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Question Number 131602 by rs4089 last updated on 06/Feb/21

	Answered by mnjuly1970 last updated on 06/Feb/21

	$\frac{ζ (3)}{8}$
	Answered by mathmax by abdo last updated on 06/Feb/21
	$Φ =∫_0 ^(π/2) ((ln(sinx).ln(cosx))/(tanx))dx let ϕ(a) =∫_0 ^(π/2) ((ln(asinx)ln(cosx))/(tanx))dx ⇒ ϕ^′ (a) =∫_0 ^(π/2) ((sinx)/(asinx)).((ln(cosx))/(tanx))dx =(1/a)∫_0 ^(π/2) ((ln(cosx))/(tanx))dx =(1/a)∫_0 ^(π/2) ((ln((√(1/(1+tan^2 x)))))/(tanx))dx =_(tanx=t) (1/a)∫_0 ^∞ ((ln((1/( (√(1+t^2 ))))))/t)×(dt/(1+t^2 )) =−(1/(2a))∫_0 ^∞ ((ln(1+t^2 ))/(t(1+t^2 )))dt and ∫_0 ^∞ ((ln(1+t^2 ))/(t(1+t^2 )))dt=∫_0 ^1 (...)dt +∫_1 ^∞ (..(√()dt)) ∫_1 ^∞ (...)dt =_(t=(1/u)) −∫_0 ^1 u((ln(1+(1/u^2 )))/((1+(1/u^2 ))))(−(du/u^2 )) =∫_0 ^1 u((ln(1+u^2 )−2lnu)/(u^2 +1))du =∫_0 ^1 (u/(u^2 +1))ln(1+u^2 )du−2∫_0 ^1 ((ulnu)/(u^2 +1))du also by parts ∫_0 ^1 (u/(u^2 +1))ln(1+u^2 )du =[(1/2)ln^2 (1+u^2 )]_0 ^1 −∫_0 ^1 (1/2)ln(1+u^2 ).((2u)/(1+u^2 )) =(1/2)ln^2 (2)−∫_0 ^1 (u/(1+u^2 ))ln(1+u^2 )du ⇒∫_0 ^1 ((uln(1+u^2 ))/(u^2 +1))du=(1/4)ln^2 (2) ∫_0 ^1 ((ulnu)/(u^2 +1))du =[(1/2)ln(1+u^2 )lnu]_0 ^1 −(1/2)∫_0 ^1 ln(1+u^2 )(du/u) =−(1/2)∫_0 ^1 ((ln(1+u^2 ))/u)du we have ln(1+x)=Σ_(n=1) ^∞ (−1)^(n−1) (x^n /n) ⇒ln(1+u^2 )=Σ_(n=1) ^∞ (−1)^(n−1) (u^(2n) /n) ⇒∫_0 ^1 ((ln(1+u^2 ))/u)du =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^1 u^(2n−1) du =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)[(1/(2n))u^(2n) ]_0 ^(1 ) =(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 ) ⇒ ∫_0 ^1 ((ulnu)/(1+u^2 ))du =(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(1/4)(2^(1−2) −1)ξ(2) =(1/4)(−(1/2))ξ(2) =−(1/8).(π^2 /6) =−(π^2 /(48)) ⇒ ϕ^′ (a) =−(1/(2a)){(1/4)ln^2 (2)+(π^2 /(24))} ⇒ ϕ(a)=−((1/8)ln^2 2+(π^2 /(48)))lna +C ⇒ϕ(1)=C rest to find C!! be continued...$
	$Φ = \int_{0}^{\frac{π}{2}} \frac{\ln (sinx) . \ln (cosx)}{tanx} dx let φ (a) = \int_{0}^{\frac{π}{2}} \frac{\ln (asinx) \ln (cosx)}{tanx} dx \Rightarrow$ $φ^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{sinx}{asinx} . \frac{\ln (cosx)}{tanx} dx = \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{\ln (cosx)}{tanx} dx$ $= \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{\ln (\sqrt{\frac{1}{1 + \tan^{2} x}})}{tanx} dx =_{tanx = t} \frac{1}{a} \int_{0}^{\infty} \frac{\ln (\frac{1}{\sqrt{1 + t^{2}}})}{t} \times \frac{dt}{1 + t^{2}}$ $= - \frac{1}{2 a} \int_{0}^{\infty} \frac{\ln (1 + t^{2})}{t (1 + t^{2})} dt and \int_{0}^{\infty} \frac{\ln (1 + t^{2})}{t (1 + t^{2})} dt = \int_{0}^{1} (. . .) dt + \int_{1}^{\infty} (. . \sqrt{) dt}$ $\int_{1}^{\infty} (. . .) dt =_{t = \frac{1}{u}} - \int_{0}^{1} u \frac{\ln (1 + \frac{1}{u^{2}})}{(1 + \frac{1}{u^{2}})} (- \frac{du}{u^{2}})$ $= \int_{0}^{1} u \frac{\ln (1 + u^{2}) - 2 lnu}{u^{2} + 1} du = \int_{0}^{1} \frac{u}{u^{2} + 1} \ln (1 + u^{2}) du - 2 \int_{0}^{1} \frac{ulnu}{u^{2} + 1} du$ $also by parts \int_{0}^{1} \frac{u}{u^{2} + 1} \ln (1 + u^{2}) du$ $= {[\frac{1}{2} \ln^{2} (1 + u^{2})]}_{0}^{1} - \int_{0}^{1} \frac{1}{2} \ln (1 + u^{2}) . \frac{2 u}{1 + u^{2}}$ $= \frac{1}{2} \ln^{2} (2) - \int_{0}^{1} \frac{u}{1 + u^{2}} \ln (1 + u^{2}) du \Rightarrow \int_{0}^{1} \frac{uln (1 + u^{2})}{u^{2} + 1} du = \frac{1}{4} \ln^{2} (2)$ $\int_{0}^{1} \frac{ulnu}{u^{2} + 1} du = {[\frac{1}{2} \ln (1 + u^{2}) lnu]}_{0}^{1} - \frac{1}{2} \int_{0}^{1} \ln (1 + u^{2}) \frac{du}{u}$ $= - \frac{1}{2} \int_{0}^{1} \frac{\ln (1 + u^{2})}{u} du we have$ $\ln (1 + x) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n}}{n} \Rightarrow \ln (1 + u^{2}) = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{u^{2 n}}{n}$ $\Rightarrow \int_{0}^{1} \frac{\ln (1 + u^{2})}{u} du = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} u^{2 n - 1} du$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {[\frac{1}{2 n} u^{2 n}]}_{0}^{1} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} \Rightarrow$ $\int_{0}^{1} \frac{ulnu}{1 + u^{2}} du = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = \frac{1}{4} (2^{1 - 2} - 1) ξ (2)$ $= \frac{1}{4} (- \frac{1}{2}) ξ (2) = - \frac{1}{8} . \frac{π^{2}}{6} = - \frac{π^{2}}{48} \Rightarrow$ $φ^{^{'}} (a) = - \frac{1}{2 a} {\frac{1}{4} \ln^{2} (2) + \frac{π^{2}}{24}} \Rightarrow$ $φ (a) = - (\frac{1}{8} \ln^{2} 2 + \frac{π^{2}}{48}) lna + C \Rightarrow φ (1) = C rest to find C!! be continued . . .$
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