Question Number 118757 by mohammad17 last updated on 19/Oct/20
Commented by mohammad17 last updated on 19/Oct/20
$${with}\:{out}\:{x}={tan}\:\theta \\ $$
Answered by mnjuly1970 last updated on 19/Oct/20
$$\frac{\pi}{\mathrm{4}}\overset{?} {=}{answer} \\ $$$${method}\:\mathrm{1}: \\ $$$$\:{euler}\:{reflection}\:{formula}. \\ $$$${x}^{\mathrm{2}} ={t}\:\Rightarrow\Omega=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\left(\frac{−\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)−\mathrm{1}} }{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\:\:\Rightarrow\:\Omega=\frac{\mathrm{1}}{\mathrm{2}}\:\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\overset{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}} {=}\frac{\pi}{\mathrm{4}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:…{m}.{n}.{july}.\mathrm{1970}… \\ $$$$………. \\ $$$${methot}\:\mathrm{2}\:: \\ $$$$\:{x}=\frac{\mathrm{1}}{{t}}\:\Rightarrow\:\Omega\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{dt}}{{t}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\left({t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}=\frac{\pi}{\mathrm{2}}\:−\Omega \\ $$$$\mathrm{2}\Omega\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\Omega=\frac{\pi}{\mathrm{4}}\:\:\checkmark\checkmark \\ $$$$\:\:\:.{m}.{n}.\mathrm{1970}.. \\ $$$$ \\ $$
Commented by mohammad17 last updated on 19/Oct/20
$${yes}\:{sir}\:{its}\:{right} \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Oct/20
$$\mathrm{excellent} \\ $$
Commented by mnjuly1970 last updated on 19/Oct/20
$${thank}\:{you} \\ $$
Commented by mnjuly1970 last updated on 19/Oct/20
$$\:{sincerely}\:{yours} \\ $$
Answered by 1549442205PVT last updated on 19/Oct/20
$$\mathrm{Put}\:\mathrm{x}=\mathrm{tan}\varphi\Rightarrow\mathrm{dx}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \varphi\right)\mathrm{d}\varphi \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{d}\varphi}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \varphi\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \varphi\mathrm{d}\varphi \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos2}\varphi}{\mathrm{2}}\mathrm{d}\varphi=\left(\frac{\varphi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin2}\varphi\right)_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{4}} \\ $$