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In my post, I found that $$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x = -\frac{\pi}{2} \ln 2 .$$

Then I try to generalize the result to the integral by the same technique. $$ J_{n}:=\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}} d x $$

Letting $x=\cos \theta $ converting $I_n$ into \begin{aligned} J_{n} &=\int_{0}^{\frac{\pi}{2} } \frac{\ln^n (\cos \theta)}{\sin \theta} \sin \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \ln ^{n}(\cos \theta) d \theta \end{aligned}

By my post, $$\begin{aligned} (-2)^{n}J_n&= 2 \ln 2(-2)^{n-1} J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k_ !} \zeta(n-k) (-2)^{k} J_k \\J_n&= -\ln 2 J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{(-1)^{n-k}}{k !} \left(1-\frac{1}{2^{n-k-1}} \right)\zeta(n-k) J_k \end{aligned} $$

which is a reduction formula for $J_n.$

For example, $$ \begin{aligned} \int_{0}^{1} \frac{\ln ^2x}{\sqrt{1-x^{2}}} d x&=-\ln 2 J_{1}+\frac{1}{2} \zeta(2) J_{0} \\ &=-\ln 2\left(-\frac{\pi}{2} \ln 2\right)+\frac{1}{2} \zeta(2) J_{0} \\ &=\frac{\pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{24}, \end{aligned} $$

which is checked by Wolframalpha.

My Question

Is there any closed form for the integral? Your comments and methods are warmly welcome.

Lai
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  • If nothing else, $J_n$ for $n \geq 3$ will involve numbers like $\zeta(3)$ which can't be simplified further. That may or may not count as a "closed form" in your book. – Michael Seifert Mar 15 '22 at 15:37
  • Malmsten determined related integrals which are essentially a derivative away from the simpler integral $\int_{0}^{1} \frac{\ln^n x}{1+x^2} , dx$ and the answers are quite non-trivial, so I wouldn’t suspect your integral to have a closed-form. – KStarGamer Mar 15 '22 at 18:47
  • Although, Mathematica does give $\int_{0}^{1} \frac{\ln^n x}{1+x^2} , dx = (-1)^n 4^{-1-n} \Gamma (1+n) \left(\zeta\left(1+n,\frac{1}{4}\right)-\zeta\left(1+n,\frac{3}{4}\right)\right)$ for $\Re (n)>-1$, but I’m not sure about your integral. – KStarGamer Mar 15 '22 at 18:55
  • You could try generalizing the Fourier series approach though evaluating the multisum in the general $n$-case will probably be very difficult. – user170231 Dec 13 '22 at 04:04

2 Answers2

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Under the substitution $x^2 \to x$ we get that $$I_{n,m} =\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x = \frac{1}{2^{n+1}}\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x $$ Recalling the definition of the Beta function we see that \begin{align*} B\left(\frac{1}{2}+t, 1-m\right) =& \int_0^{1} x^{-\frac{1}{2}+t}(1-x)^{-m}\mathrm{d}x \\ \mathbin{\color{blue}{\implies}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} =&\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x =2^{n+1}I_{n,m} \end{align*} So we get the generalization

$$\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x =\frac{1}{2^{n+1}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} \qquad \text{for }\ \ m<1,\ n \in \mathbb{N}$$

This result could technically be expanded in terms of polygamma functions using the General Leibniz rule, but I expect this will just result in a way messier expression.

Robert Lee
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This is not an answer

$$I_n=\int_{0}^{1} \frac{\log^n(x)}{\sqrt{1-x^{2}}}\,dx$$ Using the binomial expansion of the denominator $$I_n=\sum_{k=0}^\infty\frac{\Gamma \left(k+\frac{1}{2}\right)}{\sqrt{\pi }\, k!}\int_{0}^{1} x^{2k}\,\log^n(x)\,dx$$ $$\int_{0}^{1} x^{2k}\,\log^n(x)\,dx=(-1)^n \frac{ \Gamma (n+1)}{(2 k+1)^{n+1} }$$ $$I_n=(-1)^n\frac{ n!}{\sqrt{\pi }}\sum_{k=0}^\infty \frac{ \Gamma \left(k+\frac{1}{2}\right)}{(2 k+1)^{n+1}\,k!}$$ which I am unable to simplify even using special function. But, for a given $n$, this leads to the results.

Claude Leibovici
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