# super-cooles-Integral-0-dx-1-x-2-2-

Question Number 131405 by bramlexs22 last updated on 04/Feb/21
$$\\$$$$\:\:\ldots\ldots\:\:\mathrm{super}\:\mathrm{cooles}\:\mathrm{Integral}\:\iddots\iddots \\$$$$\:\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=? \\$$
Commented by bramlexs22 last updated on 04/Feb/21
$$\\$$$$\rightarrow\rightarrow\smile\rightarrow\rightarrow\mathrm{ok}\:\mathrm{alle}\:\mathrm{antworten}\:\mathrm{sind}\:\mathrm{super} \\$$
Commented by Dwaipayan Shikari last updated on 04/Feb/21
$${For}\:{General}\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{{n}} \right)^{{n}} }=\frac{\Gamma\left({n}−\frac{\mathrm{1}}{{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right)}{{n}!}\:\:\:{n}\neq\mathrm{0}\:,\mathrm{1} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}!}=\frac{\pi}{\mathrm{4}} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{3}} }=\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{6}}.\frac{\mathrm{5}}{\mathrm{3}}.\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{10}\pi}{\mathrm{27}\sqrt{\mathrm{3}}} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{4}} }=\frac{\mathrm{231}\pi}{\mathrm{256}\sqrt{\mathrm{2}}} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{5}} \right)^{\mathrm{5}} }=\frac{\mathrm{2054}\sqrt{\mathrm{2}}}{\:\mathrm{9375}\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}\pi \\$$$$… \\$$
Answered by mnjuly1970 last updated on 04/Feb/21
$$\phi\overset{{x}=\frac{\mathrm{1}}{{t}}} {=}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\ast\:\frac{{dt}}{{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} .{t}^{\mathrm{2}} }{dt} \\$$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}=\frac{\pi}{\mathrm{2}}−\phi \\$$$$\:\:\:\:\:\:\therefore\:\phi=\frac{\pi}{\mathrm{4}}\:\:….\checkmark \\$$
Answered by Ar Brandon last updated on 04/Feb/21
$$\mathrm{x}^{\mathrm{2}} =\mathrm{u}\:\Rightarrow\mathrm{2xdx}=\mathrm{du} \\$$$$\mathcal{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} }\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{3}}{\mathrm{2}}\right) \\$$$$\:\:\:=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{2}\right)}=\frac{\pi}{\mathrm{4}} \\$$
Answered by mnjuly1970 last updated on 04/Feb/21
$$\: \\$$$$\:\:\phi=\int_{\mathrm{0}} ^{\:\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\overset{{x}^{\mathrm{2}} ={t}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}}\:,\frac{\mathrm{3}}{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{4}}\:\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{4}}..\checkmark \\$$
Answered by mathmax by abdo last updated on 04/Feb/21
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\Rightarrow\mathrm{I}=_{\mathrm{x}=\mathrm{tant}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}\right)^{\mathrm{2}} }\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{t}} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{2}}\mathrm{dt}\:=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{sin}\left(\mathrm{2t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\pi}{\mathrm{4}}+\mathrm{0}\:\Rightarrow \\$$$$\mathrm{I}=\frac{\pi}{\mathrm{4}} \\$$
Answered by mathmax by abdo last updated on 04/Feb/21
$$\mathrm{parametric}\:\mathrm{method}\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\Rightarrow \\$$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2a}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2a}}\mathrm{f}^{'} \left(\mathrm{a}\right) \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}^{'} \left(\mathrm{1}\right)\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)\:=_{\mathrm{x}=\mathrm{az}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{adz}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)} \\$$$$=\frac{\mathrm{1}}{\mathrm{a}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dz}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2a}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)\:=−\frac{\pi}{\mathrm{2a}^{\mathrm{2}} }\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=−\frac{\pi}{\mathrm{2}}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{4}} \\$$