... advanced calculus... prove that:: ∫_0 ^( 1) ln(Γ(x)).cos^2 (πx)dx =((ln(2π))/4)+(1/8) m.n.1970


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Question Number 117280 by mnjuly1970 last updated on 10/Oct/20

$. . . a d v a n c e d c a l c u l u s . . .$ $p r o v e t h a t ::$ $\int_{0}^{1} l n (Γ (x)) . {c o s}^{2} (π x) d x$ $= \frac{l n (2 π)}{4} + \frac{1}{8}$ $m . n . 1970$
	Answered by AbduraufKodiriy last updated on 10/Oct/20

	$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ $I = \int_{0}^{1} l n (Γ (x)) {c o s}^{2} (π x) d x = \int_{0}^{1} l n (Γ (1 - x)) {c o s}^{2} (π (1 - x)) d x =$ $= \int_{0}^{1} l n (Γ (1 - x)) {c o s}^{2} (π x) d x$ $T h e r e f o r e : 2 I = \int_{0}^{1} (l n (Γ (x)) + l n (Γ (1 - x))) {c o s}^{2} (π x) d x \Rightarrow$ $\Rightarrow I = \frac{1}{2} \int_{0}^{1} l n (Γ (x) Γ (1 - x)) {c o s}^{2} (π x) d x = \frac{1}{2} \int_{0}^{1} l n (\frac{π}{s i n (π x)}) {c o s}^{2} (π x) d x =$ $= \frac{l n (π)}{2} \int_{0}^{1} {c o s}^{2} (π x) d x - \frac{1}{2} \int {c o s}^{2} (π x) l n (s i n (π x)) d x;$ $I_{1} = \int_{0}^{1} {c o s}^{2} (π x) l n (s i n (π x)) d x \Rightarrow I = \frac{l n (π)}{4} - \frac{1}{2} I_{1}$ $I_{1} = \| \begin{matrix} π x = t \Rightarrow x = \frac{t}{π} \\ d x = \frac{1}{π} d t \end{matrix} \| = \frac{1}{π} \int_{0}^{π} {c o s}^{2} (t) l n (s i n (t)) d t =$ $= \frac{1}{2 π} \int_{0}^{π} l n (s i n (t)) d t + \frac{1}{2 π} \int_{0}^{π} c o s (2 t) l n (s i n (t)) d t; W e k n o w : \int_{0}^{π} l n (s i n (t)) d t = - π l n 2$ $I_{1} = - \frac{1}{2} l n (2) + \frac{1}{2} \int_{0}^{π} c o s (2 t) l n (s i n (t)) d t = - \frac{1}{2} l n 2 + (\frac{s i n (2 t) l n (s i n (t))}{4 π} - \frac{t}{4 π} - \frac{s i n (2 t)}{8 π}) ∣_{0}^{π} =$ $= - \frac{1}{2} l n (2) - \frac{1}{4} \Rightarrow I = \frac{l n (π)}{4} + \frac{l n (2)}{4} + \frac{1}{8} = \frac{l n (2 π)}{4} + \frac{1}{8}$
	Commented by AbduraufKodiriy last updated on 10/Oct/20

	Commented by mnjuly1970 last updated on 10/Oct/20

	$t h a n y o u m r a b d u r a u f k . .$
	Commented by mnjuly1970 last updated on 10/Oct/20

	$j a v o b = j a v a b = a n s w e r . . .$
	Commented by mnjuly1970 last updated on 10/Oct/20

	$m e r c e y a g h a y e a b d u r a u f k$
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