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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.

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    $\begingroup$ first step is to use Wolfram alpha, realise the answer is too complicated and conclude you went wrong somewhere else earlier on in question $\endgroup$
    – Displayname
    Commented Feb 4, 2019 at 0:48
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    $\begingroup$ No, this is how it should be. I just don't know how to get to that answer. Also, it's not a textbook question. I came up with this problem myself @Displayname $\endgroup$
    – KKZiomek
    Commented Feb 4, 2019 at 0:50
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    $\begingroup$ ohh okay, I assumed it was some textbook/assignment question and wondered why they'd give such a horrendous integral to compute...but I can't help you really but im sure someone will come along soon $\endgroup$
    – Displayname
    Commented Feb 4, 2019 at 0:52

2 Answers 2

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

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  • $\begingroup$ Thank you. How did I not come up with this on my own? Thank you once more! $\endgroup$
    – KKZiomek
    Commented Feb 4, 2019 at 1:15
  • $\begingroup$ Only, you forgot about a constant one half in the front I think. Other than that, the substitution works and is much easier than I thought $\endgroup$
    – KKZiomek
    Commented Feb 4, 2019 at 1:17
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    $\begingroup$ The intuition behind the substitution comes since $\frac{2-x}{2}=t$ works great with the denominator, but also $x=2(1-t)$ allows to get $\ln(1-t)$ separately. $\endgroup$
    – user625055
    Commented Feb 4, 2019 at 1:17
  • $\begingroup$ I didn't forgot. Write as $\frac12\int \frac{\ln x}{\frac{2-x}{2}}dx$, do you see it better now? The $\frac12$ cancels with $-2$ from $dt$. $\endgroup$
    – user625055
    Commented Feb 4, 2019 at 1:18
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    $\begingroup$ Oh right. Thank you ;) I forgot about differential $\endgroup$
    – KKZiomek
    Commented Feb 4, 2019 at 1:20
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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 $(1)$

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