prove-that-0-1-1-x-2-1-x-2-dx-4-pi-G-where-G-catalan-constant-

Question Number 110118 by mathdave last updated on 27/Aug/20

p r o v e t h a t

\int_{0}^{1} Γ (1 - \frac{x}{2}) Γ (1 + \frac{x}{2}) d x = \frac{4}{π} G

w h e r e G (c a t a l a n c o n s t a n t)

Commented by Sarah85 last updated on 27/Aug/20

please show how to get

\underset{0}{\int^{π}} \sin (a \sin (x)) d x = π H_{0} (a)

where H_{0} (a) is the Struve - H - Function

Commented by maths mind last updated on 30/Aug/20

H_{t} (a) = \sum_{m ⩾ 0} \frac{{(- 1)}^{m}}{Γ (m + \frac{3}{2}) Γ (m + t + \frac{3}{2})} {(\frac{z}{2})}^{2 m + t + 1}

s i n (a s i n (x)) = \sum_{k ⩾ 0} {(- 1)}^{k} \frac{{(a s i n (x))}^{(2 k + 1)}}{(2 k + 1)!}

\int_{0}^{π} {s i n}^{m} (x) d x = \int_{0}^{\frac{π}{2}} {{s i n}^{m} (x) + {c o s}^{m} (x)} d x

= 2 \int_{0}^{\frac{π}{2}} {s i n}^{m} (x) d x = β (\frac{m + 1}{2}, \frac{1}{2}) = \frac{Γ (\frac{m + 1}{2}) Γ (\frac{1}{2})}{Γ (\frac{m}{2} + 1)}

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} Γ (k + 1) Γ (\frac{1}{2})}{Γ (2 k + 2) . Γ (k + \frac{3}{2})} a^{2 k + 1}

Γ (2 k + 2) = \underset{j = 1}{\prod^{k}} (2 j) . \underset{j = 0}{\prod^{k}} (2 j + 1)

= 2^{k} Γ (k + 1) . 2^{k + 1} Γ (\frac{k}{2} + \frac{3}{2}) = 2^{2 k + 1} Γ (k + 1) Γ (\frac{k + 3}{2})

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k} Γ (k + 1) Γ (\frac{1}{2})}{2^{2 k + 1} Γ (k + 1) Γ (\frac{k + 3}{2}) . Γ (\frac{k + 3}{2})} a^{2 k + 1}

Missing \left or extra \right

= \sqrt{π} . H_{0} (a)

Commented by Sarah85 last updated on 30/Aug/20

thank you very much!

Commented by maths mind last updated on 30/Aug/20

w i t h e p l e a s u r

Answered by maths mind last updated on 15/Sep/20

Γ (1 + \frac{x}{2}) = \frac{x}{2} Γ (\frac{x}{2})

\Leftrightarrow \int_{0}^{1} Γ (1 - \frac{x}{2}) Γ (\frac{x}{2}) . \frac{x}{2} d x

= \frac{1}{2} \int_{0}^{1} Γ (1 - \frac{x}{2}) Γ (\frac{x}{2}) x d x

Γ (1 - x) Γ (x) = \frac{π}{s i n (π x)}, x \in] 0, 1 [

w e f i n d

\frac{1}{2} \int_{0}^{1} π \frac{x}{2 s i n (\frac{π x}{2})} d x

l e t y = \frac{π x}{2} \Rightarrow d x = \frac{2}{π} d y

= \frac{1}{2} \int_{0}^{\frac{π}{2}} . π \frac{2 y}{π s i n (y)} . . \frac{2 d y}{π} = \frac{2}{π} \int_{0}^{\frac{π}{2}} \frac{y d y}{s i n (y)}

\int_{0}^{\frac{π}{2}} \frac{y d y}{s i n (y)}, l e t t g (\frac{y}{2}) = u \Leftrightarrow \int_{0}^{1} 2 \frac{a r c t a n (u)}{\frac{2 u}{1 + u^{2}}} . \frac{2}{1 + u^{2}} d u

= 2 \int_{0}^{1} \frac{a r c t a n (u)}{u} d u = 2 \int_{0}^{1} \sum_{k ⩾ 0} {(- 1)}^{k} \frac{u^{2 k + 1}}{2 k + 1} . \frac{d u}{u}

= \sum_{k ⩾ 0} \int_{0}^{1} {(- 1)}^{k} \frac{u^{2 k}}{2 k + 1} du = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} = G

\Rightarrow \int_{0}^{\frac{π}{2}} \frac{x}{s i n (x)} = 2 \int_{0}^{1} \frac{\tan^{- 1} (x)}{x} d x = 2 G

\Rightarrow \int_{0}^{1} Γ (1 + \frac{x}{2}) Γ (1 - \frac{x}{2}) d x = \frac{2}{π} \int_{0}^{\frac{π}{2}} \frac{d x}{s i n (x)} = \frac{2}{π} . 2 G = \frac{4}{π} G