nice-calculus-0-pi-2-xsin-x-cos-x-ln-sin-x-ln-cos-x-dx-pi-16-pi-3-192-

Question Number 122528 by mnjuly1970 last updated on 17/Nov/20

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

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

\overset{? ? ?}{=} \frac{π}{16} - \frac{π^{3}}{192} ✓

Answered by mindispower last updated on 18/Nov/20

\int_{a}^{b} f (x) d x = \frac{1}{2} \int_{a}^{b} (f (a + b - x) + f (x)) d x

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

Ω = \frac{π}{4} \int_{0}^{\frac{π}{2}} s i n (x) c o s (x) l n (s i n (x)) l n (c o s (x)) d x

f (a, b) = \frac{π}{4} \int_{0}^{\frac{π}{2}} s i n (x) c o s (x) {s i n}^{a} (x) {c o s}^{b} (x) d x

Ω = \frac{\partial^{2}}{\partial a \partial b} f (0, 0)

f = \frac{π}{8} . 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 (1 + \frac{a}{2}) - 1} (x) . {c o s}^{2 (1 + \frac{b}{2}) - 1} (x) d x

= \frac{π}{8} β (1 + a, 1 + b)

\frac{\partial}{\partial b} (\frac{\partial}{\partial a}) f = \frac{π}{8} . \frac{\partial}{\partial b} . \frac{1}{2} β (\frac{a}{2} + 1, \frac{b}{2} + 1) (Ψ (\frac{a}{2} + 1) - Ψ (\frac{1}{2} (a + b) + 2)

= \frac{π}{32} β (\frac{a}{2} + 1, \frac{b}{2} + 1) (Ψ (\frac{b}{2} + 1) - Ψ (\frac{1}{2} (a + b) + 2)) (Ψ (\frac{a}{2} + 1) - Ψ (\frac{1}{2} (a + b) + 2))

- \frac{π}{32} Ψ^{1} (\frac{1}{2} (a + b) + 2) β (\frac{a}{2} + 1, \frac{b}{2} + 1)

Ω = \frac{π}{32} β (1, 1) {(Ψ (1) - Ψ (2))}^{2} - \frac{π}{32} Ψ^{1} (2) β (1, 1)

= \frac{π}{32} - \frac{π}{32} Ψ^{'} (2)

Ψ^{1} (z) = \sum_{n ⩾ 0} \frac{1}{{(n + z)}^{2}}

Ψ^{1} (2) = \sum_{n ⩾ 0} \frac{1}{{(n + 2)}^{2}} = \sum_{n ⩾ 1} (\frac{1}{n^{2}}) - 1 = \frac{π^{2}}{6} - 1

= \frac{π}{32} - \frac{π}{32} (\frac{π^{2}}{6} - 1)

= \frac{π}{16} - \frac{π^{3}}{192}

Commented by mnjuly1970 last updated on 18/Nov/20

t h a n k y o u s i r p o w e r \dots

b y u s i n g t h e e u l e r b e t a f u n c t i o n

. g r a t e f u l . s i n c e r e l y y o u r s . m . n

Commented by mindispower last updated on 18/Nov/20

w i t h e p l e a s u r n i c e d a y