Menu Close

Find-a-closed-form-a-R-and-a-0-a-0-x-4-1-x-2-1-a-4-x-4-dx-

Tinku Tara 0 Comments
Pin




Question Number 150034 by mathdanisur last updated on 08/Aug/21
Find a closed form: a∈R and a≠0 Ω(a)=∫_( 0) ^( ∞) (x^4 /((1+x^2 )(1+a^4 x^4 ))) dx
$$\mathrm{Find}\:\boldsymbol{\mathrm{a}}\:\mathrm{closed}\:\mathrm{form}:\:\:\boldsymbol{\mathrm{a}}\in\mathbb{R}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{a}}\neq\mathrm{0} \\ $$$$\Omega\left(\mathrm{a}\right)=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\frac{\mathrm{x}^{\mathrm{4}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{4}} \mathrm{x}^{\mathrm{4}} \right)}\:\mathrm{dx} \\ $$
Answered by mathmax by abdo last updated on 09/Aug/21
Ψ=∫_0 ^∞ (x^4 /((x^2 +1)(a^4 x^4 +1)))dx ⇒Ψ=_(ax=t) ∫_0 ^∞ (t^4 /(a^4 ((t^2 /a^2 )+1)(t^4 +1)))(dt/a) =(1/a^5 )∫_0 ^∞ ((a^2 t^4 )/((t^2 +a^2 )(t^4 +1)))dt =(1/a^3 )∫_0 ^∞ (t^4 /((t^2 +a^2 )(t^4 +1)))dt (we suppose a>0 because Ω(−a)=Ω(a)) ⇒a^3 Ψ=(1/2)∫_(−∞) ^(+∞) (t^4 /((t^2 +a^2 )(t^4 +1)))dt let ϕ(z)=(z^4 /((z^2 +a^2 )(z^4 +1))) on ϕ(z)=(z^4 /((z−ia)(z+ia)(z^2 −i)(z^2 +i))) =(z^4 /((z−ia)(z+ia)(z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =(z^4 /((z−ia)(z+ia)(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{ Res(ϕ,ia)+Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,ia)=(((ia)^4 )/((2ia)((ia)^(4 ) +1)))=(a^4 /(2ia(1+a^4 ))) Res(ϕ,e^((iπ)/4) )=((−1)/((i+a^2 )(2e^((iπ)/4) )(2i))) =((−e^(−((iπ)/4)) )/(4i(a^2 +i))) Res(ϕ,−e^(−((iπ)/4)) ) =((−1)/((−i+a^2 )(−2e^(−((iπ)/4)) )(−2i)))=((−e^((iπ)/4) )/(4i(−i+a^2 ))) =(e^((iπ)/4) /(4i(a^2 −i))).....be continued...
$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{4}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{a}^{\mathrm{4}} \mathrm{x}^{\mathrm{4}} \:+\mathrm{1}\right)}\mathrm{dx}\:\Rightarrow\Psi=_{\mathrm{ax}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{4}} \left(\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)}\frac{\mathrm{dt}}{\mathrm{a}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{5}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{a}^{\mathrm{2}} \mathrm{t}^{\mathrm{4}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{4}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}\right)}\mathrm{dt} \\ $$$$\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{a}>\mathrm{0}\:\:\mathrm{because}\:\Omega\left(−\mathrm{a}\right)=\Omega\left(\mathrm{a}\right)\right) \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{3}} \:\Psi=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{t}^{\mathrm{4}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}\right)}\mathrm{dt}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{4}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}\right)} \\ $$$$\mathrm{on}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{4}} }{\left(\mathrm{z}−\mathrm{ia}\right)\left(\mathrm{z}+\mathrm{ia}\right)\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{i}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{4}} }{\left(\mathrm{z}−\mathrm{ia}\right)\left(\mathrm{z}+\mathrm{ia}\right)\left(\mathrm{z}−\sqrt{\mathrm{i}}\right)\left(\mathrm{z}+\sqrt{\mathrm{i}}\right)\left(\mathrm{z}−\sqrt{−\mathrm{i}}\right)\left(\mathrm{z}+\sqrt{−\mathrm{i}}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{4}} }{\left(\mathrm{z}−\mathrm{ia}\right)\left(\mathrm{z}+\mathrm{ia}\right)\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{ia}\right)+\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{ia}\right)=\frac{\left(\mathrm{ia}\right)^{\mathrm{4}} }{\left(\mathrm{2ia}\right)\left(\left(\mathrm{ia}\right)^{\mathrm{4}\:} +\mathrm{1}\right)}=\frac{\mathrm{a}^{\mathrm{4}} }{\mathrm{2ia}\left(\mathrm{1}+\mathrm{a}^{\mathrm{4}} \right)} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)=\frac{−\mathrm{1}}{\left(\mathrm{i}+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{2i}\right)}\:=\frac{−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{4i}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{i}\right)} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{1}}{\left(−\mathrm{i}+\mathrm{a}^{\mathrm{2}} \right)\left(−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2i}\right)}=\frac{−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{4i}\left(−\mathrm{i}+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{\mathrm{4i}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{i}\right)}…..\mathrm{be}\:\mathrm{continued}… \\ $$
Commented by mathdanisur last updated on 09/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Pin

Leave a Reply

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