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Question Number 162864 by mathacek last updated on 01/Jan/22

	Answered by Ar Brandon last updated on 01/Jan/22
	$I=∫_0 ^π xln(sinx)dx=∫_0 ^π (π−x)ln(sinx)dx =π∫_0 ^π ln(sinx)dx−∫_0 ^π xln(sinx)dx I=(π/2)∫_0 ^π ln(sinx)dx , x=2u ⇒dx=2du =π∫_0 ^(π/2) ln(sin(2u))dx=π∫_0 ^(π/2) ln(2sinucosu)du =π∫_0 ^(π/2) ln(2)du+π∫_0 ^(π/2) ln(sinu)du+π∫_0 ^(π/2) ln(cosu)du =((π^2 ln2)/2)+2π∫_0 ^(π/2) ln(sinu)du, {∫_0 ^(π/2) ln(sinu)du=∫_0 ^(π/2) ln(cosu)du} =((π^2 ln2)/2)−π^2 ln2=−(1/2)π^2 ln2$
	$I = \int_{0}^{π} x \ln (\sin x) d x = \int_{0}^{π} (π - x) \ln (\sin x) d x$ $= π \int_{0}^{π} \ln (\sin x) d x - \int_{0}^{π} x \ln (\sin x) d x$ $I = \frac{π}{2} \int_{0}^{π} \ln (\sin x) d x, x = 2 u \Rightarrow d x = 2 d u$ $= π \int_{0}^{\frac{π}{2}} \ln (\sin (2 u)) d x = π \int_{0}^{\frac{π}{2}} \ln (2 \sin u \cos u) d u$ $= π \int_{0}^{\frac{π}{2}} \ln (2) d u + π \int_{0}^{\frac{π}{2}} \ln (\sin u) d u + π \int_{0}^{\frac{π}{2}} \ln (\cos u) d u$ $= \frac{π^{2} \ln 2}{2} + 2 π \int_{0}^{\frac{π}{2}} \ln (\sin u) d u, {\int_{0}^{\frac{π}{2}} \ln (\sin u) d u = \int_{0}^{\frac{π}{2}} \ln (\cos u) d u}$ $= \frac{π^{2} \ln 2}{2} - π^{2} \ln 2 = - \frac{1}{2} π^{2} \ln 2$
	Commented by Ar Brandon last updated on 01/Jan/22

	$I = \int_{0}^{\frac{π}{2}} \ln (\sin x) d x = \int_{0}^{\frac{π}{2}} \ln (\cos x) d x$ $2 I = \int_{0}^{\frac{π}{2}} \ln (\sin x \cos x) d x = \int_{0}^{\frac{π}{2}} \ln (\frac{1}{2} \sin (2 x)) d x$ $= \int_{0}^{\frac{π}{2}} \ln (\frac{1}{2}) d x + \int_{0}^{\frac{π}{2}} \ln (\sin 2 x) d x$ $= - \frac{π \ln 2}{2} + \frac{1}{2} \int_{0}^{π} \ln (\sin u) d u$
	Answered by Lordose last updated on 01/Jan/22

	$I = \int_{0}^{π} xln (\sin (x)) dx \overset{{King}^{'} s Rule}{=} \int_{0}^{π} (π - x) \ln (\sin (x)) dx$ $2 I = π \int_{0}^{π} \ln (\sin (x)) dx \overset{u = 2 x}{=} \frac{π}{2} \int_{0}^{2 π} \ln (\sin (\frac{u}{2})) dx$ $I = \frac{π}{4} (- {Cl}_{2} (2 π) - 2 π \ln (2))$ $I = - \frac{π^{2} \ln (2)}{2}$
	Commented by mathacek last updated on 02/Jan/22

	${W h a t}^{'} s f o r {Cl}_{2} ?$
	Commented by Ar Brandon last updated on 02/Jan/22

	$Clausen function$ ${C l}_{2} (z) = \sum_{n ⩾ 1} \frac{\sin (n z)}{n^{2}}$
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