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In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to find the residues in the upper halfplane.

$\int_0^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \frac12\int_{-\infty}^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \pi i\sum_{\{\Im z > 0\}}$res$(\frac{x^2}{x^4 + 6x^2 + 13})$

There are two residues to calculate:

  • $z = \sqrt{-3 + 2i}$: $\frac{\sqrt{-3 + 2i}}{4(-3+ 2i) + 12} = -\frac i8\sqrt{-3 + 2i}$

  • $z = \sqrt{-3 - 2i}: \frac i8\sqrt{-3 - 2i}$

(Wolfram Alpha if you don't want to trust me)

Giving me overall for the integral:

$\frac\pi8 (\sqrt{-3 + 2i} - \sqrt{-3 - 2i}) = $1.427346... i

But the answer is clearly not meant to be imaginary.

Christian Fieldhouse
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2 Answers2

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Let us try to avoid useless computations: $x^4+6x^2+13=(x^2+\alpha)(x^2+\beta)$ for a couple of conjugated complex numbers $\alpha,\beta$ with positive real part and such that $\alpha\beta=13$ and $\alpha+\beta=6$.
By partial fraction decomposition we have

$$ \int_{0}^{+\infty}\frac{x^2}{(x^2+\alpha)(x^2+\beta)}\,dx = \frac{1}{\beta-\alpha}\int_{0}^{+\infty}\left(\frac{\beta}{x^2+\beta}-\frac{\alpha}{x^2+\alpha}\right)\,dx = \frac{\pi}{2\left(\sqrt{\beta}+\sqrt{\alpha}\right)}$$ and $$\left(\sqrt{\alpha}+\sqrt{\beta}\right)^2 = \alpha+\beta+2\sqrt{\alpha\beta} = 6+2\sqrt{13} $$ hence the wanted integral equals $\frac{\pi}{2\sqrt{6+2\sqrt{13}}}$. Similarly $$ \int_{0}^{+\infty}\frac{x^2\,dx}{x^4+Ax^2+B} = \frac{\pi}{2\sqrt{A+2\sqrt{B}}} $$

for any $A,B>0$. Lazy is good.

Jack D'Aurizio
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You need both square-roots in the upper half-plane. The difference between those two square-roots is real.

Empy2
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  • No, actually the OP does not need to compute the roots in a explicit way. (Nor to apply the residue theorem) – Jack D'Aurizio Apr 04 '18 at 14:37
  • Ok, I have discovered that in fact $\sqrt{-3 - 2i}$ has negative imaginary part, so I want the residue of its negative, (which is the negative of its residue), giving $\frac\pi8 (\sqrt{-3 + 2i} + \sqrt{-3 - 2i})$ which is real and presumably correct. – Christian Fieldhouse Apr 04 '18 at 14:42