While solving a question, I got stuck at


How should I proceed, what I want is to convert $\mathcal{N}$ as $k\mathcal{D}$.

  • $\begingroup$ Use $\lim_{b\to\infty}$ for the integral in the denominator so that it looks like $\int_1^b$. You could also split the integral in the numerator into $\int_0^1 \frac{ln(x)*ln(x)}{(1-x)(1-x)}$ $\endgroup$
    – Eric Brown
    Nov 4, 2019 at 21:31
  • $\begingroup$ How shall I proceed then? $\endgroup$
    – Zenix
    Nov 4, 2019 at 21:34
  • 1
    $\begingroup$ Do we have convergence for the integral in the denominator? $\endgroup$
    – dan_fulea
    Nov 4, 2019 at 21:36
  • 1
    $\begingroup$ "Integral equation" is not the right term for this. For one thing, it's not an equation. An integral equation does not identify a particular function and ask what its integral is; rather it involves a relationship between a not-yet-identified function and an integral involving it. $\endgroup$ Nov 4, 2019 at 22:12
  • 1
    $\begingroup$ It could be useful to re-write denominator using $t=\frac1{x}$: \begin{eqnarray} \mathcal D &=& \int_1^{+\infty}\frac{\log x}{(1-x)x}dx=\\ &=&-\int_0^1\frac{-\log t}{\left(1-\frac1{t}\right)\frac1{t}}\left(-\frac1{t^2}\right)dt=\\ &=&-\int_0^1\frac{\log t}{t-1}dt \end{eqnarray} $\endgroup$
    – dfnu
    Nov 4, 2019 at 22:28

2 Answers 2


Integrating the denominator by parts leads to

\begin{eqnarray} \mathcal D &=& \int_1^{+\infty}\frac{\log x}{(1-x)x}dx=\\ &=& \frac12\left[\frac{\log^2 x}{(1-x)}\right]_1^{+\infty}-\frac12\int_1^{+\infty}\frac{\log^2x}{(1-x)^2}dx=\\ &=&-\frac12\int_1^{+\infty}\frac{\log^2x}{(1-x)^2}dx=\\ &\stackrel{t=\frac1{x}}{=}&-\frac12\int_1^0\frac{\log^2t}{\left(1-\frac1{t}\right)^2}\left(-\frac1{t^2}\right)dt=\\ &=&-\frac12\int_0^1 \frac{\log^2 t}{(1-t)^2}dt = \\ &=&-\frac12 \cdot\mathcal N, \end{eqnarray} where $\mathcal N$ is your numerator. So $$\frac{\mathcal N}{\mathcal D}=-2.$$


We have that by parts

$$\int_{1}^{\infty}{\dfrac{\ln\left(x\right)}{\left(1-x\right)x}}dx=\left[\dfrac{\ln^2\left(x\right)}{\left(1-x\right)}\right]_1^\infty-\int_{1}^{\infty}{\frac{\ln x}{x(1-x)}+\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx$$

$$\implies \int_{1}^{\infty}{\dfrac{\ln\left(x\right)}{\left(1-x\right)x}}dx=-\frac12\int_{1}^{\infty}{\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx$$

and by $y=\frac1x$



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .