pi-2-pi-2-sin-2x-1-2-x-dx-

Question Number 87977 by Ar Brandon last updated on 07/Apr/20

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{s i n 2 x}{1 + 2^{x}} d x

Commented by abdomathmax last updated on 08/Apr/20

a t f o r m o f s e r i e

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

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

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} s i n (2 x) d x

b u t \int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} s i n (2 x) d x

= I m (\int_{- \frac{π}{2}}^{\frac{π}{2}} 2^{- (n + 1) x} e^{2 i x} d x)

\int_{- \frac{π}{2}}^{\frac{π}{2}} e^{- (n + 1) x l n 2 + 2 i x} d x

= \int_{- \frac{π}{2}}^{\frac{π}{2}} e^{(2 i - (n + 1) l n 2) x} d x

= {[\frac{1}{2 i - (n + 1) l n 2} e^{(2 i - (n + 1) l n 2) x}]}_{- \frac{π}{2}}^{\frac{π}{2}}

= \frac{1}{2 i - (n + 1) l n 2} {- e^{- (n + 1) l n 2 \frac{π}{2}} - e^{(n + 1) l n 2 \frac{π}{2}}}

= \frac{(n + 1) l n 2 - 2 i}{{(n + 1)}^{2} {l n}^{2} 2 + 4} {e^{\frac{π}{2} (n + 1) l n 2} + e^{- \frac{π}{2} (n + 1) l n 2}} \Rightarrow

I m (\dots) = - \frac{2}{{(n + 1)}^{2} {l n}^{2} 2 + 4} \times 2 c h (\frac{π}{2} (n + 1) l n 2) \Rightarrow

\Rightarrow

I = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(n + 1)}^{2} {l n}^{2} 2 + 4} \times c h (\frac{π}{2} (n + 1) l n 2)

Commented by Ar Brandon last updated on 08/Apr/20

A m a z i n g!!!!!