I need help computing $\int {\ln x\over 2-x}\, dx$

Question

While integrating $\ln(\sec x)$, at one point I managed to break the integral into two. But I wasn't able to integrate one of those parts.

The integral I am having a difficulty with is:

$$\int {\ln x\over 2-x}\,dx$$

I am aware of the fact that this integral cannot be expressed with elementary functions. In fact, my educated guess is that the answer should contain a dilogarithm term.

I have integrated similar functions and used the dilogarithm (or even trilogarithm once), but in this case, I have no idea how to get on solving this integral.

Any help will be appreciated.

Answer 1

Let $x=2-2t$ and use $\int \frac{\ln(1-x)}{x}dx=-\text{Li}_2(x)+C$ $$\int {\ln x\over 2-x}dx=-\int \frac{\ln 2+\ln(1-t)}{t}dt=-\ln 2 \ln t +\text{Li}_2(t)+C$$

Answer 2

If the power series for the dilogarithm function is easily recognized, $$ \begin{align} \int\frac{\log(x)}{2-x}\,\mathrm{d}x &=-\int\frac{\log(2-y)}y\,\mathrm{d}y\tag1\\ &=-\log(2)\log(y)-\int\frac{\log\left(1-\frac y2\right)}y\,\mathrm{d}y\tag2\\ &=-\log(2)\log(y)+\int\sum_{k=1}^\infty\frac{y^{k-1}}{k2^k}\,\mathrm{d}y\tag3\\ &=-\log(2)\log(y)+\sum_{k=1}^\infty\frac{y^k}{k^22^k}+C\tag4\\[3pt] &=-\log(2)\log(y)+\operatorname{Li}_2\left(\frac y2\right)+C\tag5\\[6pt] &=-\log(2)\log(2-x)+\operatorname{Li}_2\left(\frac{2-x}2\right)+C\tag6 \end{align} $$ Explanation:
$(1)$: substitute $x=2-y$
$(2)$: pull $-\int\frac{\log(2)}y\,\mathrm{d}y$ out front
$(3)$: apply the series for $\log\left(1-\frac y2\right)$
$(4)$: integrate the series, term by term
$(5)$: recognize the power series for $\operatorname{Li}_2$
$(6)$: undo the substitution from $(1)$