Question Number 72111 by A8;15: last updated on 24/Oct/19
Commented by mathmax by abdo last updated on 24/Oct/19
![let I=∫_0 ^(e/π) ((arctan(((πx)/e)))/(πx +e))dx changement ((πx)/e) =t give I =∫_0 ^1 ((arctant)/(π(((et)/π))+e))(e/π)dt =(e/π)∫_0 ^1 ((arctan(t))/(e(t+1)))dt =(1/π) ∫_0 ^1 ((arctan(t))/(t+1))dt by parts u^′ =(1/(t+1)) and v=arctan(t) ⇒ ∫_0 ^1 ((arctan(t))/(t+1))dt =[ln(t+1)arctant]_0 ^1 −∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt =(π/4)ln(2)−∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt but ∫_0 ^1 ((ln(t+1))/(1+t^2 ))dt =_(t=tanθ) ∫_0 ^(π/4) ((ln(tanθ +1))/(1+tan^2 θ))(1+tan^2 θ)dθ =∫_0 ^(π/4) ln(((sinθ)/(cosθ))+1)dθ =∫_0 ^(π/4) ln(cosθ +sinθ)−ln(cosθ)}dθ =∫_0 ^(π/4) ln((√2)cos((π/4)−θ))−∫_0 ^(π/4) ln(cosθ)dθ =_((π/4)−θ=u) −∫_0 ^(π/4) { ((1/2)(π/4))ln(2)+ln(cosu)}(−du)−∫_0 ^(π/4) ln(cosu)du =−(π/8)ln(2) ⇒∫_0 ^1 ((arctan(t))/(1+t^2 ))dt =(π/4)ln2−(π/8)ln(2) ⇒I=((ln(2))/8)](https://www.tinkutara.com/question/Q72121.png)
$${let}\:{I}=\int_{\mathrm{0}} ^{\frac{{e}}{\pi}} \:\frac{{arctan}\left(\frac{\pi{x}}{{e}}\right)}{\pi{x}\:+{e}}{dx}\:{changement}\:\frac{\pi{x}}{{e}}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctant}}{\pi\left(\frac{{et}}{\pi}\right)+{e}}\frac{{e}}{\pi}{dt}\:=\frac{{e}}{\pi}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctan}\left({t}\right)}{{e}\left({t}+\mathrm{1}\right)}{dt}\:=\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}\right)}{{t}+\mathrm{1}}{dt} \\ $$$${by}\:{parts}\:{u}^{'} =\frac{\mathrm{1}}{{t}+\mathrm{1}}\:{and}\:{v}={arctan}\left({t}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}\right)}{{t}+\mathrm{1}}{dt}\:=\left[{ln}\left({t}+\mathrm{1}\right){arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}+\mathrm{1}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}+\mathrm{1}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}+\mathrm{1}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=_{{t}={tan}\theta} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({tan}\theta\:+\mathrm{1}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$\left.=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\frac{{sin}\theta}{{cos}\theta}+\mathrm{1}\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\:+{sin}\theta\right)−{ln}\left({cos}\theta\right)\right\}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\sqrt{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}−\theta\right)\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\theta\right){d}\theta \\ $$$$=_{\frac{\pi}{\mathrm{4}}−\theta={u}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left\{\:\left(\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{4}}\right){ln}\left(\mathrm{2}\right)+{ln}\left({cosu}\right)\right\}\left(−{du}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cosu}\right){du} \\ $$$$=−\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\frac{\pi}{\mathrm{4}}{ln}\mathrm{2}−\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\Rightarrow{I}=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$