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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.

Alexander
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  • Here is an example where the complex integration loses efficiency to the simple Newton-Leibniz rule of anti-derivatives. – A.Γ. Aug 09 '15 at 22:02
  • 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. – Winther Aug 09 '15 at 23:12

1 Answers1

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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 $$I=\frac{\pi}{2}(\sqrt{3}-\sqrt{2})$$

Mark Viola
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  • 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 – Alexander Aug 09 '15 at 22:42
  • 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 $$\oint_C \frac{1}{z^4+5x^2+6}dz=2\pi i \sum \text{Residues}$$ – Mark Viola Aug 09 '15 at 23:57
  • the thing is when I calculate the $Res_{z_0} \left( \frac{3}{x^2 +3}\right) = 0$, and the same for $Res_{z_0} \left( \frac{2}{x^2 +}\right)$ – Alexander Aug 10 '15 at 01:15
  • Only the residues in the upper half plane are implicated when we close the contour in the upper-half plane. – Mark Viola Aug 10 '15 at 01:18
  • do you mean that only for the $z_0 = i\sqrt{3}$ and $z_0 = i\sqrt{2}$, not the negative pole ? – Alexander Aug 10 '15 at 01:25
  • oh yeah, I think I got it, it's because it is from $0$ to $\infty$ – Alexander Aug 10 '15 at 01:28
  • @Alexander Well, not quite. It is because we closed in the upper half plane. If we close in the lower half plane, then the other poles are implicated AND the contour is clockwise ... therefore, we multiply the sum of those residues by $-2\pi i$ instead of $2\pi i$. – Mark Viola Aug 10 '15 at 01:33