Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $?

Question

After reading the fantastic solutions to the integral in the post,

$$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x=-\frac{\pi^2}8,$$ and found that $$ \int_0^1 \frac{\ln ^2 x}{1-x^2} d x =\frac{7\zeta(3)}{4}. $$

Then I keep on exploring the integral in general and obtain a formula,

$$ \int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right] $$ where $n\ge 2.$

First of all, we use its partner integral $$ I(a)=\int_0^1 \frac{x^a}{\left(1-x^2\right)^2} d x $$ Using the series for $|x|<1,$ $$ \frac{1}{\left(1-x^2\right)^2}=\sum_{k=1}^{\infty} k x^{2 k-2}, $$ we have $$ I(a) =\sum_{k=1}^{\infty} k \int_0^1 x^{2 k-2+a} d x =\sum_{k=1}^{\infty} \frac{k}{2 k-1+a} $$ $$ \begin{aligned} \int_0^1 \frac{\ln ^n x}{\left(1-x^2\right)^2} d x & =\left.\sum_{k=1}^{\infty} \frac{\partial^n}{d a^n}\left(\frac{k}{2 k-1+a}\right)\right|_{a=0} \\ & =(-1)^n n ! \sum_{k=1}^{\infty} \frac{k}{(2 k-1)^{n+1}} \\ & =\frac{(-1)^n n !}{2} \sum_{k=1}^{\infty}\left[\frac{1}{(2 k-1)^n}+\frac{1}{(2 k-1)^{n+1}}\right] \\ & =\boxed{\frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right]} \end{aligned} $$ For examples, $$ I_2=\frac{3}{4} \zeta(2)+\frac{7}{8} \zeta(3)= \frac{\pi^2}{8} +\frac{7}{8} \zeta(3) $$ $$ I_3=-3\left[\frac{7}{8} \zeta(3)+\frac{15}{16} \zeta(4)\right]=-\frac{21}{8} \zeta(3)-\frac{\pi^2}{32} $$

My question: Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $?

I don't think you can do much since there is no known simple closed form for $\zeta(n)$ when $n$ is odd and is greater than $1$. — Gary, Feb 14 '23 at 09:23
Can we evaluate the integral when $n\ge 2$ is a real number ? — Lai, Feb 14 '23 at 09:27
The case $n=2$ is $\frac18\pi^2+\frac78\zeta(3)$. You can't get simpler than that. — J.G., Feb 14 '23 at 09:29
@Lai $\log x$ is negative for $0<x<1$. What do you mean by its power $n$ when $n$ is not an integer? — Gary, Feb 14 '23 at 09:30
Yes, so $n$ must be an integer greater than or equal to 2. Is there any other method? — Lai, Feb 14 '23 at 09:33
I do not know the answer but I admire. It is a very nice work. — Claude Leibovici, Feb 14 '23 at 09:42
Thank you for all your support. I just wonder if there are some other methods with answers in other form, then we may simplify it. — Lai, Feb 14 '23 at 09:58
You almost have $\eta(x)=(1-2^{1-x})\zeta(x)$ with the Dirichlet eta function — Тyma Gaidash, Feb 14 '23 at 12:49
Why downvotes for this question ? Nice work, good explanation, good presentatio from an obviously enthusiastic user. — Claude Leibovici, Feb 14 '23 at 13:28
@Lai Don't let it bother you. Your posts are always very interesting. — Martin Gales, Feb 14 '23 at 15:56
Yes, you are right. I better rewrite the question as :Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $? — Lai, Feb 15 '23 at 01:35
Martin, thank you for your support and encouragement. I’ll continue to write post. — Lai, Feb 17 '23 at 11:42

Quanto · Accepted Answer · 2023-02-16T19:37:23.190

Utilize \begin{align} J(m)=\int_0^1 \frac{\ln^m x}{1-x^2}dx =&\int_0^1 \frac{\ln^m x }{1-x}dx -\int_0^1 {\frac{x\ln^m x }{1-x^2} } \overset{x^2\to x}{dx} \\=& \ \left(1-\frac1{2^{m+1}}\right)\int_0^1\frac{\ln^mx}{1-x}dx\\ =&\ \left(1-\frac1{2^{m+1}}\right)(-1)^m m!\ \zeta(m+1) \end{align} to evaluate \begin{align} \int_0^{1} \frac{\ln^n x}{\left(1-x^2\right)^2}dx &=\int_0^{1} \frac{\ln ^n x}{2x}\ d\left(\frac{x^2}{1-x^2}\right) \overset{ibp}=\frac12J(n) -\frac n2 J(n-1)\\ &= \frac{(-1)^n n !}{2}\left[\left(1-\frac{1}{2^n}\right)\zeta(n)+\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1)\right] \end{align}

Wow, a decent solutions with a unique skill! – Lai Feb 15 '23 at 08:17 — Lai, Feb 15 '23 at 08:17

Is there any other simple method to evaluate $\int_0^{1} \frac{\ln ^n x}{\left(1-x^2\right)^2}dx $?

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