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Recently I got interested in Logarithmic Integrals from this $\int_{0}^{1}\frac{\ln^4\left(1-x^{2}\right)}{1+x}dx$ and sought out to find Higher Power variations.

Let, $$I_n=\int_{0}^{1}\frac{\ln^n\left(1-x^{2}\right)}{1+x}dx$$

Consider the following expression : $$E=\zeta(\alpha_1)\zeta(\alpha_2)...\zeta(\alpha_n)\ln^\alpha2$$ whose weight is $\alpha_1+\alpha_2+...+\alpha_n+\alpha.$

$$\boxed{\text{Conjecture: All $I_n$ can be written as a Linear Combination involving Expression $E$ of weight $n+1$.}}$$

Here are the first few examples :

$$I_1=-\frac{1}{2}\zeta\left(2\right)+\ln^{2}2$$

$$I_2=2\zeta\left(3\right)-2\zeta\left(2\right)\ln 2+\frac{4}{3}\ln^32$$

$$I_3=-\frac{27}{4}\zeta\left(4\right)+12\zeta\left(3\right)\ln2-6\zeta\left(2\right)\ln^22+2\ln^42$$

$$I_4=72ζ\left(5\right)-24ζ\left(3\right)ζ\left(2\right)-54ζ\left(4\right)\ln2+48ζ\left(3\right)\ln^{2}2-16ζ\left(2\right)\ln^{3}2+\frac{16}{5}\ln^{5}2$$

$$I_5=-\frac{1185}{4}ζ\left(6\right)+120ζ^{2}\left(3\right)+720ζ\left(5\right)\ln2-240ζ\left(3\right)ζ\left(2\right)\ln2-270ζ\left(4\right)\ln^{2}2+160ζ\left(3\right)\ln^{3}2-40ζ\left(2\right)\ln^{4}2+\frac{16}{3}\ln^{6}2$$

$$I_6=6480ζ\left(7\right)-2160ζ\left(5\right)ζ\left(2\right)-1620ζ\left(4\right)ζ\left(3\right)-3555ζ\left(6\right)\ln2+1440ζ^{2}\left(3\right)\ln2+4320ζ\left(5\right)\ln^{2}2-1440ζ\left(3\right)ζ\left(2\right)\ln^{2}2-1080ζ\left(4\right)\ln^{3}2+480ζ\left(3\right)\ln^{4}2-96ζ\left(2\right)\ln^{5}2+\frac{64}{7}\ln^{7}2$$

Could the Conjecture be Proved or Disproved.
I had made a Similar Conjecture for another Logarithmic Integral of Similar Form but it stopped working from Weight $8$.

Miracle Invoker
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  • Observe that this kind of integral is a sum of terms of "weight" equals to $n+1$. The weight of a product is the sum of weigth and $\zeta(n)$ has weight $n$, $\ln 2$ has weight $1$, $\pi$ has weight $1$,$\text{G}$ has weight $2$.... – FDP Oct 21 '23 at 09:07
  • @FDP How do we know which kind of Integral will have this pattern, also can we prove that it will always be such a sum. By the way how did you come to the conclusion of $G$ having weight $2$. – Miracle Invoker Oct 21 '23 at 10:25
  • Empirically, $\displaystyle \int_0^1 R(x)\prod_{i=1}^n(\ln(P_i(x)))^{n_i}dx$ $R$ rational function, $P_i$ rational function,,has "weight",$\displaystyle \boxed{1+\sum_{k=1}^n n_i}$. $\displaystyle \text{G}=-\int_0^1 \frac{\ln x}{1+x^2}dx$, its "weight" is $1+1=2$. $\displaystyle\pi=\int_0^1 \frac{4}{1+x^2}dx$, its "weight" is $0+1=1$ – FDP Oct 21 '23 at 11:09

2 Answers2

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Your integral can be found by using Cornel's method ( page $80$ of his first book Almost Impossible Integral , Sums, and Series):

$$\int_0^1 \frac{\ln^n(1-x^2)}{1+x}dx=\int_0^1(1-x)\frac{\ln^n(1-x^2)}{1-x^2}\ dx\overset{x^2=t}{=}\frac12\int_0^1\frac{1-\sqrt{t}}{\sqrt{t}}.\frac{\ln^n(1-t)}{1-t}\ dt$$

$$ \overset{IBP}{=}-\frac1{4(n+1)}\int_0^1 t^{-3/2}\ln^{n+1}(1-t)\ dt=-\frac{1}{4(n+1)}\lim_{y\to 1}\frac{\partial^{n+1}}{\partial y^{n+1}}\text{B}\left(-\frac 12,y\right)$$

Ali Shadhar
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Let’s continue Ali Shadhar’s journey starting with $$ I_n=-\frac{1}{4(n+1)} \lim _{y \rightarrow 1} \frac{d^{n+1}}{d y^{n+1}} B\left(-\frac{1}{2}, y\right)= -\frac{1}{4(n+1)} B^{(n+1)}\left(-\frac12, 1\right) $$ We need to investigate the high derivative of the Beta function. \begin{aligned} B\left(-\frac{1}{2}, y\right) & =\frac{\Gamma\left(-\frac{1}{2}\right) \Gamma(y)}{\Gamma\left(y-\frac{1}{2}\right)} =-2 \sqrt{\pi} \frac{\Gamma(y)}{\Gamma\left(y-\frac{1}{2}\right)} \end{aligned} By logarithmic differentiation, we have $$ B^{\prime}\left(-\frac{1}{2}, y\right)=B\left(-\frac{1}{2} ,y\right)\left[\psi(y)-\psi\left(y-\frac{1}{2}\right)\right] $$ Then we apply Leibniz’s Rule to $n^{th}$ derivative of $B’\left(-\frac{1}{2}, y\right)$ and get $$ \begin{aligned} B^{(n+1)}\left(-\frac{1}{2}, y\right) & =\left[B\left(-\frac12, y\right)\left(\psi(y)-\psi\left(y-\frac{1}{2}\right)\right)\right]^{(n)} \\ & =\sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right) B^{(n-k)}\left(-\frac12, y\right)\left[\psi^{(k)}(y)-\psi^{(k)}\left(y-\frac{1}{2}\right)\right] \end{aligned} $$ Taking limit $y$ to $1$ yields $$ B^{(n+1)}\left(-\frac12,1\right)=\sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right) B^{(n-k)}\left(-\frac12, 1\right)\left[\psi^{(k)}(1)-\psi^{(k)} \left(\frac{1}{2}\right)\right] $$ We can now hopefully prove the conjecture of OP by Mathematical Induction if we can prove that $$ \psi^{(k)}(1)-\psi^{(k)}\left(\frac{1}{2}\right) $$ is a linear combination $L_k$ of $E$.

Would you like to lend me a hand?

Latest edit after being reminded by @Mariusz Iwaniuk

By my post, I had proved that$$ \psi^{(n)}(1)-\psi^{(n)}\left(\frac{1}{2}\right)=(-1)^{n} n !\left(2^{n+1}-2\right)\zeta(n+1) $$ Therefore $$ B^{(n+1)}\left(-\frac12,1\right)=\sum_{k=0}^n\left(\begin{array}{l} n \\ k \end{array}\right) B^{(n-k)}\left(-\frac12, 1\right) (-1)^{n} n !\left(2^{n+1}-2\right)\zeta(n+1) $$ Now it is sufficient to prove the conjecture by Mathematical Induction on $$P(n):B^{(n)}\left(-\frac12,1\right) \textrm{ is a linear combination } L_k \textrm{ of } E up to \ln^{n+1} 2, \quad \textrm{ where } n\in \mathbb{N}. $$

$P(1)$ is obviously true since $B^{(1)}\left(-\frac12,1\right) =-4\ln 2.$

Now assume that $P(k) $ are true for $k=1,2,\cdots,m,$ then $$ \begin{aligned} B^{(m+1)}\left(-\frac{1}{2}, 1\right)= & \sum_{k=0}^m\left(\begin{array}{c} m \\ k \end{array}\right) B^{(m-k)}\left(-\frac{1}{2}, 1\right) (-1)^k k !\left(2^{k+1}-2\right) \zeta(k+1) \\ = & \sum_{k=0}^m\left(\begin{array}{c} m \\ k \end{array}\right) L_{m-k}(-1)^k k !\left(2^{k+1}-2\right) \zeta(k+1) \end{aligned} $$ which is a linear combination of $E_k$ up to $\ln^{m+2} 2$ Therefore $P(m+1)$ is also true and hence by M.I., $P(n)$ is true for all natural number $n$. Consequently, $I_n$ is a linear combination of $E$ up to $\ln^{n+1} 2$.

Lai
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