Question Number 31098 by abdo imad last updated on 02/Mar/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{1}} ^{\infty} \:\:\frac{{arctan}\left({x}+\mathrm{1}\right)\:−{arctanx}}{{x}^{\mathrm{2}} }{dx}. \\ $$
Commented by abdo imad last updated on 04/Mar/18
![let put I(ξ)= ∫_1 ^ξ ((arctan(x+1)−arctanx)/x^2 )dx we have I=lim_(ξ→+∞) I(ξ) I(ξ)= ∫_2 ^(ξ+1) ((arctant)/((t−1)^2 ))dt −∫_1 ^ξ ((arctanx)/x^2 )dt let integrate by psrts ∫_2 ^(1+ξ) ((arctant)/((t−1)^2 ))dt=[ ((−1)/(t−1))arctant]_2 ^(1+ξ) −∫_2 ^(1+ξ) ((−1)/(t−1)) (dt/(1+t^2 )) =arctan2 −(1/ξ)arctan(1+ξ) +∫_2 ^(1+ξ) (dt/((t−1)(1+t^2 ))) →arctan2 +∫_2 ^(+∞) (dt/((t−1)(1+t^2 ))) but ∫_2 ^(+∞) (dt/((t−1)(1+t^2 ))) =∫_1 ^(+∞) (dx/(x(1+(x+1)^2 )))=∫_1 ^(+∞) (dx/(x(x^2 +2x +2))) (1/(x(x^2 +2x+2)))= (a/x) + ((bx+c)/(x^2 +2x+2))=f(x) a=lim_(x→0) xf(x)=(1/2) ,lim_(x→∞) xf(x)=0=a+b⇒b=−(1/2)so f(x)= (1/(2x)) +((((−1)/2)x +c)/(x^2 +2x +2)) we have f(1)=(1/5)=(1/2) +(1/5)(−(1/2)+c) ⇒1=(5/2) −(1/2) +c ⇒1=2+c ⇒c=−1 and ∫_1 ^(+∞) (dx/(x(x^2 +2x+2)))=∫_1 ^(+∞) ((1/(2x)) −(1/2) ((x+2)/(x^2 +2x +2))) =∫_1 ^(+∞) (dx/(2x)) −(1/4)∫_1 ^(+∞) ((2x+4)/(x^2 +2x+2))dx =∫_1 ^(+∞) (dx/(2x)) −(1/4)∫_1 ^(+∞) ((2x+2)/(x^2 +2x+2))dx −(1/2) ∫_1 ^(+∞) (dx/(x^2 +2x+2)) =[(1/2)ln∣x∣−(1/4)ln∣x^2 +2x+2∣]_1 ^(+∞) −(1/2)∫_1 ^(+∞) (dx/((x+1)^2 +1)) =[ln(((√(∣x∣))/(√(x^2 +2x+2)))4](Q31238.png)
$${let}\:{put}\:{I}\left(\xi\right)=\:\int_{\mathrm{1}} ^{\xi} \:\frac{{arctan}\left({x}+\mathrm{1}\right)−{arctanx}}{{x}^{\mathrm{2}} }{dx} \\ $$$${we}\:{have}\:{I}={lim}_{\xi\rightarrow+\infty} {I}\left(\xi\right) \\ $$$${I}\left(\xi\right)=\:\int_{\mathrm{2}} ^{\xi+\mathrm{1}} \:\frac{{arctant}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}\:−\int_{\mathrm{1}} ^{\xi} \:\frac{{arctanx}}{{x}^{\mathrm{2}} }{dt}\:{let}\:{integrate}\:{by}\:{psrts} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{1}+\xi} \:\frac{{arctant}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}=\left[\:\frac{−\mathrm{1}}{{t}−\mathrm{1}}{arctant}\right]_{\mathrm{2}} ^{\mathrm{1}+\xi} \:\:−\int_{\mathrm{2}} ^{\mathrm{1}+\xi} \frac{−\mathrm{1}}{{t}−\mathrm{1}}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$={arctan}\mathrm{2}\:−\frac{\mathrm{1}}{\xi}{arctan}\left(\mathrm{1}+\xi\right)\:+\int_{\mathrm{2}} ^{\mathrm{1}+\xi} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\rightarrow{arctan}\mathrm{2}\:+\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:{but} \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{{x}\left(\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)}=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)}=\:\frac{{a}}{{x}}\:+\:\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}={f}\left({x}\right) \\ $$$${a}={lim}_{{x}\rightarrow\mathrm{0}} {xf}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:,{lim}_{{x}\rightarrow\infty} {xf}\left({x}\right)=\mathrm{0}={a}+{b}\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{2}}{so} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:+\frac{\frac{−\mathrm{1}}{\mathrm{2}}{x}\:+{c}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}\:{we}\:{have}\:{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{5}}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{c}\right) \\ $$$$\Rightarrow\mathrm{1}=\frac{\mathrm{5}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow\mathrm{1}=\mathrm{2}+{c}\:\Rightarrow{c}=−\mathrm{1}\:{and} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)}=\int_{\mathrm{1}} ^{+\infty} \:\left(\frac{\mathrm{1}}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}\right) \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{1}} ^{+\infty} \:\frac{\mathrm{2}{x}+\mathrm{4}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{1}} ^{+\infty} \frac{\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\mid\right]_{\mathrm{1}} ^{+\infty} \:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\left[{ln}\left(\frac{\sqrt{\mid{x}\mid}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}}\mathrm{4}\right.\right. \\ $$
Commented by abdo imad last updated on 04/Mar/18
![=−ln((^4 (√5))^(−1) ) −(1/2)∫_2 ^(+∞) (dt/(1+t^2 )) =(1/4)ln(5) −(1/2) [arctant]_2 ^(+∞) =(1/4)ln(5)−(1/2)((π/2) −arctan2) =(1/4)ln(5) −(π/4) +(1/2) arctan2 ∫_1 ^ξ ((arctanx)/x^2 ) =[((−1)/x) arctanx]_1 ^ξ − ∫_1 ^ξ ((−1)/x) (dx/(1+x^2 ))→ (π/4) +∫_1 ^(+∞) (dx/(x(1+x^2 )))=(π/4) +∫_1 ^(+∞) ((1/x) −(x/(1+x^2 )))dx =(π/4) +[ln∣(x/(√(1+x^2 )))∣]_1 ^(+∞) =(π/4) −ln((1/(√2)))=(π/4) +ln((√2)) ⇒ I=(1/4)ln(5) −(π/2) +(1/2) arctan2 −ln((√2))](Q31239.png)
$$=−{ln}\left(\left(^{\mathrm{4}} \sqrt{\mathrm{5}}\right)^{−\mathrm{1}} \right)\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{+\infty} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{5}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:\left[{arctant}\right]_{\mathrm{2}} ^{+\infty} \:\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{5}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctan}\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{5}\right)\:−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\mathrm{2} \\ $$$$\int_{\mathrm{1}} ^{\xi} \:\:\frac{{arctanx}}{{x}^{\mathrm{2}} }\:=\left[\frac{−\mathrm{1}}{{x}}\:{arctanx}\right]_{\mathrm{1}} ^{\xi} \:−\:\int_{\mathrm{1}} ^{\xi} \:\frac{−\mathrm{1}}{{x}}\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\rightarrow \\ $$$$\frac{\pi}{\mathrm{4}}\:+\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}=\frac{\pi}{\mathrm{4}}\:+\int_{\mathrm{1}} ^{+\infty} \left(\frac{\mathrm{1}}{{x}}\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\:+\left[{ln}\mid\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid\right]_{\mathrm{1}} ^{+\infty} =\frac{\pi}{\mathrm{4}}\:−{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)=\frac{\pi}{\mathrm{4}}\:+{ln}\left(\sqrt{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{5}\right)\:−\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\mathrm{2}\:−{ln}\left(\sqrt{\mathrm{2}}\right) \\ $$