Question Number 109658 by 150505R last updated on 24/Aug/20
Answered by mathmax by abdo last updated on 25/Aug/20
![A =∫_0 ^(e/π) ((arctan(((πx)/e)))/(πx +e)) dx changement ((πx)/e) =t give A =∫_0 ^1 ((arctan(t))/(et +e)).(e/π) dt =(1/π) ∫_0 ^1 ((arctan(t))/(t+1)) dt ⇒πA =∫_0 ^1 ((arctan(t))/(1+t))dt =_(by parts) [ln(1+t)arctan(t)]_0 ^1 −∫_0 ^1 ((ln(1+t))/(1+t^2 )) dt =(π/4)ln(2)−∫_0 ^1 ((ln(1+t))/(1+t^2 )) dt but ∫_0 ^(1 ) ((ln(1+t))/(1+t^2 ))dt =_(t=tanθ) ∫_0 ^(π/4) ((ln(1+tanθ))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_0 ^(π/4) ln(1+tanθ)dθ=(π/8)ln(2)(proved) ⇒πA =(π/4)ln(2)−(π/8)ln(2) =(π/8)ln(2) ⇒★ A =((ln(2))/8)★](https://www.tinkutara.com/question/Q109800.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{e}}{\pi}} \:\frac{\mathrm{arctan}\left(\frac{\pi\mathrm{x}}{\mathrm{e}}\right)}{\pi\mathrm{x}\:+\mathrm{e}}\:\mathrm{dx}\:\:\:\mathrm{changement}\:\frac{\pi\mathrm{x}}{\mathrm{e}}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{et}\:+\mathrm{e}}.\frac{\mathrm{e}}{\pi}\:\mathrm{dt}\:=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{t}+\mathrm{1}}\:\mathrm{dt}\:\Rightarrow\pi\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=_{\mathrm{by}\:\mathrm{parts}} \:\:\:\left[\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)\mathrm{arctan}\left(\mathrm{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:\:\mathrm{but}\:\int_{\mathrm{0}} ^{\mathrm{1}\:} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=_{\mathrm{t}=\mathrm{tan}\theta} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\theta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)\mathrm{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\theta\right)\mathrm{d}\theta=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)\left(\mathrm{proved}\right) \\ $$$$\Rightarrow\pi\mathrm{A}\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\bigstar\:\mathrm{A}\:=\frac{\mathrm{ln}\left(\mathrm{2}\right)}{\mathrm{8}}\bigstar \\ $$$$ \\ $$