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1-2014-2014-tan-1-x-x-dx-




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

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