Menu Close

let-f-a-0-dx-x-2-1-x-2-a-with-a-gt-0-1-determine-a-explicit-form-of-f-a-2-calculate-g-a-0-dx-x-2-1-x-2-a-2-3-give-f-n-a-at-form-of-integral-4-calcul

Tinku Tara 0 Comments
Pin




Question Number 67674 by Abdo msup. last updated on 30/Aug/19
let f(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a))) with a>0 1) determine a explicit form of f(a) 2) calculate g(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a)^2 )) 3)give f^((n)) (a) at form of integral 4)calculate ∫_0 ^∞ (dx/((x^2 +1)(x^2 +3)^2 )) and ∫_0 ^∞ (dx/((x^2 +1)^3 ))
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{a}\right)}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){give}\:{f}^{\left({n}\right)} \left({a}\right)\:{at}\:{form}\:{of}\:{integral} \\ $$$$\left.\mathrm{4}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 30/Aug/19
1)f(a) =∫_0 ^∞ (dx/((x^2 +1)(x^2 +a))) ⇒2f(a) =∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +a))) let W(z) =(1/((z^2 +1)(z^2 +a))) ⇒W(z) =(1/((z−i)(z+i)(z−i(√a))(z+i(√a)))) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,i) +Res(W,i(√a))} Res(W,i) =lim_(z→i) (z−i)W(z) =(1/(2i(a−1))) ( a≠1) Res(W,i(√a)) =lim_(z→i(√a)) (z−i(√a))W(z) =(1/(2i(√a)(1−a))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{(1/(2i(a−1))) +(1/(2i(√a)(1−a)))} =(π/(a−1)) +(π/( (√a)(1−a))) =(π/(a−1))−(π/((a−1)(√a))) =(π/(a−1))(1−(1/( (√a)))) =((π((√a)−1))/( (√a)(a−1))) =(π/( (√a)((√a)+1))) ⇒ f(a) =(π/(2(√a)((√a)+1))) =(π/(2(a+(√a)))) another way f(a) =(1/(a−1))∫_0 ^∞ {(1/(x^2 +1))−(1/(x^2 +a))}dx =(1/(a−1))×(π/2)−(1/(a−1)) ∫_0 ^∞ (dx/(x^2 +a)) changement x =(√a)t give ∫_0 ^∞ (dx/(x^2 +a)) =∫_0 ^∞ (((√a)dt)/(a(t^2 +1))) =(1/( (√a)))×(π/2) ⇒f(a)=(π/(2(a−1)))−(π/(2(√a)(a−1)))
$$\left.\mathrm{1}\right){f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}\right)}\:\Rightarrow\mathrm{2}{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}\right)} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+{a}\right)}\:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{i}\sqrt{{a}}\right)\left({z}+{i}\sqrt{{a}}\right)} \\ $$$${residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},{i}\right)\:+{Res}\left({W},{i}\sqrt{{a}}\right)\right\} \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right){W}\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\left({a}−\mathrm{1}\right)}\:\:\:\left(\:{a}\neq\mathrm{1}\right) \\ $$$${Res}\left({W},{i}\sqrt{{a}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{{a}}} \:\:\:\left({z}−{i}\sqrt{{a}}\right){W}\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{a}}\left(\mathrm{1}−{a}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2}{i}\left({a}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{a}}\left(\mathrm{1}−{a}\right)}\right\} \\ $$$$=\frac{\pi}{{a}−\mathrm{1}}\:+\frac{\pi}{\:\sqrt{{a}}\left(\mathrm{1}−{a}\right)}\:=\frac{\pi}{{a}−\mathrm{1}}−\frac{\pi}{\left({a}−\mathrm{1}\right)\sqrt{{a}}}\:=\frac{\pi}{{a}−\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right) \\ $$$$=\frac{\pi\left(\sqrt{{a}}−\mathrm{1}\right)}{\:\sqrt{{a}}\left({a}−\mathrm{1}\right)}\:=\frac{\pi}{\:\sqrt{{a}}\left(\sqrt{{a}}+\mathrm{1}\right)}\:\Rightarrow\:{f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{{a}}\left(\sqrt{{a}}+\mathrm{1}\right)}\:=\frac{\pi}{\mathrm{2}\left({a}+\sqrt{{a}}\right)} \\ $$$${another}\:{way}\:\:{f}\left({a}\right)\:=\frac{\mathrm{1}}{{a}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}}\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{{a}−\mathrm{1}}×\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{a}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{a}}\:\:{changement}\:{x}\:=\sqrt{{a}}{t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{a}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\sqrt{{a}}{dt}}{{a}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\:\sqrt{{a}}}×\frac{\pi}{\mathrm{2}}\:\Rightarrow{f}\left({a}\right)=\frac{\pi}{\mathrm{2}\left({a}−\mathrm{1}\right)}−\frac{\pi}{\mathrm{2}\sqrt{{a}}\left({a}−\mathrm{1}\right)} \\ $$
Commented by mathmax by abdo last updated on 30/Aug/19
2) we have f^′ (a) =−∫_0 ^∞ (dx/((x^2 +1)(x^2 +a)^2 )) =−g(a) ⇒ g(a) =−f^′ (a) but f(a) =(π/(2(a+(√a)))) ⇒f^′ (a) =−(π/2)×(((a+(√a))^′ )/((a+(√a))^2 )) =−(π/2)×((1+(1/(2(√a))))/((a+(√a))^2 )) =−(π/2) ((2(√a)+1)/(2(√a)(a+(√a))^2 )) =−(π/4)×((2(√a)+1)/( (√a)(a+(√a))^2 )) ⇒ g(a) =((π(2(√a) +1))/(4(√a)(a+(√a))^2 ))
$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:\:{but}\:{f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}\left({a}+\sqrt{{a}}\right)}\:\Rightarrow{f}^{'} \left({a}\right)\:=−\frac{\pi}{\mathrm{2}}×\frac{\left({a}+\sqrt{{a}}\right)^{'} }{\left({a}+\sqrt{{a}}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi}{\mathrm{2}}×\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}}{\left({a}+\sqrt{{a}}\right)^{\mathrm{2}} }\:=−\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{2}\sqrt{{a}}+\mathrm{1}}{\mathrm{2}\sqrt{{a}}\left({a}+\sqrt{{a}}\right)^{\mathrm{2}} }\:=−\frac{\pi}{\mathrm{4}}×\frac{\mathrm{2}\sqrt{{a}}+\mathrm{1}}{\:\sqrt{{a}}\left({a}+\sqrt{{a}}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${g}\left({a}\right)\:=\frac{\pi\left(\mathrm{2}\sqrt{{a}}\:+\mathrm{1}\right)}{\mathrm{4}\sqrt{{a}}\left({a}+\sqrt{{a}}\right)^{\mathrm{2}} }\:\:\: \\ $$
Commented by mathmax by abdo last updated on 30/Aug/19
4)∫_0 ^∞ (dx/((x^2 +1)(x^2 +3)^2 )) =g(3) =((π(2(√3)+1))/(4(√3)(3+(√3))^2 )) let find I =∫_0 ^∞ (dx/((x^2 +1)^3 )) ⇒ 2I =∫_(−∞) ^(+∞) (dx/((x^2 +1)^3 )) let w(z)=(1/((z^2 +1)^3 )) ⇒w(z) =(1/((z−i)^3 (z+i)^3 )) the poles of w are i and −i(triples) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,i) Res(w,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 w(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =lim_(z→i) (1/2){−3(z+i)^(−4) }^((1)) =lim_(z→i) ((−3)/2){−4(z+i)^(−5) } =6 (2i)^(−5) =(6/(2^5 i^5 )) =(6/(32i)) =((3.2)/(16.2i)) =(3/(16i)) ⇒ ∫_(−∞) ^(+∞) w(z)dz =2iπ×(3/(16i)) =((3π)/8) =2I ⇒ I =((3π)/(16))
$$\left.\mathrm{4}\right)\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:={g}\left(\mathrm{3}\right)\:=\frac{\pi\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$${let}\:{find}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\:\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:{let}\:{w}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:{the}\:{poles}\:{of}\:{w}\:{are}\:{i}\:{and}\:−{i}\left({triples}\right) \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} {w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{i}\right) \\ $$$${Res}\left({w},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} {w}\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({z}+{i}\right)^{−\mathrm{3}} \right\}^{\left(\mathrm{2}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{3}\left({z}+{i}\right)^{−\mathrm{4}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{−\mathrm{3}}{\mathrm{2}}\left\{−\mathrm{4}\left({z}+{i}\right)^{−\mathrm{5}} \right\}\:=\mathrm{6}\:\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {i}^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{32}{i}}\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{16}.\mathrm{2}{i}}\:=\frac{\mathrm{3}}{\mathrm{16}{i}} \\ $$$$\Rightarrow\:\int_{−\infty} ^{+\infty} \:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{3}}{\mathrm{16}{i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:=\mathrm{2}{I}\:\Rightarrow\:{I}\:=\frac{\mathrm{3}\pi}{\mathrm{16}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *