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Question Number 130254 by Eric002 last updated on 23/Jan/21

p r o v e

\int_{0}^{\infty} \frac{{l n}^{2} (x) l n (1 + x^{2})}{1 + x^{2}} d x = \frac{7 π}{4} ζ (3) + \frac{π^{3} l n (2)}{4}

w h e r e ζ (3) i s a {p e r y}^{,} s c o n s t a n t

Answered by mindispower last updated on 26/Jan/21

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

= - 2 \int_{0}^{\frac{π}{2}} {l n}^{2} (s i n (x)) l n (c o s (x)) d x - 2 \int_{0}^{\frac{π}{2}} {l n}^{3} (c o s (x)) d x

+ 4 \int_{0}^{\frac{π}{2}} {l n}^{2} (c o s (x)) l n (s i n (x)) d x = - 2 A - 2 B + 4 C

r e c a l l

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

\frac{\partial^{3}}{\partial^{2} b \partial a} β (\frac{1}{2}, \frac{1}{2}) = 8 A, \frac{\partial^{3}}{\partial^{2} a \partial b} β (\frac{1}{2}, \frac{1}{2}) = 8 C

\frac{\partial^{3} β (\frac{1}{2}, \frac{1}{2})}{\partial^{3} a} = 8 B

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

\partial_{a}^{2} β (a, b) = β (a, b) (Ψ (a) - Ψ {(a + b)}^{2} + (Ψ^{'} (a) - Ψ^{'} (a + b)) β (a, b)

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

- 2 Ψ^{'} (a + b) (Ψ (a) - Ψ (a + b)) (Ψ (a) - Ψ (a + b)) β (a, b)

+ β (a, b) (Ψ (b) - Ψ (a + b))) (Ψ^{'} (a) - Ψ^{'} (a + b))

- Ψ^{″} (a + b) β (a, b))

\partial^{2} a \partial b β (\frac{1}{2_{}}, \frac{1}{2})

i w i l l f i n i s h ? i t l a t e r

Ψ (\frac{1}{2}) c a n e a s l y v a l u t e d

Ψ^{'} (1) = ζ (2) = \frac{π^{2}}{6} \dots =