Question Number 106098 by redmiiuser last updated on 02/Aug/20
$$\int_{\frac{\mathrm{1}}{\mathrm{2014}}} ^{\mathrm{2014}} \frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}}\:{dx} \\ $$
Answered by abdomathmax last updated on 02/Aug/20
![let find f(a) =∫_(1/a) ^a ((arctanx)/x)dx with a>0 changement x=(1/t) give f(a) =−∫_(1/a) ^a ((arctan((1/t)))/t^2 )(−t)dt =∫_(1/a) ^a (((π/2)−arctan(t))/t)dt =(π/2) ∫_(1/a) ^a (dt/t)−f(a) ⇒ 2f(a) =(π/2)[ln∣t∣]_(1/a) ^(a ) =(π/2){ln(a)+ln(a)} =πln(a) f(a) =(π/2)ln(a) ⇒∫_(1/(2014)) ^(2014) ((arctanx)/x)dx =(π/2)ln(2014)](Q106113.png)
$$\mathrm{let}\:\mathrm{find}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\mathrm{arctanx}}{\mathrm{x}}\mathrm{dx}\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0} \\ $$$$\mathrm{changement}\:\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=−\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{t}}\right)}{\mathrm{t}^{\mathrm{2}} }\left(−\mathrm{t}\right)\mathrm{dt} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{t}\right)}{\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}} \:\frac{\mathrm{dt}}{\mathrm{t}}−\mathrm{f}\left(\mathrm{a}\right)\:\Rightarrow \\ $$$$\mathrm{2f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}\left[\mathrm{ln}\mid\mathrm{t}\mid\right]_{\frac{\mathrm{1}}{\mathrm{a}}} ^{\mathrm{a}\:} \:=\frac{\pi}{\mathrm{2}}\left\{\mathrm{ln}\left(\mathrm{a}\right)+\mathrm{ln}\left(\mathrm{a}\right)\right\}\:=\pi\mathrm{ln}\left(\mathrm{a}\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{a}\right)\:\Rightarrow\int_{\frac{\mathrm{1}}{\mathrm{2014}}} ^{\mathrm{2014}} \:\:\frac{\mathrm{arctanx}}{\mathrm{x}}\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2014}\right) \\ $$
Commented by redmiiuser last updated on 02/Aug/20
$${perfect} \\ $$
Commented by Aziztisffola last updated on 02/Aug/20
$$\mathrm{Nice}\:\mathrm{method}\:\mathrm{Sir}. \\ $$
Commented by mathmax by abdo last updated on 02/Aug/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}. \\ $$