How to solve $\int_0^1\frac{\operatorname{Ti_6}(x)}{1+x^2}\mathrm{d}x$

Question

I was trying to solve $\displaystyle\int_0^1\frac{\operatorname{Ti_3^2}(x)}{x}\mathrm{d}x$, where $\displaystyle\operatorname{Ti_n}(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)^n}$ denotes the inverse tangent integral of order $n$. and $\displaystyle\operatorname{Ti_n}(x)=\int_0^x\frac{\operatorname{Ti_{n-1}}(t)}{t}\mathrm{d}t$

I started by using integration by parts, $$\displaystyle\int_0^1\frac{\operatorname{Ti_3^2}(x)}{x}\mathrm{d}x\overset{IBP}{=}[\operatorname{Ti_3}(x)\operatorname{Ti_4}(x)]_0^1-\int_0^1\frac{\operatorname{Ti_2}(x)\operatorname{Ti_4}(x)}{x}\mathrm{d}x$$ $$\displaystyle\overset{IBP}{=}\beta(3)\beta(4)-[\operatorname{Ti_2}(x)\operatorname{Ti_5}(x)]+\int_0^1\frac{\arctan(x)\operatorname{Ti_5}(x)}{x}\mathrm{d}x$$ $$\displaystyle\overset{IBP}{=}\beta(3)\beta(4)-\beta(2)\beta(5)+[\arctan(x)\operatorname{Ti_6}(x)]_0^1-\int_0^1\frac{\operatorname{Ti_6}(x)}{1+x^2}\mathrm{d}x$$ $$=\beta(3)\beta(4)-\beta(2)\beta(5)+\frac{\pi}{4}\beta(6)-\int_0^1\frac{\operatorname{Ti_6}(x)}{1+x^2}\mathrm{d}x$$ $$=\frac{\pi^3}{32}\beta(4)-\frac{5\pi^5}{1536}G+\frac{\pi}{4}\beta(6)-\int_0^1\frac{\operatorname{Ti_6}(x)}{1+x^2}\mathrm{d}x$$ The integral $\displaystyle\int_0^1\frac{\operatorname{Ti_6}(x)}{1+x^2}\mathrm{d}x$ has to evaluate to $\displaystyle-\frac{\pi^3}{32}\beta(4)-\frac{5\pi^5}{1536}G-\frac{\pi}{4}\beta(6)+\frac{381}{128}\zeta(7)$, because one closed form for the initial integral is $\displaystyle\frac{\pi}{2}\beta(6)+\frac{\pi^3}{16}\beta(4)-\frac{381}{128}\zeta(7)$, but I have no idea how to solve it.

I tried to use some identities for $\operatorname{Ti_6}(x)$, but I the only one I found was $\operatorname{Ti_6}(x)=\frac{1}{2i}\left(\operatorname{Li_6}(ix)-\operatorname{Li_6}(-ix)\right)$, and I think that only makes it worse.

There's probably a better approach than mine.

Answer 1

Note the following identities used in this proof $$\operatorname{Ti}_2(x)-\operatorname{Ti}_2\left(\frac1x\right)=\frac\pi2\ln x$$ $$\operatorname{Ti}_3(x)+\operatorname{Ti}_3\left(\frac1x\right)=\frac\pi4\ln ^2x+\frac{\pi^3}{16}$$ Use integration by parts 3 times on the target integral $$I=\int_0^1\frac{\operatorname{Ti}_3^2(x)}{x}dx=-\int_0^12\operatorname{Ti}_3(x)\operatorname{Ti}_2(x)\frac{\ln x}xdx$$$$=\int_0^1\left(\operatorname{Ti}_3(x){\arctan x}+\operatorname{Ti}_2^2(x)\right)\frac{\ln^2 x}xdx=-\int_0^1\left(\operatorname{Ti}_2(x)\frac{\arctan x}x+\frac{\operatorname{Ti}_3(x)}{3(1+x^2)}\right)\ln^3(x)dx$$$$=-\int_0^1\operatorname{Ti}_2(x)\arctan(x)\frac{\ln^3(x)}{x}dx-\frac13\int_0^1\frac{\operatorname{Ti}_3(x)\ln^3x}{1+x^2}dx$$ Use integration by parts on the left integral $$=\frac{1}{4}\int_0^1\left(\frac{\arctan^2(x)}x+\frac{\operatorname{Ti}_2(x)}{1+x^2}\right)\ln^4(x)dx-\frac13\int_0^1\frac{\operatorname{Ti}_3(x)\ln^3x}{1+x^2}dx=\frac J4+\frac K4-\frac L3$$ Start with $J$ $$J=\int_0^1\frac{\arctan^2(x)\ln^4x}xdx=-\frac25\int_0^1\frac{\arctan(x)\ln^5x}{1+x^2}dx=-\frac25\int_1^\infty\frac{(\arctan(x)-\pi/2)\ln^5x}{1+x^2}dx$$$$=\frac\pi{10}\int_1^\infty\frac{\ln^5x}{x^2+1}dx-\frac15\int_0^\infty\frac{\arctan (x)\ln^5x}{x^2+1}dx$$$$=12\pi\beta(6)-\frac15\int_0^\infty\frac{\arctan (x)\ln^5x}{x^2+1}dx$$$$=12\pi\beta(6)-\frac1{320}\int_0^\infty\frac{\arctan (\sqrt x)\ln^5x}{\sqrt x(1+x)}dx$$ Using the following Taylor series, invoke Ramanujan's Master Theorem $$\frac{\arctan(\sqrt x)}{\sqrt x(1+x)}=\frac12\sum_{n=0}^\infty(-x)^n\left(\psi(n+3/2)-\psi(1/2)\right)$$ $$\int_0^\infty x^{s-1}\frac{\arctan(\sqrt x)}{\sqrt x(1+x)}dx=\frac\pi2\csc(\pi s)(\psi(3/2-s)-\psi(1/2))$$ $$\int_0^\infty \frac{\arctan(\sqrt x)\ln^5 x}{\sqrt x(1+x)}dx=\left(\frac d{ds}\right)^5\left[\frac\pi2\csc(\pi s)(\psi(3/2-s)-\psi(1/2))\right]\Big|_{s=1}=\frac{49\pi^4}6\zeta(3)+310\pi^2\zeta(5)+7620\zeta(7)$$ Therefore $$J=-\frac{49\pi^4}{1920}\zeta(3)-\frac{31\pi^2}{32}\zeta(5)-\frac{381}{16}\zeta(7)+12\pi\beta(6)$$ Continue with K $$K=\int_0^1\frac{\operatorname{Ti}_2(x)\ln^4(x)}{1+x^2}dx$$$$=\frac12\int_0^1\frac{\operatorname{Ti}_2(x)\ln^4(x)}{1+x^2}dx+\frac12\int_1^\infty\frac{(\operatorname{Ti}_2(x)-\frac\pi2\ln (x))\ln^4(x)}{1+x^2}dx$$$$=\frac12\int_0^\infty\frac{\operatorname{Ti}_2(x)\ln^4(x)}{1+x^2}dx-\frac\pi4\int_1^\infty\frac{\ln^5(x)}{1+x^2}dx$$$$=-30\pi\beta(6)+\frac1{64}\int_0^\infty\frac{\operatorname{Ti}_2(\sqrt x)\ln^4(x)}{\sqrt x(1+x)}dx$$ Use the following Taylor series and invoke Ramanujan's Master Theorem $$\frac{\operatorname{Ti}_2(\sqrt x)}{\sqrt x(1+x)}=\sum_{n=0}^\infty(-x)^n\left(\frac{\pi^2}8-\frac14\psi_1\left(n+\frac32\right)\right)$$ $$\int_0^\infty x^{s-1}\frac{\operatorname{Ti}_2(\sqrt x)}{\sqrt x(1+x)}dx=\pi\csc(\pi s)\left(\frac{\pi^2}8-\frac14\psi_1\left(\frac32-s\right)\right)$$ $$\int_0^\infty \frac{\operatorname{Ti}_2(\sqrt x)\ln^4(x)}{\sqrt x(1+x)}dx=\left(\frac d{ds}\right)^4\left[\pi\csc(\pi s)\left(\frac{\pi^2}8-\frac14\psi_1\left(\frac32-s\right)\right)\right]\Big|_{s=1}=\frac{49\pi^4}{30}\zeta(3)+124\pi^2\zeta(5)+4572\zeta(7)$$ Therefore $$K=\frac{49\pi^4}{1920}\zeta(3)+\frac{31\pi^2}{16}\zeta(5)+\frac{1143}{16}\zeta(7)-30\pi\beta(6)$$

Continue with L $$L=\int_0^1\frac{\operatorname{Ti}_3(x)\ln^3(x)}{1+x^2}dx$$$$=\frac12\int_0^1\frac{\operatorname{Ti}_3(x)\ln^3(x)}{1+x^2}dx+\frac12\int_1^\infty\frac{(\operatorname{Ti}_3(x)-\frac\pi4\ln^2(x)-\frac{\pi^3}{16})\ln^3(x)}{1+x^2}dx$$$$=\frac12\int_0^\infty\frac{\operatorname{Ti}_3(x)\ln^3(x)}{1+x^2}dx-\frac\pi8\int_1^\infty\frac{\ln^5(x)}{1+x^2}dx-\frac{\pi^3}{32}\int_1^\infty\frac{\ln^3(x)}{1+x^2}dx$$$$=-15\pi\beta(6)-\frac{3\pi^3}{16}\beta(4)+\frac1{32}\int_0^\infty\frac{\operatorname{Ti}_3(\sqrt x)\ln^3(x)}{\sqrt x(1+x)}dx$$ Use the following Taylor series and invoke Ramanujan's Master Theorem $$\frac{\operatorname{Ti}_3(\sqrt x)}{\sqrt x(1+x)}=\sum_{n=0}^\infty(-x)^n \frac{\psi_2(n+3/2)-\psi_2(1/2)}{16}$$ $$\int_0^\infty x^{s-1} \frac{\operatorname{Ti}_3(\sqrt x)}{\sqrt x(1+x)}dx=\frac\pi{16}\csc(\pi s)\left(\psi_2(3/2-s)-\psi_2(1/2)\right)$$ $$\int_0^\infty \frac{\operatorname{Ti}_3(\sqrt x)\ln^3(x)}{\sqrt x(1+x)}dx=\left(\frac d{dx}\right)^3\left[\frac\pi{16}\csc(\pi s)\left(\psi_2(3/2-s)-\psi_2(1/2)\right)\right]\Big|_{s=1}=\frac{93\pi^2}4\zeta(5)+\frac{5715}4\zeta(7)$$ Therefore $$L=\frac{93\pi^2}{128}\zeta(5)+\frac{5715}{128}\zeta(7)-\frac{3\pi^3}{16}\beta(4)-15\pi\beta(6)$$ Finally $$I=\frac14\left(-\frac{49\pi^4}{1920}\zeta(3)-\frac{31\pi^2}{32}\zeta(5)-\frac{381}{16}\zeta(7)+12\pi\beta(6)\right)+\frac14\left(\frac{49\pi^4}{1920}\zeta(3)+\frac{31\pi^2}{16}\zeta(5)+\frac{1143}{16}\zeta(7)-30\pi\beta(6)\right)-\frac13\left(\frac{93\pi^2}{128}\zeta(5)+\frac{5715}{128}\zeta(7)-\frac{3\pi^3}{16}\beta(4)-15\pi\beta(6)\right)$$ We have obtained the closed form of the target integral, in addition to 3 others $$I=\int_0^1\frac{\operatorname{Ti}_3^2(x)}{x}dx=-\frac{381}{128}\zeta(7)+\frac{\pi^3}{16}\beta(4)+\frac\pi2\beta(6)$$ $$J=\int_0^1\frac{\arctan^2(x)\ln^4x}xdx=-\frac{49\pi^4}{1920}\zeta(3)-\frac{31\pi^2}{32}\zeta(5)-\frac{381}{16}\zeta(7)+12\pi\beta(6)$$ $$K=\int_0^1\frac{\operatorname{Ti}_2(x)\ln^4(x)}{1+x^2}dx=\frac{49\pi^4}{1920}\zeta(3)+\frac{31\pi^2}{16}\zeta(5)+\frac{1143}{16}\zeta(7)-30\pi\beta(6)$$ $$L=\int_0^1\frac{\operatorname{Ti}_3(x)\ln^3(x)}{1+x^2}dx=\frac{93\pi^2}{128}\zeta(5)+\frac{5715}{128}\zeta(7)-\frac{3\pi^3}{16}\beta(4)-15\pi\beta(6)$$ Additionally, a generalized version of the Taylor series used here exists $$\frac{\operatorname{Ti}_{k+1}(\sqrt x)}{\sqrt x(1+x)}=\frac{(-1/2)^{k+1}}{k!}\sum_{n=0}^\infty(-x)^n\left(\psi_k\left(\frac12\right)-\psi_k\left(\frac{2n+3}2\right)\right)$$ which leads to the following integral $$\int_0^\infty\frac{x^{s-1}\operatorname{Ti}_{k+1}(x)}{1+x^2}dx=\frac{(-1)^{k+1}}{2^{k+2}k!}\frac\pi{\cos(\pi s/2)}\left(\psi_k\left(\frac12\right)-\psi_k\left(1-\frac s2\right)\right)$$ In hindsight, this alone leads to a more direct approach to the integral in the title

Answer 2

Too long for a comment:

Let the integral be denoted by $I$: $$I = \int_0^1 \frac{\operatorname{Ti_6}(x)}{1+x^2} \mathrm{d}x$$

Let $x = \tan\theta$, then $\mathrm{d}x = \sec^2\theta \mathrm{d}\theta$ and we have $$I = \int_0^{\pi/4} \frac{\operatorname{Ti_6}(\tan\theta)}{1+\tan^2\theta} \sec^2\theta \mathrm{d}\theta = \int_0^{\pi/4} \operatorname{Ti_6}(\tan\theta) \mathrm{d}\theta$$ With the provided property in OP's post (cf. also here for $Ti_2$), we have $$ \operatorname{Ti}_n(x) = \Im(\operatorname{Li}_n(ix)) $$ and $$ \operatorname{Li}_n(z) = \sum_{k=1}^\infty \frac{z^k}{k^n}. $$

Therefore we have $$ I = \int_0^{\pi/4} \operatorname{Ti_6}(\tan\theta) \mathrm{d}\theta = \int_0^{\pi/4} \Im(\text{Li}_6(i \tan (x))) \, dx = \Im\left(\sum_{k=1}^\infty \frac{i^k}{k^6}\int_0^{\frac{\pi }{4}} \tan ^k(x) \, dx\right) $$

We have $$ \int_0^{\frac{\pi }{4}} \tan ^k(x) \, dx =\frac{1}{4} \left(\psi ^{(0)}\left(\frac{k+3}{4}\right)-\psi ^{(0)}\left(\frac{k+1}{4}\right)\right) $$ where $\psi ^{(0)}$ is the Digamma function. Also $$ \Im(i^k)=\sin\left(\frac{\pi}{2}k\right)=\begin{cases} 0,\quad k \text{ is even}\\ (-1)^k, \quad k \text{ is odd} \end{cases} $$ And so we obtain: $$ I = \frac 1 4\sum _{k=0}^{\infty } \frac{(-1)^k \left(\psi ^{(0)}\left(1+\frac{k}{2}\right)-\psi ^{(0)}\left(\frac{1+k}{2}\right)\right)}{(2 k+1)^6 } $$

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Using Legendre´s duplication formula, I was able to deduce $$ \psi ^{(0)}\left(z+\frac 12\right)-\psi ^{(0)}(z)=2 \psi ^{(0)}(2 z)-2\psi ^{(0)}(z)-2\log(2) $$ for which I thought could be used to break down the series with $z=\frac{k+1}{2}$, but it didn't help me come to the closed form provided by OP.