Differentiation Questions

Question Number 151118 by mnjuly1970 last updated on 18/Aug/21

$$\:\: \\$$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+{arctan}\left(\mathrm{2}{x}\right)}{{x}^{\:\mathrm{2}} +\mathrm{1}}\overset{?} {=}\frac{\pi^{\:\mathrm{2}} }{\mathrm{4}} \\$$$$\\$$

Answered by Olaf_Thorendsen last updated on 18/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{arctan}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\$$$$\mathrm{Let}\:{u}\:=\:\frac{\mathrm{1}}{{x}}\:: \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{u}}\right)+\mathrm{arctan}\left(\frac{\mathrm{2}}{{u}}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}\:{du} \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{2}{u}\right)+\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\frac{{u}}{\mathrm{2}}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}\:{du} \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\pi}{{u}^{\mathrm{2}} +\mathrm{1}}\:{du}−\mathrm{I} \\$$$$\mathrm{2I}\:=\:\pi\left[\mathrm{arctan}{u}\right]_{\mathrm{0}} ^{\infty} \:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\$$$$\mathrm{I}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\$$

Commented by puissant last updated on 18/Aug/21

$${monsieur}\:{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+{arctan}\left(\mathrm{2}{x}\right)=\frac{\pi}{\mathrm{2}} \\$$$${avec}\:{ca}\:{on}\:{avait}\:{directement}\:{le}\: \\$$$${resultat}.. \\$$

Commented by Olaf_Thorendsen last updated on 18/Aug/21

$$\mathrm{Non}. \\$$$$\mathrm{arctan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{arctan}\left(\frac{\mathrm{2}}{{x}}\right)\:=\:\frac{\pi}{\mathrm{2}}. \\$$$$\mathrm{Deja}\:\mathrm{quand}\:{x}\:\mathrm{tend}\:\mathrm{vers}\:\mathrm{l}'\mathrm{infini} \\$$$$\mathrm{par}\:\mathrm{exemple},\:\mathrm{vous}\:\mathrm{voyez}\:\mathrm{bien}\:\mathrm{que} \\$$$$\mathrm{votre}\:\mathrm{formule}\:\mathrm{est}\:\mathrm{fausse}. \\$$

Commented by mnjuly1970 last updated on 18/Aug/21

$$\:\:\:\:\:\:{very}\:{nice}\:{mr}\:{olaf}... \\$$

Commented by puissant last updated on 18/Aug/21

$${oui}\:{monsieur}\:{vous}\:{avez}\:{raison}\:{car} \\$$$${arctan}\left({x}\right)+{arctan}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\pi}{\mathrm{2}}{sgn}\left({x}\right).. \\$$