Find close form for $\int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2}dx$

Question

I am trying to find a closed form for the integral,$$\int_0^1 \frac{\log(x)\log(1+x^2)}{1+x^2}dx$$ I tried using the sub, $x=\frac{1-x}{1+x}$ but it was to no avail. I also tried the trig sub, $x=\tan(y)$ but was stuck with the integral, $$\int_0^{\pi/4} \log^2(\sin(y))dy$$ I need help with any of the two integrals.

Answer 1

\begin{gather*} \int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\ =\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\underbrace{\int_1^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x}_{x\to 1/x}\\ =\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\ +2\int_0^1\frac{\ln^{2a+1}(x)}{1+x^2}\mathrm{d}x\\ \left\{\text{add $\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x$ to both sides then divide by 2}\right\}\\ =\frac12\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\ln^{2a+1}(x)}{1+x^2}\mathrm{d}x \end{gather*} By the definition of Dirichlet beta function, the second integral is

$$\int_0^1\frac{\ln^{2a+1}(x)}{1+x^2}\mathrm{d}x=-(2a+1)!\beta(2a+2).$$

To get the first integral, we need following result

$$\int_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x$$ $$=2{\pi^{2a+1}}\ln2|E_{2a}|+(2a)!{\pi^{2a+1}}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1)$$

which was found by @user178256 here and I managed to provide a rigorous proof:

we follow the same technique as in here. Replace $n$ by $n-m$ in the beta function in: \begin{equation*} \int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\mathrm{d}x=\operatorname{B}(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}, \end{equation*} we get \begin{equation*} \int_0^\infty\frac{x^{m-1}}{(1+x)^{n}}\mathrm{d}x=\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}. \end{equation*} Differentiate $2a$ times with respect to $m$ and once with respect to $n$, \begin{equation*} \int\limits_0^\infty \frac{\ln^{2a}(x)x^{m-1}\ln(1+x)}{(1+x)^n}\mathrm{d}x=-\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}. \end{equation*}

Now take the limit on both sides letting $m\to 1/2$ and $n\to1$,

\begin{gather*} \int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\lim_{\substack{m\to 1/2\\n\to 1}}\frac{\partial^{2a}}{\partial m^{2a}} \operatorname{\Gamma}(m)\frac{\partial}{\partial n} \frac{\operatorname{\Gamma}(n-m)}{\operatorname{\Gamma}(n)}\\ \{\text{use $\Gamma'(x)=\Gamma(x)\psi(x)$}\}\\ =-\lim_{m\to 1/2}\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\left(\lim_{n\to 1}\frac{\Gamma(n-m)[{\psi}(n-m) -\psi(n)]}{\Gamma(n)}\right)\\ \{\text{use $\psi(1)=-\gamma$ }\}\\ =-\lim_{m\to 1/2 }\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\Gamma(1-m)[\psi(1-m) + \gamma]\\ \left\{\text{use $\Gamma(m)\Gamma(1-m)=\pi \csc(m\pi)$ }\right\}\\ =-\pi\lim_{m\to \frac12 }\frac{d^{2a}}{dm^{2a}}\csc(m\pi)\, (\psi(1-m)+\gamma).\\ \left\{\text{use $\frac{d^a}{dm^a}(f*g)=\sum_{k=0}^a \binom{a}{k} \frac{d^{a-k}}{dm^{a-k}} f*\frac{d^k}{dm^k}g$}\right\}\\ =-\pi\sum_{k=0}^{2a} \binom{2a}{k}\lim_{m\to \frac12}\frac{d^{2a-k}}{dm^{2a-k}} \csc(m\pi) \lim_{m\to \frac12}\frac{d^{k}}{dm^k} (\psi(1-m)+\gamma)\\ \left\{\text{use $\sum_{k=0}^{2a}f(k)=\sum_{k=0}^a f(2k+1)+\sum_{k=0}^a f(2k)$}\right\}\\ =-\pi\sum_{k=0}^{a} \binom{2a}{2k+1}\lim_{m\to \frac12}\frac{d^{2a-2k-1}}{dm^{2a-2k-1}} \csc(m\pi)\, \lim_{m\to \frac12}\frac{d^{2k+1}}{dm^{2k+1}} (\psi(1-m)+\gamma)\\ =-\pi\sum_{k=0}^{a} \binom{2a}{2k}\lim_{m\to \frac12}\frac{d^{2a-2k}}{dm^{2a-2k}} \csc(m\pi)\, \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)\\ \left\{\text{ignore first sum since $\lim_{m\to \frac12}\frac{d^{a}}{dm^{a}}\csc(m\pi)=0$ for odd $a$}\right\}\\ =-\pi\sum_{k=0}^{a} \binom{2a}{2k}|E_{2a-2k}|\pi^{2a-2k}\, \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)\\ \{\text{separate the first term using $\psi(1/2)+\gamma=-2\ln(2)$}\}\\ =2\ln(2)|E_{2a}|\pi^{2a+1}\\ -\pi\sum_{k=1}^{a} \binom{2a}{2k}|E_{2a-2k}|\pi^{2a-2k}\, \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)\\ \{\text{use the definition of the polygamma function}\}\\ =2\ln(2)|E_{2a}|\pi^{2a+1}\\ -\pi\sum_{k=1}^{a} \binom{2a}{2k}|E_{2a-2k}|\pi^{2a-2k} \psi^{(2k)}\left(\frac12\right)\\ =2\ln(2)|E_{2a}|\pi^{2a+1}\\ +\pi\sum_{k=1}^{a} \binom{2a}{2k}|E_{2a-2k}|\pi^{2a-2k}\, (2k)!(2^{2k+1}-1)\zeta(2k+1), \end{gather*}

Setting $\sqrt{x}=y$ in the integral gives \begin{equation*} \int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=2^{2a+1}\int_0^\infty \frac{\ln^{2a}(y)\ln(1+y^2)}{1+y^2}\mathrm{d}y. \end{equation*}

Thus,

\begin{gather} \int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x=-(2a+1)!\beta(2a+2)+\ln(2)|E_{2a}|\left(\frac{\pi}{2}\right)^{2a+1}\nonumber\\ +\frac{\left(\frac{\pi}{2}\right)^{2a+1}}{2}\sum_{k=1}^{a} \binom{2a}{2k}(2k)!|E_{2a-2k}|\pi^{-2k}(2^{2k+1}-1)\zeta(2k+1). \end{gather}

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Used results:

\begin{equation} \lim_{z\to \frac12}\frac{d^{2a}}{dz^{2a}}\csc(z\pi)=|E_{2a}|\pi^{2a} \end{equation} where $E_r$ is the Euler number.

\begin{equation} \psi^{(a)}\left(\frac12\right)=(-1)^{a}a!(1-2^{a+1})\zeta(a+1). \end{equation}

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Answer 2

Using Feynman's trick $$I(a)=\int_0^1 \frac{\log(x)\log(1+ax^2)}{1+x^2}dx$$ $$I'(a)=\int_0^1 \frac{x^2 \log (x)}{\left(x^2+1\right) \left(a x^2+1\right)} \,dx$$ $$I'(a)=\frac{C}{1-a}-\frac{i \left(\text{Li}_2\left(i \sqrt{a}\right)-\text{Li}_2\left(-i \sqrt{a}\right)\right)}{2 (a-1) \sqrt{a}}$$ which, integrated for $a$ between $0$ and $1$, leads to what @David G. Stork gave in comments.

Now, for an approximation, you could use the value of $I'(a)$ and its derivatives at the boudaries and obtain $$I'(0)=C-1\qquad \qquad I'(1)=\frac{\pi }{8}-\frac{C}{2}$$ $$I''(0)=C-\frac{8}{9}\qquad \qquad I''(1)=\frac{3 C}{8}+\frac{1}{16}-\frac{\pi }{8}$$ $$I'''(0)=2 C-\frac{418}{225}\qquad \qquad I'''(1)=-\frac{5 C}{8}-\frac{3}{16}+\frac{23 \pi }{96}$$ and build the polynomial $$I'(a)=\sum_{n=1}^5 b_n \,a^n$$ $$(C-1)+ \left(C-\frac{8}{9}\right)a+ \left(C-\frac{209}{225}\right)a^2+\frac{ (-123900 C+85326+8975 \pi )}{4800}a^3+$$ $$\frac{ (264600 C-174764-21525 \pi )}{7200}a^4+\frac{ (-207900 C+134126+17925 \pi )}{14400}a^5$$ which, integrated, leads to $$I(1)=\frac{311 C}{960}-\frac{264463}{432000}+\frac{887 \pi }{11520}=-0.0735565$$ to be compared to the value of $-0.0735540$ given by @David G. Stork' expression.

This could be much improved increasing the degree of the polynomial approximant.

For example, pushing the expansion to degree $9$ would give $$I(1)=\frac{27697 C}{80640}-\frac{10258077467}{16003008000}+\frac{222527 \pi }{2764800}=-0.0735539700$$ to be compared to the exact $-0.0735539567$

This is a kind of two point Taylor series.