show-that-0-pi-2-ln-sec-x-ln-csc-x-dx-pi-2-ln-2-2-2-pi-4-48-

Question Number 89593 by M±th+et£s last updated on 19/Apr/20

s h o w t h a t

\int_{0}^{\frac{π}{2}} l n (s e c (x)) l n (c s c (x)) d x = \frac{π^{2} {l n}^{2} (2)}{2} - \frac{π^{4}}{48}

Commented by maths mind last updated on 19/Apr/20

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

= \frac{1}{2} \partial_{a} \partial_{b} 2 \int_{0}^{\frac{π}{2}} {s i n}^{a} (x) {c o s}^{b} (x) d x ∣_{(a, b) = (0, 0)}

= \frac{1}{2} \partial_{a} \partial_{b} β {(\frac{a + 1}{2}, \frac{b + 1}{2})}_{(a, b) = (0, 0)}

\partial_{x} β (x, y) = β (x, y) (ψ (x) - ψ (x + y))

\partial_{a} \partial_{b} β (\frac{a + 1}{2}, \frac{b + 1}{2})

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

= \frac{1}{4} β (\frac{a + 1}{2}, \frac{b + 1}{2}) (Ψ (\frac{a + 1}{2}) - Ψ (\frac{a + b}{2} + 1)) (Ψ (\frac{b + 1}{2}) - Ψ (\frac{a + b}{2} + 1))

- \frac{1}{4} Ψ^{'} (\frac{a + b}{2} + 1) β (\frac{a + 1}{2}, \frac{b + 1}{2})

(a = b = 0)

= \frac{β (\frac{1}{2}, \frac{1}{2})}{4} ({(Ψ (\frac{1}{2}) - 1)}^{2} - Ψ^{'} (1))

= \frac{π}{8} (4 {l n}^{2} (2) - \frac{π^{2}}{6}) = \frac{π {l n}^{2} (2)}{2} - \frac{π^{3}}{48}

Commented by M±th+et£s last updated on 19/Apr/20

t h a n k y o u v e r r y m u c h i t s g r e a t s o l u t i o n

Commented by maths mind last updated on 19/Apr/20

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

m i n d i s p o w e r!!

Commented by M±th+et£s last updated on 19/Apr/20

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i n t h e f o r u m

Commented by M±th+et£s last updated on 20/Apr/20

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t h a t ?