Evaluate $\int_0^\infty \frac{x^2\operatorname{Ti}_2(x^2)}{x^4+1} \space dx$

Question

At first, I wanted to evaluate this integral [this is not the integral in question] $$I=\int_0^{\frac{\pi}{2}}x\sqrt{\tan x}\space dx=\int_0^\infty\frac{2x^2\arctan{x^2}}{x^4+1}dx $$ I substituted $u=\sqrt{\tan x}$ and I followed up by parametrizing $$F(t)=\int_0^\infty\frac{x^2\arctan{tx^2}}{x^4+1}dx$$$$I=2F(1)$$ $F(t)$ can be evaluated with Leibniz's rule fairly easily. Differentiate, evaluate, then integrate, as normal. $$F(t)=\int_0^\infty\frac{x^2\arctan{tx^2}}{x^4+1}dx=\frac{\pi}{2\sqrt2}\ln{(\sqrt t+1)}-\frac{\pi}{4\sqrt2}\ln{(t+1)}+\frac{\pi}{2\sqrt2}\arctan(t)$$

Here, I noticed that $\int_0^{y^2} \frac{F(t)}{t}dt$ yields a nice result. $$\int_0^\infty\frac{x^2}{x^4+1}\int_0^y\ \frac{\arctan{(tx^2)}}{t}\space dt\space dx$$$$=\frac{\pi}{2\sqrt2}\int_0^y\frac{\ln{(\sqrt t+1)}}{t}dt-\frac{\pi}{4\sqrt2}\int_0^y\frac{\ln{(t+1)}}{t}dt+\frac{\pi}{2\sqrt2}\int_0^y\frac{\arctan(t)}{t}dt$$ This results in the following $$\int_0^\infty\frac{x^2\operatorname{Ti}_2(x^2y^2)}{x^4+1}dx=-\frac{\pi}{\sqrt2}\operatorname{Li}_2(-y)+\frac{\pi}{4\sqrt2}\operatorname{Li}_2(-y^2)+\frac{\pi}{\sqrt2}\operatorname{Ti}_2(y)$$ A special case is $y=1$, which is the integral in question. $$\int_0^\infty\frac{x^2\operatorname{Ti}_2(x^2)}{x^4+1}dx=\frac{\pi^3}{16\sqrt2}+\frac{\pi G}{\sqrt2}$$ Where $G$ is Catalan's constant

How else can this result be achieved?

Answer 1

\begin{align}J&=\int_0^\infty \frac{x^2\operatorname{Ti}_2(x^2)}{x^4+1}dx\\ &=\int_0^\infty \frac{x^2}{1+x^4}\left(\int_0^{x^2}\frac{\arctan t}{t}dt\right)dx\\ &=\int_0^\infty \frac{x^2}{1+x^4}\left(\int_1^{x^2}\frac{\arctan t}{t}dt\right)dx+\int_0^\infty \frac{x^2}{1+x^4}\left(\int_0^1\frac{\arctan t}{t}dt\right)dx\\ &=\text{G}\underbrace{\int_0^\infty \frac{x^2}{1+x^4}dx}_{=A}+\underbrace{\int_0^\infty \frac{x^2}{1+x^4}\left(\int_1^{x^2}\frac{\arctan t}{t}dt\right)dx}_{u=\frac{1}{x}}\\ &=\text{G}A+\int_0^\infty \frac{1}{1+x^4}\left(\underbrace{\int_{1}^{\frac{1}{x^2}}\frac{\arctan t}{t}dt}_{y=\frac{1}{t}}\right)dx\\ &=\text{G}A+\int_0^\infty \frac{1}{1+x^4}\left(\int_{x^2}^1 \left(\frac{\frac{\pi}{2}-\arctan y}{y}\right)\right)dx\\ &=\text{G}A-\pi\int_0^\infty \frac{\ln x}{1+x^4}dx+\int_0^{\infty}\frac{1}{1+x^4}\left(\int_1^{x^2}\frac{\arctan y}{y}dy\right)\\ &=\int_0^{\infty}\frac{1}{1+x^4}\left(\underbrace{\int_0^{x^2}\frac{\arctan y}{y}dy}_{\text{IBP}}\right) -\pi\underbrace{\int_0^\infty \frac{\ln x}{1+x^4}dx}_{=B}\\ &=\int_0^\infty \frac{1}{1+x^4}\left(\Big[\arctan y\ln y\Big]_0^{x^2}-\underbrace{\int_0^{x^2}\frac{\ln y}{1+y^2}dy}_{z=\sqrt{y}} \right)dx-\pi B\\ &=2\underbrace{\int_0^\infty \frac{\arctan\left(x^2\right)\ln x}{1+x^4}dx}_{=L}-4\underbrace{\int_0^\infty \frac{1}{1+x^4}\left(\int_0^x\frac{z\ln z}{1+z^4}dz\right)dx}_{=K}-\pi B\\ K&=\int_0^\infty \frac{1}{1+x^4}\left(\int_0^1\frac{x^2z\ln(xz)}{1+x^4z^4}\right)dx\\ &=\int_0^1 \left(\frac{z^2}{z^4-1}\underbrace{\int_0^\infty\frac{z(xz)^2\ln(xz)}{1+x^4z^4}dz}_{u(x)=xz}-\frac{z\ln z}{z^4-1}\int_0^\infty \frac{x^2dx}{1+x^4}-\frac{z}{z^4-1}\underbrace{\int_0^\infty \frac{x^2\ln x dx}{1+x^4}}_{=C}\right)dz\\ &=C\int_0^1 \frac{z}{(1+z^2)(1+z)}dz-A\underbrace{\int_0^1\frac{z\ln z}{z^4-1}dz}_{t=z^2}\\ &=\frac{1}{2}C\left[\frac{1}{2}\ln(1+z^2)-\ln(1+z)+\arctan z\right]_0^1-\frac{1}{4}A\int_0^{1}\frac{\ln t}{t^2-1}dt\\ &=\frac{\pi C}{8}-\frac{C\ln 2}{4}-\frac{A}{4}\int_0^1\frac{\ln t}{t-1}dt+\frac{A}{4}\underbrace{\int_0^1\frac{t\ln t}{t^2-1}dt}_{v=t^2}\\ &=\frac{\pi C}{8}-\frac{C\ln 2}{4}-\frac{3A}{16}\int_0^1\frac{\ln t}{t-1}dt\\ &=\boxed{\frac{\pi C}{8}-\frac{C\ln 2}{4}-\frac{A\pi^2}{32}}\\ L&=\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln t}{1+t^4}dt\right)\arctan\left(x^2\right)\right]_0^\infty-\int_0^\infty \frac{2x}{1+x^4}\left(\int_0^x\frac{\ln t}{1+t^4}dt\right)dx\\ &=\frac{B\pi}{2}-\int_0^\infty \frac{2x^2}{1+x^4}\left(\int_0^1\frac{\ln(tx)}{1+t^4x^4}dt\right)\\ &=\frac{B\pi}{2}+2\times\\&\int_0^1\left(\frac{\ln t}{t^4-1}\int_0^\infty \frac{x^2dx}{1+x^4}+\frac{1}{t^4-1}\int_0^\infty \frac{x^2dx\ln x}{1+x^4}-\frac{t}{t^4-1}\underbrace{\int_0^\infty \frac{t(xt)^2\ln(tx)}{1+(tx)^4}dx}_{u(x)=tx}\right)dt\\ &=\frac{B\pi}{2}+2A\int_0^1 \frac{\ln t}{t^4-1}dt-2C\int_0^1\frac{1}{(1+t)(1+t^2)}dt\\ &=\boxed{\frac{B\pi}{2}+\frac{A\pi^2}{8}+A\text{G}-\frac{C\ln 2}{2}-\frac{C\pi}{4}} \end{align} Therefore, \begin{align}\boxed{J=\frac{3A\pi^2}{8}+2A\text{G}-C\pi}\end{align}

\begin{align}A&\overset{u=\frac{1}{x}}=\int_0^\infty\frac{1}{1+u^4}du\\ 2A&=\int_0^\infty\frac{1+\frac{1}{x^2}}{\frac{1}{u^2}+u^2}du=\int_0^\infty\frac{1+\frac{1}{x^2}}{\left(u-\frac{1}{u}\right)^2+2}du\\ &\overset{z=u-\frac{1}{u}}=\int_{-\infty}^{+\infty}\frac{1}{2+z^2}dz\\ A&=\int_{0}^{+\infty}\frac{1}{2+z^2}dz=\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_0^\infty=\frac{\pi}{2\sqrt{2}}\\ C&\overset{u=\frac{1}{x}}=-\int_0^\infty \frac{\ln u}{1+u^4}du\\ R&=\int_0^\infty\int_0^\infty\frac{\ln(tu)}{(1+t^4)(1+u^4)}dtdu\\&\overset{z(u)=tu}=\int_0^\infty\int_0^\infty \frac{t^3\ln z}{(z^4+t^4)(1+t^4)}dtdz\\ &=\frac{1}{4}\int_0^\infty \left[\ln\left(\frac{1+t^4}{z^4+t^4}\right)\right]_{t=0}^{t=\infty} \frac{\ln z}{z^4-1} dz\\ &=\int_0^\infty \frac{\ln^2 z}{z^4-1}dz\\ &=\frac{1}{2}\underbrace{\int_0^\infty \frac{\ln^2 z}{z^2-1}dz}_{=0}-\frac{1}{2}\int_0^\infty \frac{\ln^2 z}{z^2+1} dz\\ &=-\frac{1}{2}\int_0^\infty \frac{\ln^2 z}{z^2+1} dz=-\frac{\pi^3}{16} \end{align}

On the other hand, \begin{align}R&=-2AC\end{align} therefore, \begin{align}C&=-\frac{R}{2A}=\frac{\frac{\pi^3}{16}}{2\times \frac{\pi}{2\sqrt{2}}}=\frac{\pi^2}{8\sqrt{2}}\\ J&=\frac{3\pi^2}{8}\times\frac{\pi}{2\sqrt{2}}+2\text{G}\times\frac{\pi}{2\sqrt{2}}-\pi\times\frac{\pi^2}{8\sqrt{2}}\\ J&=\boxed{\frac{\pi^3}{16\sqrt{2}}+\frac{\pi\text{G}}{\sqrt{2}}} \end{align}

NB: I assume, \begin{align}\int_0^\infty \frac{\ln^2 x}{1+x^2}dx&=\frac{\pi^3}{8}\\ \int_0^\infty \frac{\ln x}{x-1}dx&=\frac{\pi^2}{6}\\ \end{align}

Answer 2

Too long for comments

The more general integral$$I_a=\int_0^\infty\frac{2x^2}{x^4+a^4}\tan^{-1}(x^2)\,dx$$ is quite easy to compute after partial fraction decomposition $$I_a=\int_0^\infty\Bigg[\frac{\tan^{-1}(x^2)}{x^2+i\,a^2}+\frac{\tan^{-1}(x^2)}{x^2-i\,a^2}\Bigg]\,dx$$

$CAS$ provide $$\int_0^\infty \frac{\tan^{-1}(x^2)}{x^2+i\,a^2}=\frac{\sqrt[4]{-1} \pi \left(-\tanh ^{-1}\left(a^2\right)-i \tan ^{-1}(a)+\tanh ^{-1}(a)\right)}{a}$$ $$\int_0^\infty \frac{\tan^{-1}(x^2)}{x^2-i\,a^2}=\frac{\sqrt[4]{-1} \pi \left(i \tanh ^{-1}\left(a^2\right)+\tan ^{-1}(a)-i \tanh ^{-1}(a)\right)}{a}$$

This gives $$\color{blue}{I_a=\frac{\pi\sqrt{2} }{a} \left(\tanh ^{-1}\left(\frac{a}{a^2+a+1}\right)+\tan ^{-1}(a)\right)}$$

Answer 3

As was done in the answer HERE by the user Setness Ramesory, we can use the duplication formula for the dilogarithm to put the integrand into a form that lends itself to an evaluation using contour integration.

The key point, as explained in the linked answer, is that the branch cut for $\operatorname{Li}_{2}(e^{i \pi/4}z)$ is entirely in the lower half-plane, while the branch cut for $\operatorname{Li}_{2}(-e^{i \pi/4}z) $ is entirely in the upper half-plane.

And since $\operatorname{Li}_{2}(z) \sim -\frac{\ln^{2}(-z)}{2}$ as $|z| \to \infty$ (see here), we have

$$ \begin{align} &\int_{0}^{\infty} \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{2}\operatorname{Ti}_{2}(x^{2})}{1+x^{4}} \, \mathrm dx \\ &= \frac{1}{2} \, \Im \int_{-\infty}^{\infty} \frac{x^{2}\operatorname{Li}_{2}(ix^{2})}{1+x^{4}} \, \mathrm dx \\ &= \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(e^{\pi i /4} x)}{1+x^{4}} \, \mathrm dx + \Im \int_{-\infty}^{\infty} \frac{x^{2} \operatorname{Li}_{2}(-e^{\pi i /4} x)}{1+x^{4}} \, \mathrm dx \\ &= \Im \int_{-\infty}^{\infty} f(x) \, \mathrm dx + \Im \int_{-\infty}^{\infty} g(x) \, \mathrm dx \\ &= \small\Im \left( 2 \pi i \left(\operatorname{Res} \left[f(z), e^{\pi i/4}\right] + \operatorname{Res} \left[f(z), e^{3 \pi i/4}\right]\right) \right) - \Im \left( \, 2 \pi i \left(\operatorname{Res} \left[g(z), e^{-\pi i/4}\right] + \operatorname{Res} \left[g(z), e^{-3 \pi i/4}\right]\right) \right) \\ &= \small\Im \left( 2 \pi i \left(\frac{i e^{-3 \pi i/4} \operatorname{Li}_{2}(i)}{4} - \frac{i e^{- \pi i/4}\operatorname{Li}_{2}(-1)}{4} \right)\right)- \Im \left( 2 \pi i \left(- \frac{ie^{3 \pi i /4}\operatorname{Li}_{2}(-1)}{4} + \frac{ie^{\pi i /4} \operatorname{Li}_{2}(i)}{4} \right) \right) \\ &= \Im \left( \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(i) \right) + \Im \left(\frac{\pi i }{\sqrt{2}} \, \operatorname{Li}_{2}(i) \right)- \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &= \frac{\pi }{\sqrt{2}} \, \operatorname{Ti}_{2}(1) + \frac{ \pi}{4 \sqrt{2}} \, \operatorname{Li}_{2}(-1) - \frac{\pi}{\sqrt{2}} \, \operatorname{Li}_{2}(-1) \\ &= \frac{\pi G}{\sqrt{2}} - \frac{\pi^{3}}{48 \sqrt{2}}+ \frac{\pi^{3}}{12 \sqrt{2}} \\ &= \frac{\pi G}{\sqrt{2}} + \frac{\pi^{3}}{16 \sqrt{2}} . \end{align}$$

Answer 4

Just like it was mentioned in the comments, the integrand in question is strictly positive. I don't know how Wolfram Alpha is interpreting your input. (Maybe it is interpreting it as the Generalized Inverse Tangent Function?)

Your $F'(t)$ is correct. I simplified it down to

$$F'(t) = \frac{\pi}{2\sqrt{2}}\frac{\sqrt{t}-1}{\sqrt{t}\left(t^{2}-1\right)}.$$

Using the Fundamental Theorem of Calculus, we can integrate both sides with respect to $t$ on the interval $\left[0,1\right]$ to get

$$\frac{\pi}{\sqrt{2}}\int_{0}^{1}\frac{\sqrt{t}-1}{\sqrt{t}\left(t^{2}-1\right)}dt.$$

Solving this integral should be straightforward. The final answer is

\begin{align} \dfrac{\pi}{2\sqrt{2}}\left(\ln\left(2\right)+\dfrac{\pi}{2}\right). \end{align}

Hopefully, this answer helps. Please let me know if there are any questions.