∫_0 ^1 log(tanθ)dθ


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Question Number 105940 by Dwaipayan Shikari last updated on 01/Aug/20

$\int_{0}^{1} l o g (t a n θ) d θ$
	Answered by mathmax by abdo last updated on 01/Aug/20
	$A =∫_0 ^1 ln(tanθ)dθ by parts A = [θln(tanθ)]_0 ^1 −∫_0 ^1 θ×((1+tan^2 θ)/(tanθ)) dθ =ln((π/4))−∫_0 ^1 θ{(1/(tanθ)) +tanθ}dθ changement tanθ =x give ∫_0 ^1 θ{(1/(tanθ)) +tanθ }dθ =∫_0 ^(tan(1)) arctanx{(1/x) +x}(dx/(1+x^2 )) =∫_0 ^(tan(1)) ((1+x^2 )/x) ×((arctanx)/(1+x^2 ))dx =∫_0 ^(tan(1)) ((arctan(x))/x)dx let f(a) =∫_0 ^(tan(1)) ((arctan(αx))/x)dx ⇒f^′ (a) =∫_0 ^(tan(1)) (x/((1+α^2 x^2 )x))dx =∫_0 ^(tan(1)) (dx/(1+α^2 x^2 )) =_(αx =z) ∫_0 ^(αtan(1)) (dz/(α(1+z^2 ))) =(1/α)[arctanz]_0 ^(αtan(1)) =((arctan(αtan(1)))/α) ⇒f(α) =∫_0 ^α ((arctan(utan(1)))/u) du + C and ∫_0 ^(tan(1)) ((arctanx)/x)dx =f(1) =∫_0 ^1 ((arctan(utan(1)))/u) du +C ..be continued...$
	$A = \int_{0}^{1} \ln (\tan θ) d θ by parts A = {[θ \ln (\tan θ)]}_{0}^{1} - \int_{0}^{1} θ \times \frac{1 + \tan^{2} θ}{\tan θ} d θ$ $= \ln (\frac{π}{4}) - \int_{0}^{1} θ {\frac{1}{\tan θ} + \tan θ} d θ changement \tan θ = x give$ $\int_{0}^{1} θ {\frac{1}{\tan θ} + \tan θ} d θ = \int_{0}^{\tan (1)} arctanx {\frac{1}{x} + x} \frac{dx}{1 + x^{2}}$ $= \int_{0}^{\tan (1)} \frac{1 + x^{2}}{x} \times \frac{arctanx}{1 + x^{2}} dx = \int_{0}^{\tan (1)} \frac{\arctan (x)}{x} dx$ $let f (a) = \int_{0}^{\tan (1)} \frac{\arctan (α x)}{x} dx \Rightarrow f^{^{'}} (a) = \int_{0}^{\tan (1)} \frac{x}{(1 + α^{2} x^{2}) x} dx$ $= \int_{0}^{\tan (1)} \frac{dx}{1 + α^{2} x^{2}} =_{α x = z} \int_{0}^{α \tan (1)} \frac{dz}{α (1 + z^{2})} = \frac{1}{α} {[arctanz]}_{0}^{α \tan (1)}$ $= \frac{\arctan (α \tan (1))}{α} \Rightarrow f (α) = \int_{0}^{α} \frac{\arctan (utan (1))}{u} du + C$ $and \int_{0}^{\tan (1)} \frac{arctanx}{x} dx = f (1) = \int_{0}^{1} \frac{\arctan (utan (1))}{u} du + C$ $. . be continued . . .$
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