Prove $ \int_0^{\infty} \frac{\ln ^3 x}{(1+x)^{n}} d x =A+B \pi^2$ for some rational numbers $A$ and $B$, where $n\neq 1 $

Question

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.

Answer 1

By differentiation

We may start our journey with the integral $$ I(a)=\int_0^{+\infty} \frac{\ln ^3 x}{(a+x)^2} d x $$ Using substitution $ax\mapsto x$ yields $$ \begin{aligned} I(a) & =\frac{1}{a} \int_0^{+\infty} \frac{\ln ^3(a x)}{(1+x)^2} d x \\ & =\frac{1}{a} \int_0^{\infty} \frac{\ln ^3 a+3 \ln ^2 a\ln x+3 \ln a \ln ^2 x+\ln ^3 x}{(1+x)^2} dx \end{aligned} $$ Noting that $\int_0^{+\infty} \frac{\ln ^{2n+1} x}{(1+x)^2} d x=0$, we have $$ I(a)=\frac{\ln ^3 a}{a} \underbrace{ \int_0^{\infty} \frac{1}{\left(1+x\right)^2}d x}_{=1} + \frac{3 \ln a}{a} \underbrace{ \int_0^{\infty} \frac{\ln ^2 x}{(1+x)^2} d x}_{\frac{\pi^2}{3} } =\frac{\ln ^3 a}{a}+\frac{\pi^2 \ln a}{a} $$ By differentiating $I(a)$ w.r.t. $a$ by $n-2$ times, we get

$$\boxed{I_n=\left.\frac{(-1)^n}{(n-1) !} \frac{\partial^{n-2}}{\partial a^{n-2}} \left(\frac{\ln ^3 a}{a}+\frac{\pi^2 \ln a}{a} \right)\right|_{a=1}=A_n+B_n\pi^2 } $$ where $ \left. \displaystyle A_n= \frac{(-1)^n}{4(n-1) !} \frac{\partial^{n-1}}{\partial a^{n-1}} \left(\ln ^4 a\right) \right|_{a=1} $ and $\displaystyle \left. B_n= \frac{(-1)^n}{2(n-1) !} \frac{\partial^{n-1}}{\partial a^{n-1}} \left(\ln^2 a\right) \right|_{a=1} $ are rational.

For examples, $$ \left. I_5=-\frac{1}{24} \frac{\partial^4}{\partial a^4}\left(\frac{\ln ^4 a}{4}+\frac{ \pi^2 \ln^2 a}{2} \right)\right|_{a=1} =-\frac{1}{24}\left(6+11 \pi^2\right) $$ $$I_6=\left.\frac{1}{120} \frac{\partial^5}{\partial a^5}\left(\frac{\ln ^4 a}{4}+\frac{\pi^2 \ln^2 a}{2}\right)\right|_{a=1}= -\frac{1}{12}\left(6+5 \pi^2\right)$$

Answer 2

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 $n$, $$ \boxed{ I_n=A_n+B_n \pi^2} \tag*{} $$

Answer 3

Using the results in one of the answers, the coefficients $A_n$ and $B_n$ may be expressed in terms of the Nörlund polynomials. Indeed \begin{align*} A_n & = \frac{{( - 1)^n }}{{(n - 1)!}}\left[ {\frac{{{\rm d}^{n - 2} }}{{{\rm d}z^{n - 2} }}\frac{{\log ^3 z}}{z}} \right]_{z = 1} = \frac{{( - 1)^n }}{{n - 1}}\frac{1}{{2\pi {\rm i}}}\oint_{(1 + )} {\frac{{\log ^3 z}}{z}\frac{{{\rm d}z}}{{(z - 1)^{n - 1} }}} \\ & = \frac{{( - 1)^n }}{{n - 1}}\frac{1}{{2\pi {\rm i}}}\oint_{(0 + )} {\frac{{t^3 }}{{({\rm e}^t - 1)^{n - 1} }}{\rm d}t} = \frac{{( - 1)^n }}{{n - 1}}\frac{1}{{2\pi {\rm i}}}\oint_{(0 + )} {\left( {\frac{t}{{{\rm e}^t - 1}}} \right)^{n - 1} \frac{{{\rm d}t}}{{t^{n - 4} }}} \\ & = \frac{{( - 1)^n }}{{n - 1}}\frac{{B_{n - 5}^{(n - 1)} }}{{(n - 5)!}}, \end{align*} provided $n\ge 5$. In a similar manner $$ B_n = \frac{{( - 1)^n }}{{n - 1}}\frac{{B_{n - 3}^{(n - 1)} }}{{(n - 3)!}}, $$ provided $n\ge 3$. Also $A_2=A_3=A_4=0$ and $B_2=0$. To leading order $$ A_n \sim - \frac{{\log ^3 n}}{n},\quad B_n \sim - \frac{{\log n}}{n} $$ as $n\to+\infty$ (cf. Section $3.3$ in this paper).

Answer 4

Note that \begin{align} &\int_0^{\infty} \frac{\ln ^3 x}{(1+x)^n} d x\\ =&\ \frac{d^3}{da^3} \bigg(\int_0^{\infty} \frac{x^a}{(1+x)^n} d x\bigg)_{a=0}= \frac{d^3}{da^3}\bigg( \frac{\pi a\csc\pi a}{(n-1)!} \prod_{k=1}^{n-2}(k-a)\bigg)_{a=0}\\ =& \ \frac1{n-1}\bigg( 3a_1a_2-a_1^3-2a_3 -a_1\pi^2 \bigg) =A+B\pi^2 \end{align} where $a_i=\sum_{k=1}^{n-2}\frac1{k^i}$. Explicitly \begin{align} &A=-\frac1{n-1}\left(\bigg( \sum_{k=1}^{n-2}\frac1{k}\bigg)^3+2 \sum_{k=1}^{n-2}\frac1{k^3} -3\sum_{k=1}^{n-2}\frac1{k} \cdot\sum_{k=1}^{n-2}\frac1{k^2} \right)\\ &B= -\frac1{n-1}\sum_{k=1}^{n-2}\frac1{k} \end{align}