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

Question

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

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 $\tan(x)=t$;

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

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

Using the first functional equation,

$$I=\int_0^{1}\frac{4\operatorname{Li}_3(x^2)}{1+x^2}\,dx+\int_0^{1}\frac{4\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=I_1+I_2$$

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 $n=3$,

$$I_1=\int_0^{1}\frac{4\operatorname{Li}_3(x^2)}{1+x^2}\,dx=2\int_0^1\int_0^1\frac{1}{(1+x^2)}\frac{x^2\ln^{2}t}{(1-x^2t)}\,dx\,dt$$

Edit 2:

$$I_1=2\int_0^1\int_0^1\frac{x^2\ln^{2}t}{(1-x^2t)(1+x^2)}\,dx\,dt$$

$$I_1=\int_0^1\int_0^1\frac{\ln^{2}t}{(1-x^2t)(1+x^2)}\,dx\,dt-\int_0^1\int_0^1\frac{\ln^{2}t}{(1+t)(1+x^2)}\,dx\,dt$$

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 $$\boxed{I=\frac{1}{8}\left(\zeta\left(4,\frac{1}{4}\right)-\zeta\left(4,\frac{3}{4}\right)\right)-\pi \left(\frac{2\pi G}{3}+\frac{27\zeta(3)}{4}\right)}$$

$G$ is Catalan's constant.

Edit 4 : There isn’t a nice closed form for $\zeta(3)$, like the other even zeta functions, why is that so for odd valued zeta functions?

Found a valid response - here

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Edit 5 : Finally figured out the solution,

Picking up from where I left it,

$$I=\int_0^{1}\frac{4\operatorname{Li}_3(x^2)}{1+x^2}\,dx+\int_0^{1}\frac{4\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=4I_1+4I_2$$

Considering the first integral,

$$I_1=\int_0^{1}\frac{\operatorname{Li}_3(x^2)}{1+x^2}\,dx=\int_0^1\int_0^1\frac{1}{2(1+x^2)}\frac{x^2\ln^{2}t}{(1-x^2t)}\,dx\,dt$$

$$I_1=\int_0^1\int_0^1\frac{\ln^{2}t}{2(1-x^2t)(1+x^2)}\,dx\,dt-\int_0^1\int_0^1\frac{\ln^{2}t}{2(1+t)(1+x^2)}\,dx\,dt$$

Using these standard results for both our integrals,

$$\boxed{\eta(3)\Gamma(3)=2\eta(3)=\int_0^1 \frac{(\ln(x))^2}{x+1}\,dx}$$

$$\boxed{\frac{\pi}{4}=\int_0^1 \frac{1}{x^2+1}\,dx}$$

$$\boxed{\int \frac{1}{a^2-x^2}\,dx=\frac{1}{a}\tanh^{-1}\left(\frac{x}{a}\right)}$$

And using the Eta function to Zeta function conversion given below, found from here

$$\boxed{\eta(z)=\zeta(z)(1-2^{1-z})}$$

The above integral $I_1$ simplifies to,

$$I_1=\int_0^1 \frac{\ln^2(t)\tanh^{-1}(t)}{2(1+t)\sqrt{t}}\,dt-\frac{3\pi\zeta(3)}{16}$$

I solved this first integral using the techniques mentioned here,

$$\int_0^1 \frac{\ln^2(t)\tanh^{-1}(t)}{2(1+t)\sqrt{t}}\,dt=4\beta(4)-\frac{\pi^2G}{3}$$

From above result,

$$\color{red}{4\int_0^{1}\frac{\operatorname{Li}_3(x^2)}{1+x^2}\,dx=16\beta(4)-\frac{4\pi^2G}{3}-\frac{3\pi\zeta(3)}{4}\tag1}$$

Now, similarly trying to evaluate $I_2=\int_0^{1}\frac{\operatorname{Li}_3(-x^2)}{1+x^2}\,dx$,

I used this cool identity I found, $$\color{blue}{\int_0^\infty \frac{\operatorname{Li}_a(-x^2)}{1+x^2}\,dx=\pi(1-2^{a-1})\zeta(a)\underset{a=3}\implies \int_0^\infty \frac{\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=-3\pi\zeta(3)}$$

But the limits were not matching my integral so rewriting my integral as,

$$\color{red}{\int_0^1\implies\int_0^\infty-\int_1^\infty}$$

In the above split, the first part is evaluated using the above derived result, where as in the second part, after substituting $x=\frac{1}{x}$, we get,

$$\int_0^{1}\frac{\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=-3\pi\zeta(3)-\int_0^{1}\frac{\operatorname{Li}_3\left(\frac{-1}{x^2}\right)}{1+x^2}\,dx\tag2$$

Using another useful result,

Now from here,

$$\color{green}{\operatorname{Li}_3(-t)-\operatorname{Li}_3\left(\frac{-1}{t}\right)=\frac{-(\ln(t))^3}{6}+2\operatorname{Li}_2(-1)\ln(t)\implies -\left({\zeta(2)\ln(t)}+\frac{(\ln(t))^3}{6}\right)\tag3}$$ Also used the below result in our above result,

$$\color{brown}{\operatorname{Li}_2(-1)=-\frac{\zeta(2)}{2}}$$

Also, found the generalized version of our first boxed result as follows,

$$\boxed{\Gamma(z)\eta(z)=(-1)^{z-1}\int_0^1 \frac{\ln^{z-1}(x)}{1+x}\,dx}$$

Using the above three results in $(2)$ on the two integrals it splits into (as per the boxed result $(2)$) , we get,

$$\color{purple}{4\int_0^{1}\frac{\operatorname{Li}_3(-x^2)}{1+x^2}\,dx=16\beta(4)+\frac{2\pi^2G}{3}-6\pi\zeta(3)\tag4}$$

Therefore combining our results $(1),(4)$,

$$\color{brown}{\int_{0}^{\frac{\pi}{4}}\operatorname{Li}_3(\tan^4 x) \, dx=32\beta(4)-\frac{2\pi^2G}{3}-\frac{27\pi\zeta(3)}{4}}$$

Perhaps the closed form for $\beta(4)$ isn't straight forward like $\beta(2)$ and hence using, $$\beta(z)=\frac{1}{4^z}\zeta\left(z,\frac{1}{4}\right)-\frac{1}{4^z}\zeta\left(z,\frac{3}{4}\right)$$

Using above result, we can match our original (alternate) closed form,

$$\int_{0}^{\frac{\pi}{4}}\operatorname{Li}_3(\tan^4 x) \, dx=\frac{1}{8}\left(\zeta\left(4,\frac{1}{4}\right)-\zeta\left(4,\frac{3}{4}\right)\right)-\pi \left(\frac{2\pi G}{3}+\frac{27\zeta(3)}{4}\right)$$

Edit 6:

Found this cool identity,

$$\int_0^\infty \frac{\operatorname{Li}_a(-x)}{1+x^2}=-\frac{\pi}{2^{a+1}}\eta(a)-a\beta(a+1)$$

Answer 1

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:

\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}

which follow from setting $(a,b)=(-1,\pm1)$ in Eq (27) of this preprint.

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

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

Answer 2

Not really an answer.Using Mathematica with MultipleZetaValues packages:

$$\int_0^{\frac{\pi }{4}} \text{Li}_3\left(\tan ^4(x)\right) \, dx=\int_0^1 \frac{\text{Li}_3\left(x^4\right)}{1+x^2} \, dx=-\frac{1}{3} \left(2 C \pi ^2\right)+32 L_{4,2}(4)-\frac{27 \pi \zeta (3)}{4}$$

where:$L_{4,2}(4)$ is Dirichlet L-function, $C$ is Catalan constant.

MMA code:-((2 Catalan \[Pi]^2)/3) + 32 DirichletL[4, 2, 4] - 27/4 \[Pi] Zeta[3]

Answer 3

This is not an answer.

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 $$\frac{\psi ^{(0)}\left(n+\frac{3}{4}\right)-\psi ^{(0)}\left(n+\frac{1}{4}\right)}{n^3}=$$ $$\frac{1}{2 n^4}-\frac{1}{32 n^6}+\frac{5}{512 n^8}-\frac{61}{8192 n^{10}}+\frac{1385}{131072 n^{12}}-\frac{50521}{2097152 n^{14}}+O\left(\frac{1}{n^{16}}\right)$$