nice-calculus-prove-that-0-pi-4-ln-sin-x-d-pi-4-log-2-G-2-log-2sin-x-n-1-1-n-cos-2nx-0-pi-4-log-2-n-1-cos-2nx-n

Question Number 128244 by mnjuly1970 last updated on 05/Jan/21

\dots n i c e c a l c u l u s \dots

p r o v e t h a t :: Ω = \int_{0}^{\frac{π}{4}} l n (s i n (x)) d = \frac{- π}{4} l o g (2) - \frac{G}{2}

l o g (2 s i n (x)) = \underset{n = 1}{\sum^{\infty}} \frac{- 1}{n} c o s (2 n x)

Ω = \int_{0}^{\frac{π}{4}} {- l o g (2) - \underset{n = 1}{\sum^{\infty}} \frac{c o s (2 n x)}{n}} d x

= \frac{- π}{4} l o g (2) - \underset{n = 1}{\sum^{\infty}} \int_{0}^{\frac{π}{4}} \frac{c o s (2 n x)}{n} d x

= \frac{- π}{4} l o g (2) - \underset{n = 1}{\sum^{\infty}} {[\frac{1}{2 n^{2}} s i n (2 n x)]}_{0}^{\frac{π}{4}}

= \frac{- π}{4} l o g (2) - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{s i n (\frac{n π}{2})}{n^{2}}

= \frac{- π}{4} l o g (2) - \frac{1}{2} {\frac{1}{1^{2}} - \frac{1}{3^{2}} + \frac{1}{5^{2}} - . .}

= \frac{- π}{4} l o g (2) - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{{(2 n - 1)}^{2}}

∴ Ω = \frac{- π}{4} l o g (2) - \frac{G}{2} ✓

G := c a t a l a n c o n s t a n t \dots