Question Number 82755 by mathmax by abdo last updated on 23/Feb/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by Ajao yinka last updated on 24/Feb/20
0
Commented by mathmax by abdo last updated on 24/Feb/20
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{{a}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0}\:\:{we}\:{can}\:{verify}\:{eazly}\:{that}\:{f}\:{is} \\ $$$$\left.{derivable}\:{on}\:\right]\mathrm{0},+\infty\left[\:{and}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\Rightarrow\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{lnx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=−{f}^{'} \left({a}\right)\:{let}\:{explicit}\:{f}\:{changement}\:{x}=\sqrt{{a}}{t}\:{give} \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\sqrt{{a}}{t}\right)}{{a}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\sqrt{{a}}{dt}\:=\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}{lna}\:+{lnt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{{lna}}{\:\sqrt{{a}}}×\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} \:\frac{{lnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{but}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:=\mathrm{0}\left(\left({put}\:{t}=\frac{\mathrm{1}}{{u}}\right)\:\Rightarrow\right. \\ $$$${f}\left({a}\right)\:=\frac{\pi{lna}}{\mathrm{4}\sqrt{{a}}}\:\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\pi}{\mathrm{4}}\left\{\frac{\frac{\sqrt{{a}}}{{a}}−{lna}\:×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}}{{a}}\right\} \\ $$$$=\frac{\pi}{\mathrm{4}{a}\sqrt{{a}}}−\frac{\pi}{\mathrm{4}}×\frac{{lna}}{\mathrm{2}{a}\sqrt{{a}}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{lnx}}{\left({a}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=−\frac{\pi}{\mathrm{4}{a}\sqrt{{a}}}+\frac{\pi{lna}}{\mathrm{8}{a}\sqrt{{a}}} \\ $$$${a}=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=−\frac{\pi}{\mathrm{4}} \\ $$
Answered by mind is power last updated on 24/Feb/20
$$\int_{\mathrm{0}} ^{+\infty} \frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }{dt}={A}\left({n}\right) \\ $$$${ln}\left({t}\right)=\frac{\partial}{\partial{x}}{t}^{{x}} \mid_{{x}=\mathrm{0}} \\ $$$${A}=\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{+\infty} \frac{{t}^{{x}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{{n}} }{dt}\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tg}^{{x}} \left({t}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right)^{{n}} }\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){dt}\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tg}^{{x}} \left({t}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right)^{{n}} }{dt}\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\partial}{\partial{x}}.\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{x}} \left({t}\right){cos}^{−{x}+\mathrm{2}{n}} \left({t}\right){dt}\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\partial}{\partial{x}}\beta\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}};−\frac{{x}}{\mathrm{2}}\:+{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\partial}{\partial{x}}\frac{\Gamma\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{1}\right)}\mid_{{x}=\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\frac{\partial}{\partial{x}}\frac{\Gamma\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}{{n}!}\mid_{{x}=\mathrm{0}} \\ $$$$\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\partial}{\partial{x}}\frac{\Gamma\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)}{{n}!}.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\mid_{{x}=\mathrm{0}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}!\right)}\frac{\partial}{\partial{x}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\left({x}+\mathrm{1}\right)\right)}.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\mid_{{x}=\mathrm{0}} \\ $$$${f}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$${ln}\left({f}\left({x}\right)\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{{f}'\left(\mathrm{0}\right)}{{f}\left(\mathrm{0}\right)}=−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{k}}−\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{{k}}=\frac{{H}_{{n}} }{\mathrm{2}}−{H}_{\mathrm{2}{n}} \\ $$$${f}\left(\mathrm{0}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} .{n}!} \\ $$$$\Rightarrow{f}'\left(\mathrm{0}\right)=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} {n}!}.\left(\frac{{H}_{{n}} }{\mathrm{2}}−{H}_{\mathrm{2}{n}} \right) \\ $$$${A}\left({n}\right)=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left({n}!\right)^{\mathrm{2}} }\left(\frac{{H}_{{n}} }{\mathrm{2}}−{H}_{\mathrm{2}{n}} \right)\pi=\int_{\mathrm{0}} ^{+\infty} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx} \\ $$$${n}=\mathrm{2}\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}={A}\left(\mathrm{2}\right) \\ $$$${A}\left(\mathrm{2}\right)=\frac{\mathrm{4}!}{\mathrm{2}^{\mathrm{5}} \left(\mathrm{2}!\right)^{\mathrm{2}} }\left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\pi=\frac{\mathrm{24}}{\mathrm{32}.\mathrm{4}}.\frac{−\mathrm{4}}{\mathrm{3}}\pi=\frac{−\mathrm{8}}{\mathrm{32}}.\pi=−\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$