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

Question Number 150034 by mathdanisur last updated on 08/Aug/21

Find a closed form : a \in R and a \neq 0

Ω (a) = \underset{0}{\int^{\infty}} \frac{x^{4}}{(1 + x^{2}) (1 + a^{4} x^{4})} dx

Answered by mathmax by abdo last updated on 09/Aug/21

Ψ = \int_{0}^{\infty} \frac{x^{4}}{(x^{2} + 1) (a^{4} x^{4} + 1)} dx \Rightarrow Ψ =_{ax = t} \int_{0}^{\infty} \frac{t^{4}}{a^{4} (\frac{t^{2}}{a^{2}} + 1) (t^{4} + 1)} \frac{dt}{a}

= \frac{1}{a^{5}} \int_{0}^{\infty} \frac{a^{2} t^{4}}{(t^{2} + a^{2}) (t^{4} + 1)} dt = \frac{1}{a^{3}} \int_{0}^{\infty} \frac{t^{4}}{(t^{2} + a^{2}) (t^{4} + 1)} dt

(we suppose a > 0 because Ω (- a) = Ω (a))

\Rightarrow a^{3} Ψ = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{t^{4}}{(t^{2} + a^{2}) (t^{4} + 1)} dt let φ (z) = \frac{z^{4}}{(z^{2} + a^{2}) (z^{4} + 1)}

on φ (z) = \frac{z^{4}}{(z - ia) (z + ia) (z^{2} - i) (z^{2} + i)}

= \frac{z^{4}}{(z - ia) (z + ia) (z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})}

= \frac{z^{4}}{(z - ia) (z + ia) (z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, ia) + Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})}

Res (φ, ia) = \frac{{(ia)}^{4}}{(2 ia) ({(ia)}^{4} + 1)} = \frac{a^{4}}{2 ia (1 + a^{4})}

Res (φ, e^{\frac{i π}{4}}) = \frac{- 1}{(i + a^{2}) ({2 e}^{\frac{i π}{4}}) (2 i)} = \frac{- e^{- \frac{i π}{4}}}{4 i (a^{2} + i)}

Res (φ, - e^{- \frac{i π}{4}}) = \frac{- 1}{(- i + a^{2}) (- {2 e}^{- \frac{i π}{4}}) (- 2 i)} = \frac{- e^{\frac{i π}{4}}}{4 i (- i + a^{2})}

= \frac{e^{\frac{i π}{4}}}{4 i (a^{2} - i)} \dots . . be continued \dots

Commented by mathdanisur last updated on 09/Aug/21

Thank you Ser