Ways to tackle the integral $\int_{0}^{\frac{\pi}{4}}\operatorname{Li}_3(\tan^4 x) \, dx$

$$\boxed{J = \int_0^{\frac{\pi}{4}}\operatorname{Li}_3(\tan^4(x)) \, dx}$$

Since I had no clue about trilogarithms, tried some searching to get enough understanding to solve the above integral, I found this general relation;

$$\operatorname{Li}_s(z)=\frac{\Gamma(1-s)}{2\pi^{1-s}}\left(i^{1-s}\zeta\left(1-s,\frac{1}{2}+\frac{\ln(-z)}{2\pi i}\right)+i^{s-1}\zeta\left(1-s,\frac{1}{2}-\frac{\ln(-z)}{2\pi i}\right)\right)$$

Also, some general functional equations from here, specifically,

$$\operatorname{Li}_3(z)+\operatorname{Li}_3(-z)=\frac{1}{4}\operatorname{Li}_3(z^2)$$

$$\operatorname{Li}_3(z)-\operatorname{Li}_3(-z^{-1})=\frac{-1}{6}\left(\ln^3 z+\pi^2 \ln z\right)$$

Or rewriting the above as;

$$\operatorname{Li}_3(z)-\operatorname{Li}_3(-z^{-1})=\frac{-1}{6}\ln^3 z-\zeta(2)\ln z$$

Understanding any other aspects about trilogarithms (or polylogarithm in general) required knowledge was beyond my scope. So I started solving the integral as follows;

Using some prior experience in solving some basic dilogarithmic integrals, I substituted $$\tan(x)=t$$;

$$J=\int_0^{1}\frac{\operatorname{Li}_3(t^4)}{1+t^2}\,dt$$

$$J=\int_0^{1}\frac{\operatorname{Li}_3(x^4)}{1+x^2}\,dx$$

Using the first functional equation,

$$J=\int_0^{1}\frac{4\operatorname{Li}_3(x^2)}{1+x^2}\,dx+\int_0^{1}\frac{4\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=J_1+J_2$$

Edit 1: $$\operatorname{Li}_n(z)=\frac{(-1)^{n-1}}{(n-1)!}\left[\int_0^1\frac{z\ln^{n-1}x}{-zx+1}\,dx\right]$$

Found the above here in the 5th integral representation.

Putting $$n=3$$,

$$J_1=\int_0^{1}\frac{4\operatorname{Li}_3(x^2)}{1+x^2}\,dx=2\int_0^1\int_0^1\frac{1}{(1+x^2)}\frac{x^2\ln^{2}t}{(1-x^2t)}\,dx\,dt$$

Edit 2:

$$J_1=2\int_0^1\int_0^1\frac{x^2\ln^{2}t}{(1-x^2t)(1+x^2)}\,dx\,dt$$

$$J_1=\int_0^1\int_0^1\frac{\ln^{2}t}{(1-x^2t)(1+x^2)}\,dx\,dt-\int_0^1\int_0^1\frac{\ln^{2}t}{(1+t)(1+x^2)}\,dx\,dt$$

Could take this further but it seems like I'm missing some identity or formula to move forward. I am interested in understanding how to solve this integral.

Edit 3: The answer is $$\boxed{J=\frac{1}{8}\left(\zeta\left(4,\frac{1}{4}\right)-\zeta\left(4,\frac{3}{4}\right)\right)-\pi \left(\frac{2\pi G}{3}+\frac{27\zeta(3)}{4}\right)}$$

$$G$$ is Catalan's constant.

• Why do you expect there is a nice solution? Commented Oct 28, 2023 at 7:56
• Search for the coefficients in OEIS Commented Oct 28, 2023 at 9:06
• Ayan see this $\int_0^\infty \frac{\mathrm{Li}_3(x)}{1+x^2} \, dx$ Commented Nov 4, 2023 at 15:36

A more generalized integral:

For $$q,p\in\mathbb{Z}_{\ge1}$$ with $$q+p$$ is even number, we have

$$\begin{gather} \int_0^{\frac{\pi}{4}}\ln^{q-1}(\tan(x))\operatorname{Li}_p(\tan^4(x))\mathrm{d}x=-(1-(-1)^q)2^{2p-3}(1+2^{-p})|E_{q-1}|\left(\frac{\pi}{2}\right)^{q}\eta(p)\\ -(q-1)!2^{2p-2}\sum_{k=0}^{\lfloor{\frac{q-2}{2}}\rfloor}\binom{q+p-2k-2}{p-1}\frac{|E_{2k}|}{(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}\lambda(q+p-2k-1)\\ -(q-1)!2^{2p-1}\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{q+p-2k-1}{p-1}\lambda(2k)\beta(q+p-2k)\\ -(q-1)!2^{2p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{q+p-2k-1}{q-1}2^{-4k}\zeta(2k)\beta(q+p-2k), \end{gather}$$

where $$\lfloor{\cdot}\rfloor$$ is the floor function, $$E$$ is the Euler number, $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s)$$ is the Dirichlet eta function, $$\lambda(s)=\sum_{n=0}^\infty\frac{1}{(2n+1)^s}=(1-2^{-s})\zeta(s)$$ is the lambda function, and $$\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}$$ is the Dirichlet beta function.

Proof: Let $$\tan(x)=t$$, use $$\operatorname{Li}_p(t^2)=2^{p-1}(\operatorname{Li}_p(t)+\operatorname{Li}_p(-t))$$, and then substitute the two generalizations $$(36)$$ and $$(37)$$ given in this preprint:

$$\begin{gather} \int_0^1\frac{\ln^{q-1}(x)\operatorname{Li}_p(x^2)}{1+x^2}\mathrm{d}x=-\frac14(1-(-1)^q)|E_{q-1}|\left(\frac{\pi}{2}\right)^q\eta(p)\\ -(q-1)!2^{p}\sum_{k=0}^{\lfloor{\frac{q}{2}}\rfloor}\binom{q+p-2k-1}{p-1}\lambda(2k)\beta(q+p-2k)\\ -(q-1)!2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{q+p-2k-1}{q-1}2^{-2k}\zeta(2k)\beta(q+p-2k);\\ \\ \\ \int_0^1\frac{\ln^{q-1}(x)\operatorname{Li}_p(-x^2)}{1+x^2}\mathrm{d}x=-\frac14(1-(-1)^q)2^{p}|E_{q-1}|\left(\frac{\pi}{2}\right)^{q}\eta(p)\\ -(q-1)!2^{p-1}\sum_{k=0}^{\lfloor{\frac{q-2}{2}}\rfloor}\binom{q+p-2k-2}{p-1}\frac{|E_{2k}|}{(2k)!}\left(\frac{\pi}{2}\right)^{2k+1}\lambda(q+p-2k-1)\\ +(q-1)!2^{p}\sum_{k=0}^{\lfloor{\frac{p}{2}}\rfloor}\binom{q+p-2k-1}{q-1}2^{-2k}\eta(2k)\beta(q+p-2k).\\ \\ \end{gather}$$

To find your integral, let $$q=1$$ and $$p=3$$, we get

$$\int_0^{\frac{\pi}{4}}\operatorname{Li}_3(\tan^4(x))\mathrm{d}x=32\beta(4)-4\beta(2)\zeta(2)-\frac{27}{4}\pi\zeta(3).$$

• (+110) Great answer! Commented Apr 4 at 22:11
• Thank you for the bonus @user97357329 ..much appreciated Commented Apr 4 at 22:54

If you use series $$I=\int_0^{\frac \pi 4} \text{Li}_3\left(\tan ^4(x)\right)\,dx=\frac14\sum_{n=1}^\infty \frac{\psi ^{(0)}\left(n+\frac{3}{4}\right)-\psi ^{(0)}\left(n+\frac{1}{4}\right)}{n^3}$$ and the summation is given in terms of multiple derivatives of regularized hypergeometric functions (I prefer to not type it).
$$I=\frac{256103 \pi }{864000}-\frac{45039820769}{56119635000}+\frac14\sum_{n=7}^\infty \frac{\psi ^{(0)}\left(n+\frac{3}{4}\right)-\psi ^{(0)}\left(n+\frac{1}{4}\right)}{n^3}$$ Using series $$\frac{\psi ^{(0)}\left(n+\frac{3}{4}\right)-\psi ^{(0)}\left(n+\frac{1}{4}\right)}{n^3}=$$ $$\frac{1}{2 n^4}-\frac{1}{32 n^6}+\frac{5}{512 n^8}-\frac{61}{8192 n^{10}}+\frac{1385}{131072 n^{12}}-\frac{50521}{2097152 n^{14}}+O\left(\frac{1}{n^{16}}\right)$$