How can we show that $\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\mathrm{d}t=\frac{\pi}{2a^2}\cdot\ln^2(1+a)$ without complex analysis?

Question

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\Res}{\operatorname{Res}}\newcommand{\Arg}{\operatorname{Arg}}$Consider the function: $$\begin{align}I:(0,\infty)&\to(0,\infty)\\a&\mapsto\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t\end{align}$$

Using complex analysis we can evaluate this integral:

Let $R\gt a$ be any real number. Note that $\int_0^R\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=\frac{1}{2}\int_{-R}^R\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=:\frac{1}{2}J_R(a)$ by evenness.

Define the complex contour $\gamma_R$ to be the union of the line segment $[-R,R]$ and the arc $[-\pi,0]\ni t\mapsto Re^{it}\in\Bbb C$, $C_R$. We may rewrite (with the principal branch of the logarithm, modified to $\Arg(z)\in[-\pi,\pi)$): $$\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}=\Im\left[\frac{\ln^2(1+it)}{t(t^2+a^2)}\right]=:\Im(f(t))$$Then: $$\begin{align}\left|\int_{C_R}f(z)\d z\right|&\le\int_0^\pi\frac{\ln^2(|1+iRe^{it}|)+\Arg^2\left(\frac{R\cos(t)}{1-R\sin(t)}\right)}{|(Re^{it})^2+a^2|}\d t\\&\le\frac{1}{4}\int_0^\pi\frac{\ln^2(1+R^2-2R\sin(t))+4\pi^2}{R^2-a^2}\d t\\&\le\frac{\pi}{4}\cdot\frac{\ln^2(1+R^2)+4\pi^2}{R^2-a^2}=:\varepsilon_R\end{align}$$Due to the clockwise orientation, and the fact that there is only one enclosed pole at $z=-ia$, we find: $$\begin{align}\int_{\gamma_R}f(z)\d z&=-2\pi i\cdot\Res_{z=-ia}f(z)\\&=-2\pi i\cdot\frac{\ln^2(1+a)}{-2a^2}\\&=i\cdot\frac{\pi}{a^2}\cdot\ln^2(1+a)\end{align}$$Then, taking imaginary parts: $$\begin{align}0\le\frac{1}{2}\lim_{R\to\infty}\left|J_R(a)-\frac{\pi}{a^2}\cdot\ln^2(1+a)\right|&\le\frac{1}{2}\lim_{R\to\infty}\varepsilon_R=0\\\left|I(a)-\frac{\pi}{2a^2}\cdot\ln^2(1+a)\right|&=0\end{align}$$And we can conclude: $$I(a)=\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=\frac{\pi}{2a^2}\cdot\ln^2(1+a),\,a\gt0$$

I am interested in ways to evaluate this integral using real-analytic techniques. I had two ideas (well, tips from someone else...) involving differentiating under the integral sign with $(1+t^2)^s$, to produce $\ln(1+t^2)$, and letting also $s\to-1$ to obtain the $\arctan$ term via integration by parts. That failed quite miserably... A different attempt, also using differentiating under the integral sign, was a bit more successful:

Put $I(a,b)=\int_0^\infty\frac{\arctan(bt)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t$. Then: $$\begin{align}\frac{\partial}{\partial b}I(a,b)&=\int_0^\infty\frac{\ln(1+t^2)}{(1+b^2t^2)(t^2+a^2)}\d t\\&=\frac{1}{a^2-\frac{1}{b}}\left[\int_0^\infty\frac{\ln(1+t^2)}{1+b^2t^2}\d t-\frac{1}{b^2}\int_0^\infty\frac{\ln(1+t^2)}{t^2+a^2}\d t\right]\\&=\frac{1}{a^2-\frac{1}{b}}\left[\frac{1}{b}\int_0^\infty\frac{\ln(b^2+t^2)-2\ln(b)}{1+t^2}\d t-\frac{\pi}{ab^2}\cdot\ln(1+a)\right]\\&=\frac{\pi}{a^2b-1}\cdot\ln(1+1/b)-\frac{\pi}{a^3b^2-ab}\cdot\ln(1+a)\end{align}$$And clearly $I(0)=0$, so we obtain (abusing notation slightly): $$I(a,1)=\pi\int_0^1\frac{\ln(1+1/x)}{a^2x-1}\d x-\frac{\pi}{a}\cdot\ln(1+a)\int_0^1\frac{1}{a^2x^2-x}\d x$$

And all numerical experiments suggests that the above equation is wrong.

I don't know what went wrong with my real analytic approach, hopefully someone can point it out! I used this result.

Question: Can we evaluate $I$ in any (hopefully clean...) way without using complex analysis? Motivation: I need to learn to get better at "normal" integration :)

Pointing out what went wrong with my approach is just a P.S., it is not my main interest.

Answer 1

Utilize the integral \begin{align} &\int_0^\infty\frac{\ln x}{t^2+(x+1)^2}\overset{x\to\frac{1+t^2}x} {dx}\\ =&\int_0^\infty \frac{\ln(1+t^2)-\ln x}{t^2+(x+1)^2}dx =\frac12 \int_0^\infty\frac{\ln (1+t^2)}{t^2+(x+1)^2}dx =\frac1{2t}\tan^{-1}t\ln(1+t^2) \end{align} to integrate

\begin{align} &\int_0^\infty\frac{\tan^{-1}t\ln(1+t^2)}{t(t^2+a^2)}dt\\ =& \ 2\int_0^\infty \int_0^\infty \frac{\ln x}{(t^2+a^2)(t^2+(x+1)^2)}dxdt =\frac{\pi}a \int_0^\infty \frac{\ln x}{a(x+1)+(x+1)^2} \overset{x\to \frac{1+a}x}{dx}\\ =& \ \frac{\pi}a \int_0^\infty \frac{\ln(1+a)-\ln x}{a(x+1)+(x+1)^2}dx = \frac{\pi \ln(1+a)}{2a} \int_0^\infty \frac{1}{a(x+1)+(x+1)^2}dx\\ =& \ \frac{\pi \ln(1+a)}{2a} \cdot \frac{\ln(1+a)}{a} = \frac{\pi}{2a^2}\ln^2(1+a) \end{align}