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Question Number 63782 by mathmax by abdo last updated on 09/Jul/19
let f(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^3 )) with a>0 1) calculate f(a) 2)calculste also g(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^4 )) 3) find the values of integrals ∫_0 ^∞ (dx/((x^2 +1)^3 )) ∫_0 ^∞ (dx/((x^2 +2)^4 ))
$${let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculste}\:{also}\:{g}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{4}} } \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
1) let h(a) =∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^2 )) we have h^′ (a) =−∫_(−∞) ^(+∞) ((2(2a)(a^2 +x^2 ))/((a^2 +x^2 )^4 )) =−∫_(−∞) ^(+∞) ((4a)/((a^2 +x^2 )^3 ))dx =−4a f(a) ⇒f(a)=−(1/(4a))h^′ (a) let find h(a) we have h(a) =_(x=at) ∫_(−∞) ^(+∞) ((adt)/(a^4 (1+t^2 )^2 )) =(1/a^3 ) ∫_(−∞) ^(+∞) (dt/((t^2 +1)^2 )) let w(z) =(1/((z^2 +1)^2 )) ⇒f(z) =(1/((z−i)^2 (z+i)^2 )) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,i) Res(w,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 w(z)}^((1)) =lim_(z→i) {(z+i)^(−2) }^((1)) =lim_(z→i) −2(z+i)^(−3) =((−2)/((2i)^3 )) =((−2)/(−8i)) =(1/(4i)) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ(1/(4i)) =(π/2) ⇒h(a) =(π/(2a^3 )) ⇒ h^′ (a)=(π/2) ((−3a^2 )/a^6 ) =((−3π)/(2a^4 )) ⇒f(a) =−(1/(4a))(((−3π)/(2a^4 ))) =((3π)/(8a^5 )) ★f(a) =((3π)/(8a^5 )) ★
$$\left.\mathrm{1}\right)\:{let}\:\:{h}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{we}\:{have}\:{h}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{2}\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{4}{a}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:=−\mathrm{4}{a}\:{f}\left({a}\right)\:\Rightarrow{f}\left({a}\right)=−\frac{\mathrm{1}}{\mathrm{4}{a}}{h}^{'} \left({a}\right)\:{let}\:{find}\:{h}\left({a}\right) \\ $$$${we}\:{have}\:{h}\left({a}\right)\:=_{{x}={at}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{adt}}{{a}^{\mathrm{4}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{f}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{i}\right) \\ $$$${Res}\left({w},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} {w}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\left({z}+{i}\right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:−\mathrm{2}\left({z}+{i}\right)^{−\mathrm{3}} =\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−\mathrm{8}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow{h}\left({a}\right)\:=\frac{\pi}{\mathrm{2}{a}^{\mathrm{3}} }\:\Rightarrow \\ $$$${h}^{'} \left({a}\right)=\frac{\pi}{\mathrm{2}}\:\frac{−\mathrm{3}{a}^{\mathrm{2}} }{{a}^{\mathrm{6}} }\:=\frac{−\mathrm{3}\pi}{\mathrm{2}{a}^{\mathrm{4}} }\:\Rightarrow{f}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}{a}}\left(\frac{−\mathrm{3}\pi}{\mathrm{2}{a}^{\mathrm{4}} }\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$$\bigstar{f}\left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} }\:\bigstar \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
2) we have by derivation f^′ (a) =−∫_(−∞) ^(+∞) ((3(2a)(a^2 +x^2 )^2 )/((a^2 +x^2 )^6 ))dx =−6a ∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^4 )) =−6a g(a) ⇒g(a) =−(1/(6a))f^′ (a) but f(a) =((3π)/(8a^5 )) ⇒f^′ (a) =((3π)/8)(−((5a^4 )/a^(10) )) =−((15π)/(8 a^6 )) ⇒ g(a) =−(1/(6a))(−((15π)/(8a^6 ))) =((15π)/(48 a^7 ))
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{by}\:{derivation}\:{f}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\frac{\mathrm{3}\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{6}} }{dx} \\ $$$$=−\mathrm{6}{a}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:=−\mathrm{6}{a}\:{g}\left({a}\right)\:\Rightarrow{g}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}{a}}{f}^{'} \left({a}\right) \\ $$$${but}\:{f}\left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} }\:\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\left(−\frac{\mathrm{5}{a}^{\mathrm{4}} }{{a}^{\mathrm{10}} }\right)\:=−\frac{\mathrm{15}\pi}{\mathrm{8}\:{a}^{\mathrm{6}} }\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}{a}}\left(−\frac{\mathrm{15}\pi}{\mathrm{8}{a}^{\mathrm{6}} }\right)\:=\frac{\mathrm{15}\pi}{\mathrm{48}\:{a}^{\mathrm{7}} } \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
⇒g(a) =((3.5π)/(3.16a^7 )) ⇒g(a) =((5π)/(16a^7 )) .
$$\Rightarrow{g}\left({a}\right)\:=\frac{\mathrm{3}.\mathrm{5}\pi}{\mathrm{3}.\mathrm{16}{a}^{\mathrm{7}} }\:\Rightarrow{g}\left({a}\right)\:=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} }\:. \\ $$
Commented by mathmax by abdo last updated on 09/Jul/19
3) we have proved that ∫_(−∞) ^(+∞) (dx/((a^2 +x^2 )^3 )) =((3π)/(8a^5 )) a=1 ⇒∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) =((3π)/8) ⇒2 ∫_0 ^∞ (dx/((x^2 +1)^3 )) =((3π)/8) ⇒ ∫_0 ^∞ (dx/((x^2 +1)^3 )) =((3π)/(16)) . also we have proved that ∫_(−∞) ^(+∞) (dx/((x^2 +a^2 )^4 )) =((5π)/(16a^7 )) a=(√2) ⇒∫_(−∞) ^(+∞) (dx/((x^2 +2)^4 )) =((5π)/(16 ((√2))^7 )) ⇒∫_0 ^∞ (dx/((x^2 +2)^4 )) =((5π)/(32((√2))^7 )) .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{proved}\:{that}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$${a}=\mathrm{1}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}}\:.\:{also}\:{we}\:{have}\:{proved}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} } \\ $$$${a}=\sqrt{\mathrm{2}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)^{\mathrm{4}} }\:=\frac{\mathrm{5}\pi}{\mathrm{32}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{7}} }\:. \\ $$$$ \\ $$
Answered by MJS last updated on 09/Jul/19
reduction formula ∫(dx/((px^2 +q)^n ))=(x/(2q(n−1)(px^2 +q)^(n−1) ))+((2n−3)/(2q(n−1)))∫(dx/((px^2 +q)^(n−1) )) ∫(dx/((x^2 +a^2 )^3 ))= p=1; q=a^2 ; n=3 =(x/(4a^2 (x^2 +a^2 )^2 ))+(3/(4a^2 ))∫(dx/((x^2 +a^2 )^2 ))= p=1; q=a^2 ; n=2 =(x/(4a^2 (x^2 +a^2 )^2 ))+(3/(4a^2 ))×(x/(2a^2 (x^2 +a^2 )))+(3/(4a^2 ))×(1/(2a^2 ))∫(dx/(x^2 +a^2 ))= =(x/(4a^2 (x^2 +a^2 )^2 ))+((3x)/(8a^4 (x^2 +a^2 )))+(3/(8a^4 ))×(1/a)arctan (x/a) = =((x(3x^2 +5a^2 ))/(8a^4 (x^2 +a^2 )^2 ))+(3/(8a^5 ))arctan (x/a) +C f(a)=∫_(−∞) ^(+∞) (dx/((x^2 +a^2 )^3 ))=((3π)/(8a^5 )) for g(a) we go the same path ∫(dx/((x^2 +a^2 )^4 ))=((x(15x^4 +40a^2 x^2 +33a^4 ))/(48a^6 (x^2 +a^2 )^3 ))+(5/(16a^7 ))arctan (x/2) +C g(a)=∫_(−∞) ^(+∞) (dx/((x^2 +a^2 )^4 ))=((5π)/(16a^7 ))
$$\mathrm{reduction}\:\mathrm{formula} \\ $$$$\int\frac{{dx}}{\left({px}^{\mathrm{2}} +{q}\right)^{{n}} }=\frac{{x}}{\mathrm{2}{q}\left({n}−\mathrm{1}\right)\left({px}^{\mathrm{2}} +{q}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{q}\left({n}−\mathrm{1}\right)}\int\frac{{dx}}{\left({px}^{\mathrm{2}} +{q}\right)^{{n}−\mathrm{1}} } \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }= \\ $$$${p}=\mathrm{1};\:{q}={a}^{\mathrm{2}} ;\:{n}=\mathrm{3} \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }= \\ $$$${p}=\mathrm{1};\:{q}={a}^{\mathrm{2}} ;\:{n}=\mathrm{2} \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }×\frac{{x}}{\mathrm{2}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}+\frac{\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }= \\ $$$$=\frac{{x}}{\mathrm{4}{a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}{x}}{\mathrm{8}{a}^{\mathrm{4}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}+\frac{\mathrm{3}}{\mathrm{8}{a}^{\mathrm{4}} }×\frac{\mathrm{1}}{{a}}\mathrm{arctan}\:\frac{{x}}{{a}}\:= \\ $$$$=\frac{{x}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} \right)}{\mathrm{8}{a}^{\mathrm{4}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}{a}^{\mathrm{5}} }\mathrm{arctan}\:\frac{{x}}{{a}}\:+{C} \\ $$$${f}\left({a}\right)=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{3}\pi}{\mathrm{8}{a}^{\mathrm{5}} } \\ $$$$\mathrm{for}\:{g}\left({a}\right)\:\mathrm{we}\:\mathrm{go}\:\mathrm{the}\:\mathrm{same}\:\mathrm{path} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} }=\frac{{x}\left(\mathrm{15}{x}^{\mathrm{4}} +\mathrm{40}{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{33}{a}^{\mathrm{4}} \right)}{\mathrm{48}{a}^{\mathrm{6}} \left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{16}{a}^{\mathrm{7}} }\mathrm{arctan}\:\frac{{x}}{\mathrm{2}}\:+{C} \\ $$$${g}\left({a}\right)=\underset{−\infty} {\overset{+\infty} {\int}}\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} }=\frac{\mathrm{5}\pi}{\mathrm{16}{a}^{\mathrm{7}} } \\ $$
Commented by turbo msup by abdo last updated on 09/Jul/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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