Question Number 69564 by mathmax by abdo last updated on 25/Sep/19
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}}\:\:\:{with}\:{a}\:{real}\:{and}\:{a}>\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
$$\left.\mathrm{1}\right)\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}}\:\Rightarrow\mathrm{2}{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} \:+{a}}\:\:\:\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +{a}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}\:+{a}\:=\mathrm{0}\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta^{'} =\mathrm{1}−{a}<\mathrm{0}\:\Rightarrow\Delta^{'} =\left({i}\sqrt{{a}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{{a}−\mathrm{1}} \\ $$$${t}_{\mathrm{2}} =\mathrm{1}−{i}\sqrt{{a}−\mathrm{1}}\:\Rightarrow\mid{t}_{\mathrm{1}} \mid\:=\sqrt{\mathrm{1}+{a}−\mathrm{1}}=\sqrt{{a}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \\ $$$${t}_{\mathrm{2}} =\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({z}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}^{\mathrm{2}} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}+{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left({z}+{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\right.} \\ $$$${the}\:{poles}\:{of}\:{W}\:{are}\:\overset{−} {+}\:{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \:{and}\:\overset{−} {+}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \\ $$$$\left.\int_{−\infty} ^{+\infty} \:{W}−{z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arcta}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)+{Res}\left({W},−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right\}\right. \\ $$$${Res}\left({W},{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left(\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{2}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{2}{i}\:{sin}\left({arctan}\sqrt{{a}−\mathrm{1}}\right)\right.}\:=\frac{{e}^{−\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{4}{i}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{sin}\left\{{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)\right\}} \\ $$$${Res}\left({W},−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left(\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{a}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}{isin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right)\right\}}\:=\frac{{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{4}{ia}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{sin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}} \\ $$$${but}\:{sin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right)\:=\frac{\sqrt{{a}−\mathrm{1}}}{\:\sqrt{\mathrm{1}+{a}−\mathrm{1}}}\:=\frac{\sqrt{{a}−\mathrm{1}}}{\:\sqrt{{a}}}=\sqrt{\frac{{a}−\mathrm{1}}{{a}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\sqrt{{a}−\mathrm{1}}} }{\mathrm{4}{i}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} ×\sqrt{\frac{{a}−\mathrm{1}}{{a}}}}\:+\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\sqrt{{a}−\mathrm{1}}} }{\mathrm{4}{ia}^{\frac{\mathrm{3}}{\mathrm{4}}} ×\sqrt{\frac{{a}−\mathrm{1}}{{a}}}}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \frac{\sqrt{{a}−\mathrm{1}}}{{a}^{\frac{\mathrm{1}}{\mathrm{2}}} }}\left\{\:\mathrm{2}{cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\right. \\ $$$$=\frac{\pi}{{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{{a}−\mathrm{1}}}×{cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\:\:{but} \\ $$$${cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}−\mathrm{1}}}\:=\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\frac{\pi}{\left(^{\mathrm{4}} \sqrt{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{a}}}\:=\mathrm{2}{f}\left({a}\right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{a}}}\:{with}\:\:{a}>\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)=−{f}^{'} \left({a}\right)\:\:{rest}\:{to}\:{calculate}\:{f}^{'} \left({a}\right)… \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}}\:={f}\left(\mathrm{3}\right)\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\sqrt{\mathrm{9}−\mathrm{3}}}\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\sqrt{\mathrm{6}}} \\ $$