evaluate-0-xe-x-2-4-ln-x-dx-m-pi-find-m-

Question Number 140401 by mnjuly1970 last updated on 07/May/21

e v a l u a t e ::

Φ := \int_{0}^{\infty} {x e}^{- \frac{x^{2}}{4}} l n (x) d x = m . (π γ)

f i n d^{″} m^{″} \dots \dots

Answered by qaz last updated on 07/May/21

ϕ (a) = \int_{0}^{\infty} x^{a} e^{- \frac{x^{2}}{4}} d x = \frac{Γ (\frac{a + 1}{2})}{2 {(\frac{1}{4})}^{\frac{a + 1}{2}}} = 2^{a} Γ (\frac{a + 1}{2})

Φ = ϕ {(1)}^{'} = {2^{a} l n 2 Γ (\frac{a + 1}{2}) + 2^{a} Γ (\frac{a + 1}{2}) ψ (\frac{a + 1}{2}) \cdot \frac{1}{2}}_{a = 1}

= 2 l n 2 + ψ (1)

\Rightarrow m = \frac{2 l n 2 - γ}{π γ}

Commented by mnjuly1970 last updated on 07/May/21

t h a n k s a l o t m r q a z . .

Answered by mnjuly1970 last updated on 07/May/21

{(\frac{x}{2})}^{2} = y \Rightarrow x = 2 \sqrt{y}

Φ = \int_{0}^{\infty} 2 \sqrt{y} e^{- y} l n (2 \sqrt{y}) \frac{d y}{\sqrt{y}}

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

= l n (4) - γ

l n (4) - γ = m (π γ)

m = \frac{l n (4) - γ}{π γ}

Commented by qaz last updated on 07/May/21

i e d i t e d a g a i n \dots .

Commented by mnjuly1970 last updated on 07/May/21

g r a t e f u l \dots .