Answer to Question #343591 in Complex Analysis for Not the answer

Question

Evaluate the following integrals using residue theorem: (a) ∫∞−∞ x² /(x² +9)(x² +4)²dx

Answer 1

$\int_{-\infin}^{\infin}\cfrac{x^2dx}{(x^2+9)(x^2+4)^2}$

Let $F(x) = \cfrac{x^2}{(x^2+9)(x^2+4)^2}$

$F \in C(\mathbb{R})$ because $\forall x \in \mathbb{R} \ x^2+9 \neq 0$ and $(x^2+4)^2\neq 0$

Consider $F(z) = \cfrac{z^2}{(z^2+9)(z^2+4)^2}, z \in \mathbb{C}$

Find poles of $F(z):$

$z = \pm 3i$ - simple pole

$z = \pm2i$ - order 2 pole

So, using residue theorem:

$\int_{-\infin}^{\infin}\cfrac{x^2dx}{(x^2+9)(x^2+4)^2} = 2\pi i(\mathop{\mathrm{Res}}_{z = 2i}F(z)+\mathop{\mathrm{Res}}_{z = 3i}F(z))$

Find the residuals:

$\mathop{\mathrm{Res}}_{z = 3i}F(z) = \lim_{z \to3i} [F(z)(z-3i)] = \lim_{z \to3i} [\cfrac{z^2}{(z+3i)(z^2+4)^2}] = \cfrac{-9}{6i \cdot(-9+4)} = \cfrac{-3}{50i} = \cfrac{3i}{50}$

$\mathop{\mathrm{Res}}_{z = 2i}F(z) = \lim_{z \to 2i}[(\cfrac{z^2}{(z^2+9)(z+2i)^2})']$

Find derivative:

$(\cfrac{z^2}{(z^2+9)(z+2i)^2})' = \cfrac{2z(z^2+9)(z+2i)^2-z^2(2z(z+2i)^2+2(z+2i)(z^2+9))}{(z^2+9)^2(z+2i)^4}$

Then

$\mathop{\mathrm{Res}}_{z = 2i}F(z) = \lim_{z \to 2i}[\cfrac{2z(z^2+9)(z+2i)^2-z^2(2z(z+2i)^2+2(z+2i)(z^2+9))}{(z^2+9)^2(z+2i)^4}] =\\ =\cfrac{4i(-4+9)(4i)^2+4(4i(4i)^2+2(4i)(-4+9))}{(-4+9)^2(4i)^4} = \cfrac{-320i-96i}{6400} = \cfrac{-3i}{200}$

And finally:

$\int_{-\infin}^{\infin}\cfrac{x^2dx}{(x^2+9)(x^2+4)^2} = 2\pi i(\mathop{\mathrm{Res}}_{z = 2i}F(z)+\mathop{\mathrm{Res}}_{z = 3i}F(z)) = 2\pi i(\cfrac{3i}{50}-\cfrac{13i}{200}) = \cfrac{\pi}{100}$