advanced-calculus-prove-that-0-1-x-sin-2pix-dx-pi-2-

Question Number 138041 by mnjuly1970 last updated on 09/Apr/21

\dots \dots . a d v a n c e d \dots \dots \dots c a l c u l u s \dots \dots

p r o v e t h a t :::

ϕ = \int_{0}^{1} ψ (x) . s i n (2 π x) d x = - \frac{π}{2} \dots ✓

Answered by Dwaipayan Shikari last updated on 09/Apr/21

\int_{0}^{1} ψ (x) s i n (2 π x) d x

= {[s i n (2 π x) l o g (Γ (x))]}_{0}^{1} - 2 π \int_{0}^{1} l o g (Γ (x)) c o s (2 π x)

= - π \int_{0}^{1} l o g (π) c o s (2 π x) + π \int_{0}^{1} l o g (s i n (π x)) c o s (2 π x) d x

= \int_{0}^{π} l o g (s i n (u)) c o s (2 u) d u

= {[\frac{1}{2} l o g (s i n (u)) s i n (2 u)]}_{0}^{π} - \frac{1}{2} \int_{0}^{π} \frac{s i n (2 u) c o s (u)}{s i n (u)} d u

= - \frac{1}{2} \int_{0}^{π} 1 + c o s (2 u) d u = - \frac{π}{2}

Commented by mnjuly1970 last updated on 09/Apr/21

v e r y n i c e m r

p a y a n . . t h a n k y o u s o m u c h . .

Answered by mnjuly1970 last updated on 09/Apr/21

ϕ = \int_{0}^{1} ψ (x) . s i n (2 π x) d x

ϕ = - \int_{0}^{1} ψ (1 - x) . s i n (2 π x) d x

2 ϕ = \int_{0}^{1} {(ψ (x) - ψ (1 - x)} s i n (2 π x) d x

= - π \int_{0}^{1} c o t (π x) . s i n (2 π x) d x

= - 2 π \int_{0}^{1} \frac{c o s (π x)}{s i n (π x)} {s i n (π x) . c o s (π x)} d x

∴ ϕ = - π \int_{0}^{1} \frac{1 + c o s (2 π x)}{2} d x

= - \frac{π}{2} \dots . . ✓