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

$e v a l u a t e ::$ $Φ := \int_{0}^{\infty} {x e}^{- \frac{x^{2}}{4}} l n (x) d x = m . (π γ)$ $f i n d^{″} m^{″} . . . . . .$
	Answered by qaz last updated on 07/May/21
	$φ(a)=∫_0 ^∞ x^a e^(−(x^2 /4)) dx=((Γ(((a+1)/2)))/(2((1/4))^((a+1)/2) ))=2^a Γ(((a+1)/2)) Φ=φ(1)′={2^a ln2Γ(((a+1)/2))+2^a Γ(((a+1)/2))ψ(((a+1)/2))∙(1/2)}_(a=1) =2ln2+ψ(1) ⇒m=((2ln2−γ)/(πγ))$
	$ϕ (a) = \int_{0}^{\infty} x^{a} e^{- \frac{x^{2}}{4}} d x = \frac{Γ (\frac{a + 1}{2})}{2 {(\frac{1}{4})}^{\frac{a + 1}{2}}} = 2^{a} Γ (\frac{a + 1}{2})$ $Φ = ϕ {(1)}^{'} = {2^{a} l n 2 Γ (\frac{a + 1}{2}) + 2^{a} Γ (\frac{a + 1}{2}) ψ (\frac{a + 1}{2}) \cdot \frac{1}{2}}_{a = 1}$ $= 2 l n 2 + ψ (1)$ $\Rightarrow m = \frac{2 l n 2 - γ}{π γ}$
	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

	$i e d i t e d a g a i n . . . .$
	Commented by mnjuly1970 last updated on 07/May/21

	$g r a t e f u l . . . .$
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