∫_0 ^(π/2) ln(sinx)ln(cosx)dx


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Question Number 160358 by Ar Brandon last updated on 28/Nov/21

$\int_{0}^{\frac{π}{2}} \ln (\sin x) \ln (\cos x) d x$
	Answered by amin96 last updated on 28/Nov/21
	$sin(x)=t (dt/dx)=cos(x)=(√(1−t^2 )) 𝛀=∫_0 ^1 ((ln(t)ln((√(1−t^2 ))))/( (√(1−t^2 ))))dt=(1/2)∫_0 ^1 ((ln(t)ln(1−t^2 ))/( (√(1−t^2 ))))dt t^2 =x (dx/dt)=2t=2(√x) 𝛀=(1/4)∫_0 ^1 ((ln((√x))ln(1−x))/( (√x) (√(1−x))))dx=(1/8)∫_0 ^1 ((((√(1−x)))ln(x)ln(1−x))/( (√x)×(1−x)))dx= =−(1/8)Σ_(n=1) ^∞ H_n ∫_0 ^1 x^(n−(1/2)) (1−x)^(1/2) ln(x)dx= =−(1/8)ΣH_n (Σ(((−1)^(n+1) )/n))∫_0 ^1 x^(n−(1/2)) (1−x)^(1/2) (x−1)^n dx= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^(n−(1/2)) (1−x)^(n+(1/2)) dx}= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^((n+(1/2))−1) (1−x)^((n+(3/2))−1) dx}= =(1/8)Σ_(n=1) ^∞ H_n {Σ_(n=1) ^∞ (1/n)𝛃(n+(1/2); n+(3/2))}$
	$\sin (x) = t \frac{dt}{dx} = \cos (x) = \sqrt{1 - t^{2}}$ $Ω = \int_{0}^{1} \frac{\ln (t) \ln (\sqrt{1 - t^{2}})}{\sqrt{1 - t^{2}}} dt = \frac{1}{2} \int_{0}^{1} \frac{\ln (t) \ln (1 - t^{2})}{\sqrt{1 - t^{2}}} dt$ $t^{2} = x \frac{dx}{dt} = 2 t = 2 \sqrt{x}$ $Ω = \frac{1}{4} \int_{0}^{1} \frac{\ln (\sqrt{x}) \ln (1 - x)}{\sqrt{x} \sqrt{1 - x}} dx = \frac{1}{8} \int_{0}^{1} \frac{(\sqrt{1 - x}) \ln (x) \ln (1 - x)}{\sqrt{x} \times (1 - x)} dx =$ $= - \frac{1}{8} \underset{n = 1}{\sum^{\infty}} H_{n} \int_{0}^{1} x^{n - \frac{1}{2}} {(1 - x)}^{\frac{1}{2}} \ln (x) dx =$ $= - \frac{1}{8} Σ H_{n} (Σ \frac{{(- 1)}^{n + 1}}{n}) \int_{0}^{1} x^{n - \frac{1}{2}} {(1 - x)}^{\frac{1}{2}} {(x - 1)}^{n} dx =$ $= \frac{1}{8} \underset{n = 1}{\sum^{\infty}} H_{n} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{n - \frac{1}{2}} {(1 - x)}^{n + \frac{1}{2}} dx} =$ $= \frac{1}{8} \underset{n = 1}{\sum^{\infty}} H_{n} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{(n + \frac{1}{2}) - 1} {(1 - x)}^{(n + \frac{3}{2}) - 1} dx} =$ $= \frac{1}{8} \underset{n = 1}{\sum^{\infty}} H_{n} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} β (n + \frac{1}{2}; n + \frac{3}{2})}$
	Commented by Ar Brandon last updated on 28/Nov/21

	$I wish it is further simplified .$
	Commented by amin96 last updated on 28/Nov/21

	do you know the answer?
	Commented by Ar Brandon last updated on 29/Nov/21

	$Yeah$
	Commented by Ar Brandon last updated on 29/Nov/21

	$\frac{π}{2} \ln^{2} 2 - \frac{π^{3}}{48}$
	Commented by amin96 last updated on 29/Nov/21

	$\int_{0}^{\frac{π}{2}} xln (sinx) \ln (cosx) dx$ The answer to this integral
	Commented by Ar Brandon last updated on 29/Nov/21

	$\int_{0}^{\frac{π}{2}} x \ln (\sin x) \ln (\cos x) d x = \frac{π^{2}}{8} \ln^{2} 2 - \frac{π^{4}}{192}$ $\int_{0}^{\frac{π}{2}} x \ln (\sin x) \ln (\cos x) d x = \frac{π}{4} \int_{0}^{\frac{π}{2}} \ln (\sin x) \ln (\cos x) d x$
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