<s>[INST] Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx
&= \int_0^{\frac{\pi}{2}} \frac{x}{1-\cos^2 x}\sin x\  dx \\
&= \sum_{n=0}^{\infty} \int_0^{\frac{\pi}{2}} x\sin x \ \cos^{2n}x \ dx \\
&= \sum_{n=0}^{\infty}\frac{1}{2n+1} \int_0^{\frac{\pi}{2}} \cos^{2n+1}x \ dx \\
&= \sum_{n=0}^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}
\end{align*}
which is close but not quite there.
 [/INST] As you posted:

$$ \int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx \\ \sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}$$

You can rewrite it as the following using the definition of the central binomial coefficient:
$$\sum_0^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2} = \sum_0^{\infty} \frac{4^n}{\frac{(2n)!}{n!^2}(2n+1)^2} = \sum_0^{\infty} \frac{4^n}{\frac{4^nΓ\left(n+\frac12\right)}{\sqrt{\pi}Γ(n+1)}(2n+1)^2}=\frac{\sqrt \pi}2\sum_0^\infty \frac{n!}{\left(n+\frac12\right)!(2n+1)} $$
Now let’s apply the main definition of the $\,_3\text F_2$ Hypergeometric function using Pochhammer Symbol:
$$\,_3\text F_2(a_1,a_2,a_3;b_1,b_2;z)=\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n(a_3)_n x^n}{(b_1)_n(b_2)_nn!}$$
The constant is then as seen in formula $(4)$ of this arxiv article:
$$\,_3\text F_2\left(\frac12,1,1;\frac32,\frac32;\frac{4x}{(x+1)^2}\right)=(x+1)\sum_{n=0}^\infty \frac{(-x)^n}{(2n+1)^2}$$
Therefore
$$ \int_0^{\frac{\pi}{2}} x\csc(x)= \,_3\text F_2\left(\frac12,1,1;\frac32,\frac32;1\right)=2\text C$$
Please correct me and give me feedback!
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