How do we find the exact value of $\int_{0}^{\infty} \frac{\ln ^{n}\left(1+x^{2}\right)}{1+x^{2}} d x$, where $n\in \mathbb N?$ Latest Edit
By @KStarGamer’s help, I can finally find a reduction formula for $I_n$ as below: $$\boxed{I_n= 2 \ln 2 I_{n-1}+ (n-1)!\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k!}\zeta(n-k)  I_k}
$$
where $n\geq 2.$


In my post, I started to evaluate $$I_1=\int_{0}^{\infty} \frac{\ln \left(1+x^{2}\right)}{1+x^{2}} d x =\pi \ln 2, $$
then I challenge myself on $$I_2=\int_{0}^{\infty} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x^{2}}dx$$
Again, letting $x=\tan \theta$ as for $I_1$ yields$$I_2=\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\sec ^{2} \theta\right) d \theta= 4 \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) dx $$
It’s very hard to deal with $\ln^2$ and I was stuck. Suddenly a wonderful identity came to my mind. $$
2\left(a^{2}+b^{2}\right)=(a+b)^{2}+(a-b)^{2},
\\$$
by which
$\displaystyle 2\left[\ln ^{2}(\sin x)+\ln ^{2}(\cos x)\right]=[\ln (\sin x)+\ln (\cos x)]^{2}+[\ln (\sin x)-\ln (\cos x)]^{2} ,\tag*{}\\ $
we have
$\displaystyle 4 L=\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\frac{\sin 2 x}{2}\right)}_{J} d x+\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x}_{K} \tag*{}\\ $
For the first integral, using $ \int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x=-\dfrac{\pi}{2} \ln 2 $ yields
$ \begin{aligned}J &=\int_{0}^{\frac{\pi}{2}}[\ln (\sin 2 x)-\ln 2]^{2} d x \\&=\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin 2 x) d x-2 \ln 2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x +\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{x\mapsto 2x}{=} \frac{1}{2} \int_{0}^{\pi} \ln ^{2}(\sin x) d x-\ln 2 \int_{0}^{\pi} \ln (\sin x) d x+\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{symmetry}{=} L-\ln 2(-\pi \ln 2)+\frac{\pi \ln ^{2} 2}{2} \\&=L+\frac{3 \pi \ln ^{2} 2}{2}\end{aligned}\tag*{} \\$
For the second integral, letting $ y=\tan x $ and using my post yields
$ \displaystyle K=\int_{0}^{\infty} \frac{\ln ^{2} y}{1+y^{2}} d y=\frac{\pi^{3}}{8}, \tag*{} \\$
then
$ \displaystyle 4L=L+\frac{3 \pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{8} \Rightarrow L=\frac{\pi^{3}}{24}+\frac{\pi \ln ^{2} 2}{2}\tag*{} $
Hence
$ \displaystyle \boxed{I_2=4L= \frac{\pi^{3}}{6}+2 \pi \ln ^{2} 2} \tag*{} $
My Question: Can I go further with $I_n$, where $n\geq 3$?
 A: With differentiation of $$\frac2\pi\int_{0}^{\frac\pi2} (2\cos t)^adt = \frac{\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}$$
in $a$, it can be shown that $J_k=\frac2\pi \int_{0}^{\frac\pi2}\ln^k(2\cos t)dt$ are the power-series coefficients of the exponential function below
$$
\exp\bigg( \sum_{k=1}^\infty \frac{(-1)^k}k (1-2^{1-k})\zeta(k)x^k\bigg)= \sum_{k=0}^\infty \frac{J_k}{k!}x^k
$$
which results in
\begin{align}
J_0 &= 1,\>\>\>\>\>
J_1= 0, \>\>\>\>\>
J_2=\frac12 \zeta(2),\>\>\>\>\>
J_3= -\frac32 \zeta(3),\>\>\>\>\>
J_4= \frac{57}{8} \zeta(4) \\
J_5 &= -\frac{15}2\zeta(2)\zeta(3) -\frac{45}2 \zeta(5),\>\>\>\>\>
J_6= \frac{45}2 \zeta^2(3)+ \frac{12375}{64} \zeta(6),\>\>\>\>\>
J_7= \cdots
\end{align}
Now, note that
\begin{align}
I_n & = \int_{0}^{\infty} \frac{\ln ^{n}(1+x^{2})}{1+x^{2}} d x\\
&= \int_{0}^{\frac{\pi}{2}} \ln ^{n}\sec ^{2} t\> d t= 2^{n-1}\pi\sum_{k=0}^n \binom nk (-1)^{k} \ln^{n-k}(2)\>J_{k}
\end{align}
Thus
\begin{align}
I_1 &= \pi\ln2 \\
I_2 &
= \pi [\zeta(2) +2\ln^22]\\
I_3
&=\pi [6\zeta(3) +6\ln2\> \zeta(2) +4\ln^32] \\
I_4
&= \pi[57\zeta(4)+48\ln2\>\zeta(3)+24\ln^22\>\zeta(2)+8\ln^42]\\
I_5
&= \pi[120\zeta(2)\zeta(3)+360\zeta(5)+570\ln2\>\zeta(4)\\
&\hspace{2cm}+240\ln^22\>\zeta(3)+80\ln^32\>\zeta(2)+16\ln^52]\\
I_6&=\cdots
\end{align}
A: Taking inspiration from the answer to this similar question.
$$\int_{0}^{\infty} (1+x^2)^s \, dx = \frac{\sqrt{\pi} \, \Gamma \left(-\frac{2s+1}{2}\right)}{2 \Gamma \left(-s\right)}$$
Valid for $\Re (s) < -\frac{1}{2}$.
Differentiating twice with respect to $s$ and taking the limit as $s \to -1$ gives the desired result. Differentiating once also gives the answer to your previous question.
Differentiate $n$ times to determine $I_n$.
