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Question Number 90158 by I want to learn more last updated on 21/Apr/20

Commented by I want to learn more last updated on 21/Apr/20

$The question is prove that$
Commented by abdomathmax last updated on 22/Apr/20
$we have Γ(x).Γ(1−x)=(π/(sin(πx))) ⇒ ∫_0 ^1 lnΓ(x)dx +∫_0 ^1 lnΓ(1−x)dx =∫^1 _0 ln((π/(sin(πx))))dx but ∫_0 ^1 lnΓ(1−x)dx =_(1−x=t) dt∫_0 ^1 lnΓ(t)dt ⇒ 2∫_0 ^1 ln(Γ(x))dx =ln(π)−∫_0 ^1 ln(sin(πx))dx ∫_0 ^1 ln(sin(πx))dx =_(πx=t) (1/π)∫_0 ^π ln(sint) dt =(1/π) { ∫_0 ^(π/2) ln(sint)dt +∫_(π/2) ^π ln(sint)dt→(t=(π/2)+u)} =(1/π){ ∫_0 ^(π/2) ln(sint)dt +∫_0 ^(π/2) ln(cosu)du} =(1/π){−(π/2)ln(2)−(π/2)ln(2) =(1/π)(−πln(2)) −ln(2) ⇒2∫_0 ^1 ln(Γ(x)dx =ln(π)+ln(2)=ln(2π) ⇒ ∫_0 ^1 ln(Γ(x))dx =(1/2)ln(2π)=ln((√(2π)))$
$w e h a v e Γ (x) . Γ (1 - x) = \frac{π}{s i n (π x)} \Rightarrow$ $\int_{0}^{1} l n Γ (x) d x + \int_{0}^{1} l n Γ (1 - x) d x = {\int_{0}}^{1} l n (\frac{π}{s i n (π x)}) d x$ $b u t \int_{0}^{1} l n Γ (1 - x) d x =_{1 - x = t} d t \int_{0}^{1} l n Γ (t) d t \Rightarrow$ $2 \int_{0}^{1} l n (Γ (x)) d x = l n (π) - \int_{0}^{1} l n (s i n (π x)) d x$ $\int_{0}^{1} l n (s i n (π x)) d x =_{π x = t} \frac{1}{π} \int_{0}^{π} l n (s i n t) d t$ $= \frac{1}{π} {\int_{0}^{\frac{π}{2}} l n (s i n t) d t + \int_{\frac{π}{2}}^{π} l n (s i n t) d t \to (t = \frac{π}{2} + u)}$ $= \frac{1}{π} {\int_{0}^{\frac{π}{2}} l n (s i n t) d t + \int_{0}^{\frac{π}{2}} l n (c o s u) d u}$ $= \frac{1}{π} {- \frac{π}{2} l n (2) - \frac{π}{2} l n (2) = \frac{1}{π} (- π l n (2))$ $- l n (2) \Rightarrow 2 \int_{0}^{1} l n (Γ (x) d x = l n (π) + l n (2) = l n (2 π) \Rightarrow$ $\int_{0}^{1} l n (Γ (x)) d x = \frac{1}{2} l n (2 π) = l n (\sqrt{2 π})$
	Answered by maths mind last updated on 21/Apr/20

	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ $\int_{0}^{1} l n (Γ (x)) d x = \int_{0}^{1} l n (Γ (1 - x)) d x$ $\Rightarrow \int_{0}^{1} l n (Γ (x)) d x = \frac{1}{2} \int_{0}^{1} l n (Γ (1 - x) Γ (x)) d x$ $= \frac{1}{2} \int_{0}^{1} l n (\frac{π}{s i n (π x)}) d x = l n (\sqrt{π}) - \frac{1}{2} \int_{0}^{1} l n (s i n (π x)) d x . . E$ $\int_{0}^{1} l n (s i n (π x)) d x$ $= \frac{1}{π} \int_{0}^{π} l n (s i n (r)) d r$ $\int_{0}^{π} l n (s i n (r)) d r = \int_{0}^{\frac{π}{2}} l n (\frac{s i n (2 r)}{2}) d r$ $= \int_{0}^{π} l n (s i n (x)) \frac{d x}{2} - \frac{π}{2} l n (2)$ $\Rightarrow \int_{0}^{π} l n (s i n (x)) d x = - π l n (2)$ $E = l n (\sqrt{π}) - \frac{1}{2 π} . - π l n (2) = l n (\sqrt{π}) + l n (\sqrt{2})$ $= l n (\sqrt{2 π}) = \int_{0}^{1} l n (Γ (x)) d x$
	Commented by I want to learn more last updated on 21/Apr/20

	$Thanks sir$
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