After knowing that $$
I_2=\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^{2}} d x \stackrel{x\mapsto\frac{1}{x}}{=} -\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^{2}} d x \Rightarrow \int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^{2}} d x=0
$$
I start to investigate the integrals with higher powers
$$I_n=\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^{n}} d x $$
where $n$ is a natural number greater than $2$.
We first split the interval into two as
$$
\begin{aligned}
\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^3} d x & =\int_0^1 \frac{\ln ^3 x}{(1+x)^3} d x+\int_1^{+\infty} \frac{\ln ^3 x}{(1+x)^3} d x \\
& =\int_0^1 \frac{\ln ^3 x}{(1+x)^3} d x+\int_1^0 \frac{\ln ^3\left(\frac{1}{x}\right)}{\left(1+\frac{1}{x}\right)^3} \frac{d x}{-x^2} \\
& =\int_0^1 \frac{\ln ^3 x}{(1+x)^3} d x-\int_0^1 \frac{x \ln ^3 x}{(1+x)^3} d x
\end{aligned}
$$
For any $|x|<1$, we have $$
\frac{1}{1+x}=\sum_{k \rightarrow 0}^{\infty}(-1)^k x^k
$$
Differentiating both sides w.r.t. $x$ twice yields
$$
\frac{1}{(1+x)^3}=\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k(k+2)(k+1) x^k
$$
Plugging into the integrand, we have $$
\int_0^{\infty} \frac{\ln ^3 x}{(1+x)^3}dx=\frac{1}{2}\left[\sum_{k=0}^{\infty}(-1)^k(k+2)(k+1)\left( \int_0^1 x^k \ln ^3 x d x-\int_0^1 x^{k+1} \ln ^3 x d x \right)\right]
$$
Noting that
$$
\begin{aligned}
\int_0^1 x^n \ln ^3 x d x & =\left.\frac{\partial^3}{\partial a^3} \int_0^1 x^a d x\right|_{x=w} \\
& =\left.\frac{\partial^3}{\partial a^3}\left(\frac{1}{a+1}\right)\right|_{x=n} \\
& =-\frac{6}{(n+1)^4}
\end{aligned}
$$
$$
\begin{aligned}I_3&=\frac{1}{2} \sum_{k=0}^{\infty}(-1)^k(k+2)(k+1)\left[-\frac{6}{(k+1)^4}+\frac{6}{(k+2)^4}\right]\\&= 3 \left[-\sum_{k=0}^{\infty} \frac{(-1)^k(k+2)}{(k+1)^3}+\sum_{k=0}^{\infty} \frac{(-1)^k(k+1)}{(k+2)^3}\right]\\&= -3\left[\sum_{k=0}^{\infty} \frac{(-1)^k(k+2)}{(k+1)^3}+\sum_{k=1}^{\infty} \frac{(-1)^kk}{(k+1)^3}\right]\\&=-3 \sum_{k=0}^{\infty} \frac{(-1)^k(2 k+2)}{(k+1)^3}\\&=-6 \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}\\&=-\frac{\pi^2}{2}\end{aligned}
$$
Let’s continue with $I_4$ by considering their difference
$$
\begin{aligned}
D & =\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^3} d x -\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^4} d x \\
& =\int_0^{+\infty} \frac{x \ln ^3 x}{(1+x)^4} d x \\
& =\int_0^{+\infty} \frac{\frac{1}{x} \ln ^3 \frac{1}{x}}{\left(1+\frac{1}{x}\right)^4} \frac{d x}{x^2} \\
& =-\int_0^{+\infty} \frac{x \ln ^3 x}{(1+x)^4} d x \\
& =-D\\\Rightarrow D&=0
\end{aligned}
$$
Hence $$\boxed{I_4=I_3=-\frac{\pi^2}{2} }$$
By Wolfram-alpha, we get
$$
\begin{aligned}
& I_5=-\frac{1}{24}\left(6+11 \pi^2\right) \\
& I_6=-\frac{1}{12}\left(6+5 \pi^2\right) \\
& I_7=-\frac{1}{360}(255+137 \pi^2)
\end{aligned}
$$
I guess that in general, $$
I_n=A+B \pi^2 \textrm{ for some rational numbers } A \textrm{ and }B.
$$
Can we prove it further? Your help or suggestions are highly appreciated.
Best Answer
By Mathematical Induction
I am going to prove, by Mathematical Inductio,that $$P(n):I_n= \int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^n} d x=A_n+B_n\pi^2,$$ where $n\ge 2 , A_n$ and $B_n$ are rational.
First of all, we have $$I_2=0, I_4=I_3=-\frac{\pi^2}{2} \tag*{} $$ Therefore $P(2), P(3)$ and $P(4)$ are true.
Now assume that $P(n)$ are true for all $2\le n \le k-1$.
When $n=k$, letting $x\mapsto \frac{1}{x} $ yields $$ \begin{aligned} I_{k}&=-\int_0^{+\infty} \frac{x^{k-2} \ln ^3 x}{(1+x)^{k}} d x\\&=\int_0^{+\infty} \frac{(1+x-1)^{k-2}\ln^3x}{(1+x)^{k}}dx\\ & =-\sum_{j=0}^{k-2}\left(\begin{array}{c} k-2 \\ j \end{array}\right)(-1)^{k-2-j}\int_0^{+\infty} \frac{\ln ^3 x}{(1+x)^{k-j} }d x\\& =-I_{k}+\sum_{j=1}^{k-2}\left(\begin{array}{c} k-2 \\ j \end{array}\right)(-1)^{k-1-j} I_{k-j}\\&= \frac{1}{2} \sum_{j=1}^{k-2}\left(\begin{array}{c} k-2\\ j \end{array}\right)(-1)^{k-1-j} I_{k-j}\\&= \frac{1}{2} \sum_{j=1}^{k-2}\left(\begin{array}{c} k-2\\ j \end{array}\right)(-1)^{k-1-j} (A_{k-j}+B_{k-j} \pi^2) \quad \textrm{ (By Ind. Hypo.)} \\&= A_{k}+B_{k} \pi^2, \end{aligned} $$ where $ \displaystyle A_k= \frac{1}{2} \sum_{j=1}^{k-2}\left(\begin{array}{c} k-2\\ j \end{array}\right)(-1)^{k-1-j} A_{k-j} $ and $ \displaystyle B_k=\frac{1}{2} \sum_{j=1}^{k-2}\left(\begin{array}{c} k-2\\ j \end{array}\right)(-1)^{k-1-j} B_{k-j}$ are rational too and hence $P(k)$ is also true.
By the principle of Mathematical Induction, we can conclude that for all natural numbers $n$, $$ \boxed{ I_n=A_n+B_n \pi^2} \tag*{} $$