# Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$

Recently I have encountered some integrals involving Polylogarithms like this one or closely related integrals such as this one. Hence I am quite fascinated by these kinds of definite integrals $$-$$ especially their evaluation $$-$$ I did a little bit research here one MSE. While overlooking the recent questions concerning the tag I stumbled upon this post which asks for a concrete evaluating of a polylogarithmic integral. However the post also states the equality

$$\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$$

Together with the comment "It is not difficult to show that". Since I know the author of this post $$-$$ Jack D'Aurizio $$-$$ is rather familiar with integrals of this type than me I guess indeed for himself it is easily done. Nevertheless I have my struggles showing this.

Hence we are dealing with an integral involving a Polylogarithm I thought about applying integration by parts to get rid of the the Polylogarithm. But I am not sure about the right choice of $$u$$ and $$\mathrm dv$$ respectively. My first guess was simply $$u=\operatorname{Li}_3(1-z)$$ and therefore $$\displaystyle \mathrm dv=\frac1{\sqrt{z(1-z)}}$$. From hereon the first problem occurs: integrate $$\mathrm dv$$. There are at least the two possibilities $$v=\sin^{-1}(2z-1)$$ and $$v=2\sin^{-1}(\sqrt{z})$$ which both lead to the same $$\mathrm dv$$ but on the other hand imply different results for the first IBP step. To be precise

\begin{align} \tag{1}\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz&=\left[\operatorname{Li}_3(1-z)\sin^{-1}(2z-1)\right]_0^1-\int_0^1\sin^{-1}(2z-1)\frac{\operatorname{Li}_2(z)}{z}dz\\ &=\color{red}{\frac38\pi\zeta(3)}-\int_0^1\sin^{-1}(2z-1)\frac{\operatorname{Li}_2(z)}{z}\mathrm dz\\ \tag{2}\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz&=\left[\operatorname{Li}_3(1-z)2\sin^{-1}(\sqrt{z})\right]_0^1-\int_0^12\sin^{-1}(\sqrt{z})\frac{\operatorname{Li}_2(z)}z\mathrm dz\\ &=\color{red}{0}-\int_0^12\sin^{-1}(\sqrt{z})\frac{\operatorname{Li}_2(z)}{z}\mathrm dz \end{align}

For myself I am in favor of the first option since it contains the value $$\pi\zeta(3)$$ but with the wrong coefficient. Howsoever I am not capable of solving the remaining integrals which involve a combination of the inverse sine function and the Dilogarithm. Again I thought about IBP again but I am totally confused what to choose as $$u$$ and $$\mathrm dv$$. Therefore I think I am on the wrong tack.

I have dealed with polylogarithmic and logarithmic integrals before but the square roots are causing me trouble. I tried to absorb at least the $$\sqrt{1-z}$$ within the Trilogarithm and then doing IBP which results in

\begin{align} \int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz&=\int_0^1\sum_{n=1}^{\infty}\frac{(1-z)^{n-1/2}}{n^3}\frac{dz}{\sqrt{z}}\\ &=\left[\sum_{n=1}^{\infty}\frac{(1-z)^{n-1/2}}{n^3}2\sqrt{z}\right]_0^1-\int_0^1\frac{\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)}{(1-z)^{3/2}}\sqrt{z}\mathrm dz\\ &=\color{red}{0}-\int_0^1\frac{\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)}{(1-z)^{3/2}}\sqrt{z}\mathrm dz \end{align}

I am not sure whether this is helpful at all or if it does not make the whole problem more complicated. Honestly speaking I am lost right now and do not know how to proceed of how to approach to the given equality at all.

Could someone explain me how to proceed with the given integrals including integrands combined out of inverse sine and polylogarithmic functions? Are these integrals even solvable; when yes how $$($$maybe without using the given the given integral$$)$$? Or was my whole approach nonsense and another attempt is needed here? You can also share a link or refer to another post here on MSE in case I have overlooked something.

Using the well-known identity: $${\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}}) + {\rm{L}}{{\rm{i}}_3}(1 - x) + {\rm{L}}{{\rm{i}}_3}(x) = \zeta (3) + \frac{{{\pi ^2}}}{6}\ln (1 - x) - \frac{1}{2}\ln x{\ln ^2}(1 - x) + \frac{1}{6}{\ln ^3}(1 - x)$$ we obtain (the integral on RHS can be easily evaluated by differentiating beta function): $$2\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(1 - x)}}{{\sqrt {x(1 - x)} }}dx} + \int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = - 2\pi \zeta (3) + \frac{8}{3}\pi {\ln ^3}2 - \frac{2}{3}{\pi ^3}\ln 2$$ By transformation $$u=x/(1-x)$$, we have $$\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = \int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - u)}}{{(1 + u)\sqrt u }}du}$$ I claim this integral is $$-6\pi \zeta(3)$$.

To establish this value, it suffices to show, with $$\zeta(\cdot,\cdot)$$ Hurwitz zeta function, $$\int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}}{x^{s - 1}}dx} = \frac{\pi }{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right] \qquad 0 by Mellin inversion theorem, this in turn is equivalent to, (which applies as the function tends to $$0$$ uniformly in the vertical strip $$0<\Re(s)<1$$ thanks to the $$\csc(s\pi)$$ factor) for an instance of $$c$$ with $$0: $$\tag{1} \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}} = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} \qquad x>0$$ Note that both sides of $$(1)$$ is an analytic function for $$\Re(x) > 0$$, hence it suffices to consider the case when $$0. When this is the case, we can draw a vertical semicircle on the left half-plane, with vertices $$c \pm i\infty$$, then the integral on the semicircle tends to $$0$$, calculating residues at $$-1,-2,\cdots$$ gives \begin{aligned}\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} &= \sum\limits_{n = 1}^\infty {{{( - x)}^n}\left[ {\zeta (3) - \zeta (3,1 + n)} \right]} \\ &=\sum\limits_{n = 1}^\infty {{{( - x)}^n}\sum\limits_{k = 1}^n {\frac{1}{{{k^3}}}} } = \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}} \end{aligned} where we exchanged two order of summations, completing the proof.

• How can one obtain such equalities? I am rather new to this whole field and while trying to verify a similiar equality for the dilogarithm I failed commpletely. Am I right in assuming that the second equality follows from the first one by diving throughout $\sqrt{x(1-x)}$ ensued by integrating from $0$ to $1$? Furthermore I guess the second paragraphs justifies your claim about the value of $-6\pi\zeta(3)$ to complete the above equality unless I am mistaken? – mrtaurho Oct 6 '18 at 12:19
• Yes for your last two questions. A proof of the above equality can be found in Polylogarithms and associated functions by Leonard Lewin page 155. You might also want to read the first few chapters on dilogarithm. – pisco Oct 6 '18 at 16:39
• Good to see that my intution works out at least; however I was not able to find the book as a whole - or at least the page you refered to - online. Hence I am not planning to buy book right now could you maybe share the one page? Or just add the proof within your answer? Since the other option for me would be to ask a new question here on MSE only concentrating on the equality which would not be a problem aswell. – mrtaurho Oct 6 '18 at 17:26
• I emailed an electronic version of the book to you, to the address you mention in your profile. – pisco Oct 6 '18 at 17:45
• Thank you for this. I guess I got an idea of how to attempt to the above equality after I proved for myself an equivalent version for the Dilogarithm Function. Together with the book you provided me I guess an additional post would be redundant. I appreciate your help. – mrtaurho Oct 6 '18 at 17:50

$$\int_{0}^{1}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz=\int_{0}^{1}\frac{\text{Li}_3(z)}{\sqrt{z(1-z)}}\,dz=2\int_{0}^{1}\frac{\text{Li}_3(u^2)}{\sqrt{1-u^2}}\,du=2\int_{0}^{\pi/2}\text{Li}_3(\sin^2\theta)\,d\theta$$ by the very definition of $$\text{Li}_3$$, plus the identity $$\int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}$$, equals $$\pi\sum_{n\geq 1}\frac{\binom{2n}{n}}{n^3\cdot 4^n},$$ i.e. a rather innocent hypergeometric series, namely $$2\pi\cdot\phantom{}_5 F_4\left(1,1,1,1,\frac{3}{2};2,2,2,2;1\right)$$, which can be evaluated in many ways, for instance through Fourier-Legendre series expansions, or by writing the thing above in terms of $$\int_{0}^{1}\frac{\log^2(z)\,dz}{\sqrt{1-z}},\qquad \int_{0}^{1}\frac{\log^3(z)\,dz}{\sqrt{1-z}}$$ which clearly are the second and third derivatives of a Beta function.
In "higher" terms, any chain of identities of the $$\int f(x)\omega(x)\,dx = \langle f,\omega\rangle \stackrel{\begin{array}{c}\text{series}\\[-0.2cm]\text{rearrengement}\end{array}}{=} \langle \tilde f,\tilde\omega\rangle=\int \tilde f(x)\tilde \omega(x)\,dx$$ kind induces a transformation $$f\mapsto\tilde{f}$$ which generalizes the binomial transform.
In our case $$\text{Li}_3$$ is essentially mapped into $$\log^3$$.

• First of all thank you for your quick response. I hoped that you as author of my cited post would respond to my question. I guess I understand your approach but it seems quite complicate in my mind to involve a $_5F_4$ hypergeometric function here. Anyway I will glance through your arXiv article as soon as possible. Nevertheless is there maybe an easier attempt like simply applying IBP with the right choice of $u$ and $dv$? In addition I am not quite sure how to perform the transform of $\operatorname{Li}_3$ to $\log^3$ could you maybe elaborate this process further? – mrtaurho Oct 5 '18 at 21:43
• @mrtaurho: since $\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n$, when you multiply this object by $\log^2(x)$ and integrate over $(0,1)$ you get a hypergeometric series whose structure is extremely similar to the structure of the needed $\phantom{}_5 F_4$. One should not be scared by the fact that $5$ and $4$ are "large" numbers in this context, since all the hypergeometric parameters except one are integers, so the evaluation in terms of Euler sums is expected to be simple. – Jack D'Aurizio Oct 5 '18 at 22:00
• @mrtaurho (is there maybe an easier attempt like simply applying IBP...?) Not meant to discourage you, but it is quite naive to think this kind (i.e. those of weight $\geq 3$) of integral can be tackled solely by IBP alone. Rather, hidden symmetry is often the key for a nice answer, you might be able to conceive that this mean after you have practiced a lot of them. – pisco Oct 6 '18 at 11:44
• @JackD'Aurizio To be precise I was not scared by the "large" numbers but rather by the necessity of using hypergeometric series in this context. Up to now I only worked on integrals containing a dilogarithm which was quite easy to get rid of by simply applying IBP. The complexity that arrives by going up to a trilogarithm suprised me. Therefore I was hoping for a more down to earth approach instead of involving the calculus of hypergeometric series since I am not that familiar with these kinds of evaluations. However thank you for the clarification. – mrtaurho Oct 6 '18 at 12:05
• @pisco Yeah I guessed so for myself. It is quite unsatisfactory that the standard techniques are kind of useless is this context. Talking about practice; could you maybe provide a good source for problems similiar to this one? I do not exactly know how I should search for them. Furthermore Jack's papers are a little bit to sophisticated to begin with therefore I would be glad for every form of guidance how to get started. – mrtaurho Oct 6 '18 at 12:25