Does $\int_1^2 \frac{\ln(x)}{x-1} dx$ converge and what test is used?

Question

$$\int_1^2 \frac{\ln(x)}{x-1} dx$$

How does one determine convergence of this? I am not interested in the value of it. I tried comparing to $1/(x-1)$ but the integral related to that diverges, and I know that above converges (via mathematica) so I need something that converges to compare with, but what?

Answer 1

The only potential problem is near $x=1$ (the integrand is otherwise continuous), but one may recall the standard Taylor expansion, as $u \to 0$, $$ \ln (1-u)=-u+\mathcal{O}(u^2) $$ then you may write, as $x \to 1$, $$ \ln (x)=\ln (1-(1-x))=-(1-x)+\mathcal{O}((1-x)^2) $$ giving $$ \frac{\ln (x)}{x-1}=1+\mathcal{O}(x-1) $$ near $x=1$, showing that your integral is convergent.

Answer 2

Since $$ \lim_{x\to1}\frac{\ln x}{x-1}=\lim_{x\to1}\frac{\ln x-\ln1}{x-1}=(\ln x)'\Big|_{x=1}=\frac{1}{x}\Big|_{x=1}=1, $$ therefore the function $$ f:[1,2] \to \mathbb{R},\, f(x)=\begin{cases}\frac{\ln x}{x-1} &\mbox{ for } x \in (1,2]\\ 1 & \mbox{ for } x=1\end{cases}$$ is continuous, and thus the integral $\int_1^2f(x)\,dx$ exists and is finite. Furthermore, we have \begin{eqnarray} \int_1^2\frac{\ln x}{x-1}\,dx&=&\int_1^2\frac{\ln(1+x-1)}{x-1}\,dx=\int_0^1\frac{\ln(1+t)}{t}\,dt=\int_0^1\frac1t\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}t^n\,dt\\ &=&\sum_{n=1}^\infty\int_0^1\frac{(-1)^{n-1}}{n}t^{n-1}\,dt=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2} \end{eqnarray}

Answer 3

It is easier to handle if we look at $$I = \int_1^2 \dfrac{\ln(x)}{x-1}dx = \int_0^1 \dfrac{\ln(1+t)}{t}dt$$ Note that for $t \in (0,1)$, we have $\dfrac{\ln(1+t)}t \in (0,1)$. Hence, clearly, $I$ exists and the value is in fact between $0$ and $1$. To evaluate the integrate note that $$\ln(1+t) = \sum_{k=1}^{\infty}(-1)^{k+1} \dfrac{t^k}k$$ Hence, we have \begin{align} I & = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \int_0^1 t^{k-1}dt = \sum_{k=1}^{\infty} \dfrac{(-1)^k}k \cdot \left.\dfrac{t^k}k \right \vert_0^1 = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} = \dfrac1{1^2} - \dfrac1{2^2} + \dfrac1{3^2}-\dfrac1{4^2} \pm \cdots\\ & = \left(\dfrac1{1^2} + \dfrac1{2^2} + \dfrac1{3^2} + \dfrac1{4^2} + \cdots \right) - 2\left(\dfrac1{2^2} + \dfrac1{4^2} + \cdots\right) = \zeta(2) - \dfrac12 \zeta(2) = \dfrac{\zeta(2)}2 = \dfrac{\pi^2}{12} \end{align}