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Evaluate $$\mathcal{L}_n(x)=\int_0^x \ln^n\left(2\sin\frac{t}2\right)\,dt\qquad n\in\Bbb N_0, x\in[0,\pi/2]$$

In this answer to a question of mine, @TitoPiezasIII claims that $$\frac{(-1)^n}{n!}\mathcal{L}_n\left(\frac\pi3\right)=\,_{n+2}F_{n+1}\left[{{\frac12,\dots\frac12}\atop{\frac32,\dots,\frac32}}\Bigg|\frac14\right]$$ This I do not doubt, but I am having trouble proving it. I thought that I'd try to evaluate $\mathcal{L}_n(x)$ for all $0\leq x\leq\pi/2$ instead though. What I've come up with is below.

We define $$f(s,x)=\int_0^x\left(2\sin\frac{t}{2}\right)^s dt\qquad x\in[0,\pi/2],s\geq0$$ Because $x$ is confined to the range that it is, we can easily show that $$f(s,x)=2^s\mathrm{B}\left(\sin^2(x/2);\frac{s+1}2,\frac12\right)$$ Where $$\mathrm{B}(x;a,b)=\int_0^x t^{a-1}(1-t)^{b-1}dt$$ is the incomplete Beta function. We have from here that $$\mathrm B(x;a,b)=\frac{x^a}{a}\,_2F_1(a,1-b;1+a;x)$$ So $$f(s,x)=2^s\frac{q^{s+1}}{s+1}\,_2F_1\left(\frac12,\frac{s+1}{2};\frac{s+1}2+1;q^2\right)$$ Where $q\equiv\sin\frac{x}2$. We see that $$\frac{(\frac{s+1}2)_k}{(\frac{s+1}2+1)_k}=\frac{s+1}{s+2k+1}\qquad (x)_k=\frac{\Gamma(x+k)}{\Gamma(x)}$$ so $$f(s,x)=2^sq^{s+1}\sum_{k\geq0}\frac{(1/2)_k}{s+2k+1}\frac{q^{2k}}{k!}$$ So $$f^{(n,0)}(s,x)=\left(\frac{\partial }{\partial s}\right)^n f(s,x)=\sum_{k\geq0}\frac{(1/2)_k}{k!}q^{2k+1}\left(\frac{\partial }{\partial s}\right)^n\frac{(2q)^s}{s+2k+1}$$ Then I used the Generalized Leibniz rule to get that $$\left(\frac{\partial }{\partial s}\right)^n\frac{(2q)^s}{s+2k+1}=(2q)^sn!\sum_{j=0}^{n}\frac{(-1)^j}{(n-j)!}\frac{\ln(2q)^{n-j}}{(s+2k+1)^{j+1}}$$ so $$f^{(n,0)}(s,x)=2^sq^{s+1}n!\sum_{j=0}^{n}\frac{(-1)^j}{(n-j)!}\ln(2q)^{n-j}\sum_{k\geq0}\frac{(1/2)_k}{(s+2k+1)^{j+1}}\frac{q^{2k}}{k!}$$ $$f^{(n,0)}(s,x)=2^sq^{s+1}n!\sum_{j=0}^{n}\frac{(-1)^j}{(n-j)!}\ln(2q)^{n-j}\,_{j+2}F_{j+1}\left[{{\frac12,\frac{s+1}2,\dots,\frac{s+1}2}\atop{\frac{s+3}2,\dots,\frac{s+3}2}}\Bigg|\ q^2\right]$$ So $$\mathcal L_n(x)=f^{(n,0)}(0,x)=qn!\sum_{j=0}^{n}\frac{(-1)^j}{(n-j)!}\ln(2q)^{n-j}\,_{j+2}F_{j+1}\left[{{\frac12,\dots,\frac{1}2}\atop{\frac{3}2,\dots,\frac{3}2}}\Bigg|\ q^2\right]$$ But having shown this, I am not certain that plugging in $x=\frac\pi3$ will confirm @TitoPiezasIII's result. Where did I go wrong?

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  • $\begingroup$ I guess he makes the substitution $y=2\sin\frac{t}{2}$, then expands the radical $1/\sqrt{1-y^2/4}$ into Taylor series and integrates termwise. $\endgroup$ – Nemo May 24 at 16:23

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