(1)}(1) = \sum^\infty_{n = 0} \frac{1}{(n + 1)^2} = \sum^\infty_{n = 1} \frac{1}{n^2} = \frac{\pi^2}{6},$$ where the value for the last sum comes from the well-known Basel problem. Thus $$\int^1_0 \frac{\ln x}{x - 1} \, dx = \frac{\pi^2}{6},$$ as found before.
