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Using Residue theorem to evaluate the integral: $$\int_0^{\infty} \frac{x^2}{x^4 + 5x^2 +6}dx$$

I am using partial fraction to get:

$$\int_0^{\infty} \left( \frac{3}{x^2 +3} - \frac{2}{x^2+2} \right)dx$$

Then, next step, can someone show me how to use Residue theorem to evaluate the integral.

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  • $\begingroup$ Here is an example where the complex integration loses efficiency to the simple Newton-Leibniz rule of anti-derivatives. $\endgroup$ – A.Γ. Aug 9 '15 at 22:02
  • $\begingroup$ Integrals on the form $\int_0^\infty \frac{dx}{x^n + a}$ for $a>0$ can be solved using a wedge-contour. It is the same amount of work to do one single combination of $a,n$ as the general case so the bonus is that if you solve one you solve all. $\endgroup$ – Winther Aug 9 '15 at 23:12
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You just use partial fraction expansion again, for each of the quadratic terms. We have

$$\frac{1}{x^2+3}=\frac{1}{i2\sqrt{3}}\left(\frac{1}{x-i\sqrt{3}}-\frac{1}{x+i\sqrt{3}}\right)$$

and

$$\frac{1}{x^2+2}=\frac{1}{i2\sqrt{2}}\left(\frac{1}{x-i\sqrt{2}}-\frac{1}{x+i\sqrt{2}}\right)$$

Can you finish from here?

SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO SEE THE REST OF THE ANSWER

First we note that the integrand is even and therefore we can write the integral of interest $I$ as $$I=\int_0^{\infty}\frac{x^2}{x^4+5x^2+6}dx=\frac12\int_{-\infty}^{\infty}\frac{x^2}{x^4+5x^2+6}dx$$ Next, we move to the complex plane and analyze the integral $$\oint_{C}\frac{z^2}{z^4+5z^2+6}dz$$where $C$ is the contour in the upper-half plane comprised of the real-line segment from $x=-R$ to $x=R$ and the semi-circle with radius $R$ and centered at the origin. Using the Residue Theorem, we have $$\oint_{C}\frac{z^2}{z^4+5z^2+6}dz=2\pi i \left(\frac{-2}{2i\sqrt{2}}+\frac{3}{2i\sqrt{3}}\right)=\pi(-\sqrt{2}+\sqrt{3}) $$ Notice that as as $R$ goes to infinity, the contribution from the integral over $C_R$ vanishes while the integral over the real line segment is equal to twice the integral of interest as $R\to \infty$. Therefore, we have $$I=\frac{\pi}{2}(\sqrt{3}-\sqrt{2})$$

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  • $\begingroup$ the thing is how can is how can I change the domain from $[0, \infty)$ to?. Or we can just use the the formula $\int_C f(z)dz = 2\pi i \sum Res_{z_0}f(z)$. And, my nominator is $x$ not 1 $\endgroup$ – Alexander Aug 9 '15 at 22:42
  • $\begingroup$ Well, I thought that in the hidden answer we discuss exploiting the evenness of the integrand. This allows us to write $$I=\int_0^\infty \frac{1}{x^4+5x^2+6}dx=\frac12 \int_{-\infty}^{\infty}\frac{1}{x^4+5x^2+6}dx$$Then, we relate the integral $I$ to the integral $$\oint_C \frac{1}{z^4+5x^2+6}dz=2\pi i \sum \text{Residues}$$ $\endgroup$ – Mark Viola Aug 9 '15 at 23:57
  • $\begingroup$ the thing is when I calculate the $Res_{z_0} \left( \frac{3}{x^2 +3}\right) = 0$, and the same for $Res_{z_0} \left( \frac{2}{x^2 +}\right)$ $\endgroup$ – Alexander Aug 10 '15 at 1:15
  • $\begingroup$ Only the residues in the upper half plane are implicated when we close the contour in the upper-half plane. $\endgroup$ – Mark Viola Aug 10 '15 at 1:18
  • $\begingroup$ do you mean that only for the $z_0 = i\sqrt{3}$ and $z_0 = i\sqrt{2}$, not the negative pole ? $\endgroup$ – Alexander Aug 10 '15 at 1:25

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