prove-0-e-x-ln-x-1-4-1-m-n-

Question Number 150747 by mnjuly1970 last updated on 15/Aug/21

p r o v e ::

Ω := \int_{0}^{\infty} e^{- \sqrt{x}} . l n (\sqrt[4]{x}) \overset{?}{=} 1 - γ

m . n . .

Answered by Olaf_Thorendsen last updated on 15/Aug/21

Ω = \int_{0}^{\infty} e^{- \sqrt{x}} \ln (\sqrt[4]{x}) d x

Ω = \int_{0}^{\infty} e^{- u} \ln (\sqrt{u}) (2 u d u)

Ω = \int_{0}^{\infty} {u e}^{- u} \ln u d u

ψ (z) = \frac{\int_{0}^{\infty} y^{z - 1} \ln y e^{- y} d y}{\int_{0}^{\infty} y^{z - 1} e^{- y} d y}

ψ (2) = \frac{\int_{0}^{\infty} y . \ln y e^{- y} d y}{\int_{0}^{\infty} {y e}^{- y} d y} = \frac{Ω}{\int_{0}^{\infty} {y e}^{- y} d y}

\int_{0}^{\infty} {u e}^{- u} d u = {[- {u e}^{- u}]}_{0}^{\infty} + \int_{0}^{\infty} e^{- u} d u = 1

\Rightarrow Ω = ψ (2)

ψ (z + 1) = ψ (z) + \frac{1}{z}

ψ (2) = ψ (1) + \frac{1}{1} = - γ + 1

Ω = 1 - γ = ψ (2)

Commented by mnjuly1970 last updated on 16/Aug/21

g r a t e f u l m r o l a f . .

Answered by mathmax by abdo last updated on 17/Aug/21

Ψ = \int_{0}^{\infty} e^{- \sqrt{x}} \ln (x^{\frac{1}{4}}) dx \Rightarrow Ψ =_{\sqrt{x} = t} \int_{0}^{\infty} e^{- t} \ln (t^{\frac{1}{2}}) (2 t) dt

= 2 \int_{0}^{\infty} \frac{t}{2} \ln (t) e^{- t} dt = \int_{0}^{\infty} {te}^{- t} \ln (t) dt

= \int_{0}^{\infty} (tlnt) e^{- t} dt by parts u = tlnt and v^{^{'}} = e^{- t}

= {[- e^{- t} (tlnt)]}_{0}^{\infty} - \int_{0}^{\infty} (lnt + 1) (- e^{- t}) dt

= 0 + \int_{0}^{\infty} e^{- t} (1 + lnt) dt = \int_{0}^{\infty} e^{- t} dt + \int_{0}^{\infty} e^{- t} lnt dt

= 1 - γ