n-0-2n-n-1-4-n-2n-1-3-pi-3-48-pi-4-ln-2-2-

Question Number 136733 by Ñï= last updated on 25/Mar/21

\underset{n = 0}{\sum^{\infty}} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{1}{4^{n} {(2 n + 1)}^{3}} = \frac{π^{3}}{48} + \frac{π}{4} {l n}^{2} 2

Answered by mindispower last updated on 25/Mar/21

s t a r t b y

\frac{1}{\sqrt{1 - 4 x}} = \sum_{n ⩾ 0} (\begin{matrix} 2 n \\ n \end{matrix}) x^{n}

\Rightarrow \frac{1}{\sqrt{1 - 4 x^{2}}} Σ (\begin{matrix} 2 n \\ n \end{matrix}) x^{2 n}

\Rightarrow \frac{1}{2} a r c s i n (2 x) = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{x^{2 n + 1}}{2 n + 1}

\Rightarrow \frac{1}{2} \int_{0}^{t} \frac{a r c s i n (2 x)}{x} d x = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{t^{2 n + 1}}{{(2 n + 1)}^{2}}

w e w a n t \int_{0}^{\frac{1}{2}} \frac{1}{x} \int_{0}^{x} \frac{a r c s i n (2 t)}{t} d t d x = Σ (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{1}{4^{n} {(2 n + 1)}^{3}}

b y p a r t {[l n (x) \int_{0}^{x} \frac{a r c s i n (2 t)}{t} d t]}_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \frac{l n (x) a r c s i n (2 x)}{x} d x

= l n (\frac{1}{2}) \int_{0}^{\frac{1}{2}} \frac{a r c s i n (2 x)}{x} d x - {[\frac{1}{2} {l n}^{2} (x) a r c s i n (2 x)]}_{0}^{\frac{1}{2}}

+ \int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (x)}{\sqrt{1 - 4 x^{2}}} d x

\int_{0}^{\frac{1}{2}} \frac{a r c s i n (2 x)}{x} d x = \int_{0}^{1} \frac{a r c s i n (t)}{t} d t = - \int_{0}^{1} \frac{l n (t)}{\sqrt{1 - t^{2}}} d t

= - \int_{0}^{\frac{π}{2}} l n (s i n (t)) d t = \frac{π l n (2)}{2}

\int_{0}^{\frac{1}{2}} \frac{{l n}^{2} (x)}{\sqrt{1 - 4 x^{2}}} d x = \frac{1}{2} \int_{0}^{1} \frac{{l n}^{2} (t) - 2 l n (2) l n (t) + {l n}^{2} (2)}{\sqrt{1 - t^{2}}}

= \frac{1}{2} {\int_{0}^{1} \frac{{l n}^{2} (t)}{\sqrt{1 - t^{2}}} - l n (2) . \frac{π l n (2)}{2} + {l n}^{2} (2) \frac{π}{2}}

= \frac{1}{2} A

\int_{0}^{1} \frac{{l n}^{2} (t)}{\sqrt{1 - t^{2}}} d t = \int_{0}^{\frac{π}{2}} {l n}^{2} (s i n (t)) d t = A

β (a, b) = 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 a - 1} (t) {c o s}^{2 b - 1} (t) d t

A = \frac{1}{8} \frac{\partial^{2}}{\partial a^{2}} β (\frac{3}{2}, \frac{1}{2})

\partial_{a} β = β (a, b) (Ψ (a) - Ψ (a + b)}

\partial_{a}^{2} β = β (a, b) {{(Ψ (a) - Ψ (a + b))}^{2} + Ψ^{'} (a) - Ψ^{'} (a + b)}

A = \frac{1}{8} β (\frac{3}{2}, \frac{1}{2}) {{(Ψ (\frac{1}{2}) - Ψ (2))}^{2} + Ψ^{'} (\frac{1}{2}) - Ψ^{'} (2)}

A = \frac{1}{8} (Γ (\frac{3}{2}) Γ (\frac{1}{2})) {\frac{π^{2}}{2} - \frac{π^{2}}{6} + 1 + {(- 2 l n (2) - γ - 1 + γ)}^{2}}

A = \frac{1}{8} . \frac{1}{2} . π {\frac{π^{2}}{3} + 1 + {(- 2 l n (2) - 1)}^{2}}

S = \sum_{n ⩾ 0} (\begin{matrix} 2 n \\ n \end{matrix}) . \frac{1}{4^{n} {(2 n + 1)}^{2}} = - \frac{π {l n}^{2} (2)}{2} - \frac{π {l n}^{2} (2)}{4}

+ \frac{π}{32} (4 {l n}^{2} (2) + 2 + 4 l n (2) + \frac{π^{2}}{3}) \dots c o n t i n u e d l a t e r

Commented by Ñï= last updated on 25/Mar/21

t h a n k y o u s i r!

Commented by mindispower last updated on 26/Mar/21

p l e a s u r