Question Number 75793 by ~blr237~ last updated on 17/Dec/19
$$\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{arctanx}}{\mathrm{x}\sqrt{\mathrm{log2}}}\right)^{\mathrm{2}} \mathrm{dx}=\:\pi \\ $$
Commented by mathmax by abdo last updated on 18/Dec/19
![let I =∫_0 ^∞ (((arctanx)/(x(√(ln(2))))))^2 dx ⇒ ln(2)I =∫_0 ^∞ ((arctan^2 x)/x^2 )dx by parts u^′ =(1/x^2 ) and v=arctan^2 (x) ⇒ln(2)I =[−(1/x) arctan^2 (x)]_0 ^(+∞) −∫_0 ^∞ (−(1/x))((2arctan(x))/(1+x^2 ))dx=2∫_0 ^∞ ((arcctan(x))/(x(1+x^2 )))dx ⇒ I=(2/(ln(2)))∫_0 ^∞ ((arctan(x))/(x(1+x^2 )))dx =_(arctanx=u) (2/(ln(2))) ∫_0 ^(π/2) (u/(tan(u)(1+tan^2 u)))(1+tan^2 u)du =(2/(ln(2)))∫_0 ^(π/2) (u/(tanu))du =(2/(ln(2)))∫_0 ^(π/2) u ((cosu)/(sinu))du by parts f=u and g^′ =((cosu)/(sinu)) ((ln(2))/2)I =∫_0 ^(π/2) u ((cosu)/(sinu))du=[u ln(sinu)]_0 ^(π/2) −∫_0 ^(π/2) ln(sinu)du=−(−(π/2)ln(2)) ((ln(2))/2)I =((πln(2))/2) ⇒ I =π](Q75819.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\left(\frac{{arctanx}}{{x}\sqrt{{ln}\left(\mathrm{2}\right)}}\right)^{\mathrm{2}} {dx}\:\Rightarrow\:{ln}\left(\mathrm{2}\right){I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }{dx}\:\:{by}\:{parts} \\ $$$${u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and}\:{v}={arctan}^{\mathrm{2}} \left({x}\right)\:\Rightarrow{ln}\left(\mathrm{2}\right){I}\:=\left[−\frac{\mathrm{1}}{{x}}\:{arctan}^{\mathrm{2}} \left({x}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{{x}}\right)\frac{\mathrm{2}{arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arcctan}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:\Rightarrow \\ $$$$\:{I}=\frac{\mathrm{2}}{{ln}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:\:=_{{arctanx}={u}} \frac{\mathrm{2}}{{ln}\left(\mathrm{2}\right)}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{u}}{{tan}\left({u}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right)}\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right){du} \\ $$$$=\frac{\mathrm{2}}{{ln}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{u}}{{tanu}}{du}\:=\frac{\mathrm{2}}{{ln}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {u}\:\frac{{cosu}}{{sinu}}{du}\:\:\:{by}\:{parts}\:{f}={u}\:{and}\:{g}^{'} =\frac{{cosu}}{{sinu}} \\ $$$$\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {u}\:\frac{{cosu}}{{sinu}}{du}=\left[{u}\:{ln}\left({sinu}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sinu}\right){du}=−\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right) \\ $$$$\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}{I}\:=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow\:{I}\:=\pi \\ $$
Answered by mind is power last updated on 17/Dec/19
![=[((−arctan^2 (x))/(xlog(2)))]+(1/(log(2)))∫_0 ^(+∞) ((2arctan(x))/(x(x^2 +1)))dx =(2/(log(2)))∫_0 ^(+∞) ((arcran(x))/(x(x^2 +1)))dx u=arcran(x)⇒dx=(1+tg^2 (u))du ⇔(2/(log(2)))∫_0 ^(π/2) (u/(tg(u)))du=(2/(log(2)))∫_0 ^(π/2) ((ucos(u))/(sin(u)))du by part⇔ (2/(log(2)))[ulog(sin(u))]_0 ^(π/2) −(2/(log(2)))∫_0 ^(π/2) log(sin(u))du =−(2/(log(2)))∫_0 ^(π/2) log(sin(u))du ∫_0 ^(π/2) log(sin(u))du=−log(2)(π/2) done many Times so We get −(2/(log(2))).((−log(2))/2).π=π](Q75805.png)
$$=\left[\frac{−\mathrm{arctan}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{xlog}\left(\mathrm{2}\right)}\right]+\frac{\mathrm{1}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{2arctan}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{arcran}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{arcran}\left(\mathrm{x}\right)\Rightarrow\mathrm{dx}=\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\Leftrightarrow\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{u}}{\mathrm{tg}\left(\mathrm{u}\right)}\mathrm{du}=\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ucos}\left(\mathrm{u}\right)}{\mathrm{sin}\left(\mathrm{u}\right)}\mathrm{du} \\ $$$$\mathrm{by}\:\mathrm{part}\Leftrightarrow \\ $$$$\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\left[\mathrm{ulog}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=−\mathrm{log}\left(\mathrm{2}\right)\frac{\pi}{\mathrm{2}}\:\mathrm{done}\:\mathrm{many}\:\mathrm{Times} \\ $$$$\mathrm{so}\:\mathrm{We}\:\mathrm{get}\:−\frac{\mathrm{2}}{\mathrm{log}\left(\mathrm{2}\right)}.\frac{−\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{2}}.\pi=\pi \\ $$