nice-calculus-prove-that-1-0-1-li-2-1-x-2-pi-2-2-4-2-0-1-log-1-t-t-3-4-1-t-dt-pi-3-2-2-2-3-4-log-2-pi-2-hi

Question Number 131991 by mnjuly1970 last updated on 10/Feb/21

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

p r o v e t h a t :

ϕ_{1} = \int_{0}^{1} {l i}_{2} (1 - x^{2}) = \frac{π^{2}}{2} - 4

ϕ_{2} = \int_{0}^{1} \frac{l o g (1 - t)}{t^{\frac{3}{4}} \sqrt{1 - t}} d t = π^{\frac{3}{2}} . \frac{\sqrt{2}}{Γ^{2} (\frac{3}{4})} (l o g (2) - \frac{π}{2})

h i n t 1 : ψ (\frac{3}{4}) \underset{e a s y}{\overset{w h y ? ?}{=}} - γ + \frac{π}{2} - 3 l o g (2)

h i n t 2 : ψ (\frac{1}{2}) \underset{e a s y}{\overset{w h y ? ?}{=}} - γ - 2 l o g (2)

Answered by Dwaipayan Shikari last updated on 10/Feb/21

ψ (\frac{1}{2}) = - γ + \int_{0}^{1} \frac{1 - x^{- \frac{1}{2}}}{1 - x} d x = - γ + 2 \int_{0}^{1} \frac{x - 1}{1 - x^{2}} d x

= - γ - 2 l o g (2)

Answered by Dwaipayan Shikari last updated on 10/Feb/21

I (b) = \int_{0}^{1} t^{a - 1} {(1 - t)}^{b - 1} d t = \frac{Γ (a) Γ (b)}{Γ (a + b)}

I^{'} (b) = \int_{0}^{1} t^{a - 1} {(1 - t)}^{b - 1} l o g (1 - t) d t

= \frac{Γ (a) (Γ (a + b) Γ^{'} (b) - Γ^{'} (a + b) Γ (b))}{Γ^{2} (a + b)}

p u t b = \frac{1}{2} a = \frac{1}{4}