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Question Number 80397 by M±th+et£s last updated on 02/Feb/20

s h o w t h a t

\int_{0}^{\frac{π}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{π}} + 1} d x d y = 2 c

w h r r e c d e n o t e t h a {c a t a l a n}^{,} s c o n s t a n t

Commented by mathmax by abdo last updated on 02/Feb/20

v h a n g e m e n t^{y} \sqrt{x^{π}} = t g i v e x = t^{\frac{y}{π}} \Rightarrow

\int_{0}^{\infty} \frac{d x}{(^{y} \sqrt{x^{π}}) + 1} = \int_{0}^{\infty} \frac{y}{π} t^{\frac{y}{π} - 1} \times \frac{d t}{t + 1} = \frac{y}{π} \int_{0}^{\infty} \frac{t^{\frac{y}{π} - 1}}{t + 1} d t

= \frac{y}{π} \times \frac{π}{s i n (π . \frac{y}{π})} = \frac{y}{s i n y} \Rightarrow \int_{0}^{\frac{π}{2}} \int_{0}^{\infty} \frac{d x}{(^{y} \sqrt{x^{π}} + 1)} d y

= \int_{0}^{\frac{π}{2}} \frac{y}{s i n y} d y \dots w e c a n f i n d a a p p r o x i m a t v a l u e f o r t h i s i n t e g r a l

\dots b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 03/Feb/20

c h a n g e m e n t t a n (\frac{y}{2}) = t g i v e

\int_{0}^{\frac{π}{2}} \frac{y d y}{s i n y} = \int_{0}^{1} \frac{2 a r c t a n (t)}{\frac{2 t}{1 + t^{2}}} \times \frac{2 t}{1 + t^{2}} d t = 2 \int_{0}^{1} a r c t a n (t) d t

w e h a v e {a r c t a n}^{^{'}} (t) = \frac{1}{1 + t^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{2 n} \Rightarrow

a r c t a n (t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n} + c (c = 0) \Rightarrow \int_{0}^{1} a r c t a n (t) d t

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \int_{0}^{1} t^{2 n} d t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = K (c o n s t a n t e o f c a t a l a n)

\Rightarrow \int_{0}^{\frac{π}{2}} \frac{y}{s i n y} d y = 2 K = \int_{0}^{\frac{π}{2}} \int_{0}^{\infty} \frac{d x}{(^{y} \sqrt{x^{π}} + 1)} d y

Commented by mathmax by abdo last updated on 03/Feb/20

K \sim 0, 9159 \dots