Question Number 63782 by mathmax by abdo last updated on 09/Jul/19
$${let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculste}\:{also}\:{g}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{4}} } \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{1}\right)\:{let}\:\:{h}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{have}\:{h}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{4}{a}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:=−\mathrm{4}{a}\:{f}\left({a}\right)\:\Rightarrow{f}\left({a}\right)=−\frac{\mathrm{1}}{\mathrm{4}{a}}{h}^{'} \left({a}\right)\:{let}\:{find}\:{h}\left({a}\right) \\ $$$${we}\:{have}\:{h}\left({a}\right)\:=_{{x}={at}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{adt}}{{a}^{\mathrm{4}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{f}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{i}\right) \\ $$$${Res}\left({w},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {w}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\left({z}+{i}\right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:−\mathrm{2}\left({z}+{i}\right)^{−\mathrm{3}} =\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−\mathrm{8}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow{h}\left({a}\right)\:=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }\:\Rightarrow \\ $$$${h}^{'} \left({a}\right)=\frac{\pi}{\mathrm{2}}\:\frac{−\mathrm{3}{a}^{\mathrm{2}} }{{a}^{\mathrm{6}} }\:=\frac{−\mathrm{3}\pi}{\mathrm{2}{a}^{\mathrm{4}} }\:\Rightarrow{f}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}{a}}\left(\frac{−\mathrm{3}\pi}{\mathrm{2}{a}^{\mathrm{4}} }\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$$\bigstar{f}\left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} }\:\bigstar \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{by}\:{derivation}\:{f}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{3}\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{6}} }{dx} \\ $$$$=−\mathrm{6}{a}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:=−\mathrm{6}{a}\:{g}\left({a}\right)\:\Rightarrow{g}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}{a}}{f}^{'} \left({a}\right) \\ $$$${but}\:{f}\left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} }\:\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\left(−\frac{\mathrm{5}{a}^{\mathrm{4}} }{{a}^{\mathrm{10}} }\right)\:=−\frac{\mathrm{15}\pi}{\mathrm{8}\:{a}^{\mathrm{6}} }\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}{a}}\left(−\frac{\mathrm{15}\pi}{\mathrm{8}{a}^{\mathrm{6}} }\right)\:=\frac{\mathrm{15}\pi}{\mathrm{48}\:{a}^{\mathrm{7}} } \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\Rightarrow{g}\left({a}\right)\:=\frac{\mathrm{3}.\mathrm{5}\pi}{\mathrm{3}.\mathrm{16}{a}^{\mathrm{7}} }\:\Rightarrow{g}\left({a}\right)\:=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} }\:. \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{proved}\:{that}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$${a}=\mathrm{1}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}}\:.\:{also}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} } \\ $$$${a}=\sqrt{\mathrm{2}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{32}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }\:. \\ $$$$ \\ $$
Answered by MJS last updated on 09/Jul/19
$$\mathrm{reduction}\:\mathrm{formula} \\ $$$$\int\frac{{dx}}{\left({px}^{\mathrm{2}} +{q}\right)^{{n}} }=\frac{{x}}{\mathrm{2}{q}\left({n}−\mathrm{1}\right)\left({px}^{\mathrm{2}} +{q}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{q}\left({n}−\mathrm{1}\right)}\int\frac{{dx}}{\left({px}^{\mathrm{2}} +{q}\right)^{{n}−\mathrm{1}} } \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }= \\ $$$${p}=\mathrm{1};\:{q}={a}^{\mathrm{2}} ;\:{n}=\mathrm{3} \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$${p}=\mathrm{1};\:{q}={a}^{\mathrm{2}} ;\:{n}=\mathrm{2} \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }×\frac{{x}}{\mathrm{2}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }= \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}{x}}{\mathrm{8}{a}^{\mathrm{4}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}+\frac{\mathrm{3}}{\mathrm{8}{a}^{\mathrm{4}} }×\frac{\mathrm{1}}{{a}}\mathrm{arctan}\:\frac{{x}}{{a}}\:= \\ $$$$=\frac{{x}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} \right)}{\mathrm{8}{a}^{\mathrm{4}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}{a}^{\mathrm{5}} }\mathrm{arctan}\:\frac{{x}}{{a}}\:+{C} \\ $$$${f}\left({a}\right)=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$$\mathrm{for}\:{g}\left({a}\right)\:\mathrm{we}\:\mathrm{go}\:\mathrm{the}\:\mathrm{same}\:\mathrm{path} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} }=\frac{{x}\left(\mathrm{15}{x}^{\mathrm{4}} +\mathrm{40}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{33}{a}^{\mathrm{4}} \right)}{\mathrm{48}{a}^{\mathrm{6}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{16}{a}^{\mathrm{7}} }\mathrm{arctan}\:\frac{{x}}{\mathrm{2}}\:+{C} \\ $$$${g}\left({a}\right)=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} }=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} } \\ $$
Commented by turbo msup by abdo last updated on 09/Jul/19
$${thank}\:{you}\:{sir}. \\ $$