NICE-CALCULUS-prove-that-0-x-2-ln-pix-pi-pix-dx-1-piln-pi-3-3-2-ln-ln-pi-

Question Number 127017 by mnjuly1970 last updated on 26/Dec/20

\dots N I C E C A L C U L U S \dots

p r o v e t h a t ::

\int_{0}^{\infty} (\frac{x^{2} l n (π x)}{π^{π x}}) d x

= \frac{1}{{(π l n (π))}^{3}} [(3 - 2 (γ + l n (l n (π)))]

Answered by mindispower last updated on 26/Dec/20

= \int_{0}^{\infty} x^{2} l n (π x) e^{- π x l n (π)} d x

u = π l n (π) x

\Rightarrow d x = \frac{d u}{π l n (π)}

\Leftrightarrow \int_{0}^{\infty} \frac{u^{2}}{π^{2} {l n}^{2} (π)} l n (\frac{u}{l n (π)}) e^{- u} . \frac{d u}{π l n (π)}

= \frac{1}{π^{3} {l n}^{3} (π)} [\int_{0}^{\infty} u^{2} l n (u) e^{- u} d u - l n (l n π) \int_{0}^{\infty} u^{2} e^{- u}]

= \frac{1}{π^{3} {l n}^{3} (π)} {Γ^{'} (3) - l n (l n (π) . Γ (3)}

= \frac{Γ (3) . Ψ (3)}{π^{3} {l n}^{3} (π)} - \frac{1}{π^{3} {l n}^{3} (π)} 2 l n (l n π)

= 2 (\frac{1}{2} + 1 - γ) . \frac{1}{π^{3} {l n}^{3} (π)} - \frac{2 l n (l n π)}{π^{3} {l n}^{3} (π)}

= \frac{3 - 2 γ}{π^{3} {l n}^{3} (π)} - \frac{2 l n (l n π)}{π^{3} {l n}^{3} (π)} = \frac{1}{{(π l n π)}^{3}} [(3 - 2 (γ + l n (l n π))]

Commented by mnjuly1970 last updated on 26/Dec/20

p e a c e b e u p o n y o u

s i r p o w e r . . m e r c e y . .

Answered by Dwaipayan Shikari last updated on 26/Dec/20

\int_{0}^{\infty} \frac{x^{2} l o g (π x)}{π^{π x}} d x = \frac{1}{π^{3}} \int_{0}^{\infty} \frac{u^{2} l o g (u)}{π^{u}} d u π x = u

= \frac{1}{π^{3}} \int_{0}^{\infty} e^{- u l o g (π)} u^{2} l o g (u) d u u l o g (π) = t

= \frac{1}{{(π l o g (π))}^{3}} \int_{0}^{\infty} e^{- t} t^{2} l o g (t) - \int_{0}^{\infty} e^{- t} t^{2} l o g (l o g (π))

= \frac{1}{{(π l o g (π))}^{3}} (Γ^{'} (3) - Γ (3) l o g (l o g (π)))

Γ^{'} (3) = Γ (3) ψ (3) = 2 (- γ + \sum^{\infty} \frac{1}{n} - \frac{1}{n + 2}) = - 2 γ + 3

= \frac{1}{{(π l o g (π))}^{3}} (- 2 γ + 3 - 2 l o g (l o g (π))