Some interesting integrals with dilogarithm

Calculating without techniques involving the contour integration

$$a) \ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^2-(\operatorname{Li}_2(e^{i x}))^2}{e^{-i x}-e^{i x}}\textrm{d}x;$$

$$b) \ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^3-(\operatorname{Li}_2(e^{i x}))^3}{e^{-i x}-e^{i x}}\textrm{d}x.$$

I'm working now on such a method. What would your real method inspiration be here?

Supplementary question: Calculate

$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^4-(\operatorname{Li}_2(e^{i x}))^4}{e^{-i x}-e^{i x}}\textrm{d}x.$$

Moreover, may we hope for a generalization of the type below?

$$ I(n)=\int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^n-(\operatorname{Li}_2(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x.$$

Preparing another two generalizations: $$ i) \ J(n,m)=\int_0^{2\pi} \frac{(\operatorname{Li}_m(e^{-i x}))^n-(\operatorname{Li}_m(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x;$$

$$ ii) \ K(n)=\int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\cdots \operatorname{Li}_n(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\cdots \operatorname{Li}_n(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x.$$

Solution 1:

$ \large \text{ Hooray!!!}$ The closed-form of the integral $a)$ is impressive. According to my calculations,

$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^2-(\operatorname{Li}_2(e^{i x}))^2}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^5}{48}.$$

Including also the trivial case, $n=1$,

$$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})-\operatorname{Li}_2(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^3}{4}.$$

$ \large \text{ Second Hooray!!!}$

$$ \int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^3-(\operatorname{Li}_2(e^{i x}))^3}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^7}{192}.$$

$ \large \text{Third Hooray!!!}$

I think I have found a first generalization!

$$ I(n)=\int_0^{2\pi} \frac{(\operatorname{Li}_2(e^{-i x}))^n-(\operatorname{Li}_2(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{\pi^{2n+1}}{6^n}\left(1-\left(-\frac{1}{2}\right)^n\right).$$

$ \large \text{Fourth Hooray!!!}$

Guess what?! I'm also done with the generalization $J(n,m)$ $$\ J(n,m)=\int_0^{2\pi} \frac{(\operatorname{Li}_m(e^{-i x}))^n-(\operatorname{Li}_m(e^{i x}))^n}{e^{-i x}-e^{i x}}\textrm{d}x=\pi(\zeta(m)^n-((2^{1-m}-1)\zeta(m))^n).$$

$ \large \text{Fifth Hooray!!!}$

I computed $2$ cases of the generalization in $K(n)$ and I approach the solution of the generalization. So, $$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{5}{48}\pi^3\zeta(3);$$ $$ \int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\operatorname{Li}_4(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\operatorname{Li}_4(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x=\frac{17}{6912}\pi^7 \zeta(3).$$

$ \large \text{Sixth Hooray!!!}$

Looks like I have been lucky today! Let me put the last generalization I just proved in a nice form

$$K(n)=\int_0^{2\pi} \frac{\operatorname{Li}_2(e^{-i x})\operatorname{Li}_3(e^{-i x})\cdots \operatorname{Li}_n(e^{-i x})-\operatorname{Li}_2(e^{i x})\operatorname{Li}_3(e^{i x})\cdots \operatorname{Li}_n(e^{i x})}{e^{-i x}-e^{i x}}\textrm{d}x$$

$$=\pi \left(\zeta(2)\zeta(3)\cdots \zeta(n)+(-1)^{n-1} \eta(2)\eta(3)\cdots\eta(n))\right).$$

Extra information:

https://en.wikipedia.org/wiki/Riemann_zeta_function

https://en.wikipedia.org/wiki/Dirichlet_eta_function

https://en.wikipedia.org/wiki/Polylogarithm

Solution 2:

Getting an idea by going trough an example

Consider the Integral $$ I=\int_0^{2\pi}dx\frac{\text{Li}^2_2(e^{i x})-\text{Li}^2_2(e^{-i x})}{2 i \sin(x)} $$ using the series representation for the dilogarithm this can be rewritten as

$$ I=\int_0^{2\pi}dx\sum_{n,m>0}\frac{1}{n^2 m^2}\frac{\sin(m+n)x}{\sin(x)} $$ exchange summation and integration and using the simple fact that $\int_0^{2\pi}dx \frac{\sin(lx)}{\sin(x)}=2 \pi$ for $l \in 2\mathbb{N}+1$ we get

$$ I=2 \pi s^{(2)}_2 =2 \pi\sum_{\substack{n,m>0 \\n+m=odd}}\frac{1}{n^2 m^2} $$

The closed form solution to $s^{(2)}_2$ is pretty simple to obtain. Observe to fullfil the condition $n+m=odd$ either $n$ has to be odd and $m$ to be even, or vice versa. This means we have $2$ possible combinations of even and odd which yield a contribution to our sum.

$$ s^{(2)}_2 =2\sum_{\substack{n>0,m\geq 0}}\frac{1}{(2n)^2 (2m+1)^2}=2\frac{\zeta(2)}{4}\frac{3\zeta(2)}{4}=\frac{3}{8}\zeta^2(2) $$

the strategy for providing a closed form solution will follow the same arguments, except that we additonally need a combinatoric lemma proven in the appendix

The General Case

We now want to investigate

$$ I^{(r)}_n=\int_0^{2\pi}dx\frac{\text{Li}^n_r(e^{i x})-\text{Li}^n_r(e^{-i x})}{2 i \sin(x)} $$

Going through the same procedure then in the motivating example we might show that

$$ I^{(r)}_n=2 \pi s^{(r)}_n $$

This means we are interested in a family of Euler like sums, since

$$ s^{(r)}_n=\sum_{\substack{ k_i \geq 1, \\ \sum_{n\geq i \geq 1} k=odd }}\frac{1}{\prod_{ n \geq i\geq1}{k^r_i}} $$

we now have to take care that we account for all possible partitions of the integers such that the constraint $\sum_{n\geq i \geq 1} k=odd$ is fulfilled. As shown in the Appendix, we have to choose $2l-1$numbers to be odd and $n-2l+1$ to be even. Each of this partitions contains $N_{l,n}=\binom{n}{2l-1}$ equivalent combinations. This means that

$$ s^{(r)}_n=\sum_{l_{max}(n)\geq l\geq1}N_{l,n}\sum_{k_i\geq 1, K_i \geq 0} \prod_{2l-1 \geq i\geq1}\frac{1}{{(2K_i+1)^r}} \prod_{n- 2l+1 \geq i\geq1}\frac{1}{{(2k_i)^r}} $$

using now the well known identity $\sum_{k\geq 0}(2m+1)^{-r}=(1-1/2^{-r})\zeta(r)$ we can carry out the infinite summations $$ s^{(r)}_n=\sum_{l_{max}(n)\geq l\geq1}\frac{N_{l,n}}{2^{r(n-2l+1)}}(1-\frac{1}{2^{r}})^{2l-1}\zeta^n(r)=\sum_{l_{max}(n)\geq l\geq1}c_{l,n}\zeta(r)^n $$

Furthermore the sum over coefficents can be done in closed form by virtue of the Binomial identiy:

$$ s^{(r)}_n=C_{n,r}\zeta(r)^n\,\,,\,\,C_{n,r}=\begin{cases} \frac{1}{2}\left(1+\frac{2^{n-r}}{4^{r n/2}}(2^{r-1}-1)^n\right)\,\, \text{if} \,\,n \,\, even\\ \frac{1}{2}\left(1+\frac{2^{n-r}}{4^{r(n-1)/2}}(2^{r-1}-1)^n\right)\,\, \text{if} \,\,n \,\, odd \\ \end{cases} $$

Note that we get the sums with the constraint $\sum_{n\geq i\geq1} k_i=even$ for free: $$ \bar{s}_n^{r}=\left(1-C_{n,r}\right)\zeta(r)^n $$

It is also interesting to note, that $\lim_{n\rightarrow\infty}\frac{s^{(r)}_n}{\zeta(r)^n}=\frac{1}{2}$ which can be traced back to fact that for very large $n$ we have to choose roughly $n/2$ odd factors out ouf $\sum_{n\geq 1 i\geq1} k_i$ due to concentration of $N_{l,n}$ around $n/2$.

Last but not least a few examples:

\begin{align*} s^{(2)}_2=\frac{3}{8}\zeta^2(2)\,\, ,\,\,\bar{s}^{(2)}_2=\frac{5}{8}\zeta^2(2) \\ s^{(3)}_3=\frac{91}{128}\zeta^3(3)\,\, ,\,\,s^{(3)}_5=\frac{1267}{2048}\zeta^5(3), \end{align*}

Appendix: A small detour to combinatorics

Consider the sum of integers

$$ c_m=n_1+n_2+...+n_m $$ how can we partion $c_m$ into odd and even elements, such that $c_m$ is odd? Since the odd and even numbers furnish a representation of the group $\mathbb{Z}_2$ it follows trivially that we need always an odd number $2l-1$ of the $n_m$'s to be odd. For any fixed $l$ we then have

$$ N_{l,m}=\binom{m}{2l-1}\,\, ,\,\, l \in \begin{cases} \{1,m/2\} \,\, \text{if} \,\,m \,\, \text{even}\\ \{1,\lceil m/2 \rceil\} \,\, \text{if} \,\,m \,\, \text{odd}\\ \end{cases} $$ equivalent admissible partitions of $c_m$.