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8

\begin{align}\int_0^{1} \frac{\arcsin^2 x^2}{\sqrt{1-x^2}}\,dx &= \frac{1}{2}\int_0^{1} \frac{\arcsin^2 x}{\sqrt{x}\sqrt{1-x}}\,dx \tag{1}\\&= \frac{1}{2}\int_0^{\pi/2} \frac{\theta^2\cos \theta}{\sqrt{\sin \theta - \sin^2 \theta}}\,d\theta \tag{2}\\&= \frac{1}{\sqrt{2}}\int_0^{\pi/2} \frac{\left(\frac{\pi}{2} - \theta\right)^2\cos ...


2

We have: $$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence: $$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$


2

There is a closed form of this integral: $$\int_0^{\pi/2} (\log \sin x)^n\text{ d}x=\frac{1}{2^{n+1}}B^{(n)}\left(\frac{1}{2},\frac{1}{2}\right)$$


2

Let $\chi_{\Bbb Q}(x):=1 $ if $x \in \Bbb Q$ and $0$ otherwise.


2

Hint The number of elements of length $k$ is $3^k$, so the number of elements of length $\leq k$ is $$3 + 3^2 + \cdots 3^k = \frac{3}{2} (3^k - 1) .$$ Thus, the $n$th element in the sequence $A, B, C, AA, AB, \ldots$ has length $k$ iff $$\tfrac{3}{2} (3^{k - 1} - 1) < n \leq \tfrac{3}{2}(3^k - 1) .$$


1

Assuming you mean the version defined by Landau, namely $$\xi(s) = \frac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta\left(s\right)$$ then any computer algebra system will be able to do this for you. For example, in Maple, xi := s -> s/2*(s-1)*Pi^(-s/2)*GAMMA(s/2)*Zeta(s): nth := proc(n) limit(diff(xi(s),s$n),s=0) end will give you what ...


1

With your unusual notation (normally the Legendre polynomals are denoted with $P_n$ and $P_n^{(\alpha, \beta)}$ are the Jacobi polynomials), you have (see http://dlmf.nist.gov/18.7.E9 or Abramowitz/Stegun 22.5.35) $$L_n(x) = J_n^{(0,0)}(x)$$ Use the recurrence relations for Jacobi polynomials (http://dlmf.nist.gov/18.9.E5 or A/S 22.7.18 and 22.7.19) ...


1

The change of variable $a\,t=s$ changes the integral into $$\int_{a\gamma_0}^\infty \frac{1}{s}e^{-s}\,ds = \Gamma(0,a\,\gamma_0),$$ where $\Gamma(z,s)$ is the incomplete $\gamma$ function. To solve the equation $\Gamma(0,z)=12.1$ you can use Wolfram's function FindRoot. $z$ is of the order of $10^{-6}$, which makes computations a little bit shaky. The ...


1

This is probably the hypergeometric function.


1

In mathematics, the Gaussian or ordinary hypergeometric function 2F1(a,b;c;z) is a special function represented by the hypergeometric series, that includes many other special functions as specific or limiting cases. It is a solution of a second-order linear ordinary differential equation (ODE). Every second-order linear ODE with three regular ...



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