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Question Number 85721 by M±th+et£s last updated on 24/Mar/20

s h o w t h a t

\int_{0}^{\infty} \frac{e^{- x} l n (x)}{\sqrt{x}} d x = - \sqrt{π} (γ + l n (4))

Answered by mind is power last updated on 24/Mar/20

\int_{0}^{+ \infty} \frac{e^{- x} l n (x)}{\sqrt{x}} d x

\sqrt{x} = u

\Rightarrow \int_{0}^{+ \infty} 4 e^{- u^{2}} l n (u) d u = 4 \int_{0}^{+ \infty} e^{- u^{2}} l n (u) d u

\int_{0}^{+ \infty} t^{x - 1} e^{- t} d t = Γ (x)

t = u^{2} \Rightarrow d t = 2 u d u

2 \int_{0}^{+ \infty} u^{2 x - 1} e^{- u^{2}} = Γ (x) \Rightarrow Γ^{'} (x) = 2 \int 2 l n (u) u^{2 x - 1} e^{- u^{2}} d u = Γ^{'} (x)

\Rightarrow 4 \int_{0}^{+ \infty} l n (u) e^{- u^{2}} d u = Γ^{'} (\frac{1}{2}) = Ψ (\frac{1}{2}) Γ (\frac{1}{2})

Ψ (\frac{1}{2}) = - γ - l n (2.2) = - γ - l n (4)

Γ (\frac{1}{2}) = \sqrt{π} \Rightarrow

\int_{0}^{+ \infty} \frac{e^{- x} l n (x)}{\sqrt{x}} d x = 4 \int_{0}^{+ \infty} l n (u) e^{- u^{2}} d u = \sqrt{π} (- γ - l n (4)) = - \sqrt{π} (γ + l n (4))

Commented by M±th+et£s last updated on 24/Mar/20

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