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I want to find the integral

$$I=\int_0^\infty \frac{dx}{(x^2+2x+12)^2}$$

using contour integration; I am familiar with the trigonometric substitution in real analysis.

There are no branch cuts, there are two second order poles at $z=-1\pm\sqrt{11}i$. I cannot use $\int_{-\infty}^\infty=2\int_0^\infty$ since the function isn't even.

Clearly, the choice of contour is important here (compare with this question). I tried half-circle and quarter-circle contours, but they just give some weird rationals. Is there some better contour here? Am I missing something?

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    $\begingroup$ I think you can do this integral by calculating instead $\int_\Gamma \frac{\log z}{(z^2+2x+12)^2}\,dz$ where $\Gamma$ is a curve something like: from origin out to $(R,\epsilon)$, then a circle to $(R,-\epsilon)$ and then back to the origin. Let $R\to+\infty$ and $\epsilon\to 0+$. Also, you use the $\log$ with a branch cut along the positive real axis. The reason this works (if I don't make a mistake) is that the logarithms cancel... $\endgroup$
    – mickep
    Commented Jan 11, 2015 at 19:00

1 Answer 1

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Yes. There is a standard trick for integrating functions over $[0,\infty)$, and that is to exploit the multi-valuedness of the logarithm about a branch cut. Thus, consider

$$\oint_C dz \frac{\log{z}}{(z^2+2 z+12)^2} $$

where $C$ is a keyhole contour as pictured below:

enter image description here

The radius of the small circle is $\epsilon$ and that of the bigger circle is $R$. The contour integral over $C$ is then

$$\int_{\epsilon}^R dx \frac{\log{x}}{(x^2+2 x+12)^2} + i R \int_0^{2 \pi} d\theta \, e^{i \theta}\frac{\log{\left ( R e^{i \theta} \right )}}{\left (R^2 e^{i 2 \theta} + 2 R e^{i \theta}+12 \right )^2} \\ + \int_R^{\epsilon} dx \frac{\log{x} + i 2 \pi}{(x^2+2 x+12)^2} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\left ( \epsilon e^{i \phi} \right )}}{\left (\epsilon^2 e^{i 2 \phi}+ 2 \epsilon e^{i \phi}+12 \right )^2}$$

It should be evident that, as $R \to \infty$, the second integral vanishes as $\log{R}/R^3$, and that as $\epsilon \to 0$, the fourth integral vanishes as $\epsilon \log{\epsilon}$. Thus, in this limit, the contour integral is

$$-i 2 \pi \int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} $$

By the residue theorem, this contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z_{\pm} = -1 \pm i \sqrt{11}$. Therefore we have

$$\begin{align}\int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} &= - \left [\frac{d}{dz} \frac{\log{z}}{(z-z_-)^2} \right ]_{z=z_+} - \left [\frac{d}{dz} \frac{\log{z}}{(z-z_+)^2} \right ]_{z=z_-}\\ &= -\frac1{z_+ (z_+-z_-)^2} + \frac{2 \log{z_+}}{ (z_+-z_-)^3}\\ &-\frac1{z_- (z_--z_+)^2} + \frac{2 \log{z_-}}{(z_--z_+)^3} \\ &= -\frac{z_++z_-}{z_- z_+ (z_+-z_-)^2} + 2 \frac{\log{z_+}-\log{z_-}}{(z_+-z_-)^3} \end{align}$$

To get this right, it is crucial that we understand how to compute the logs. Understand that, in defining the contour, we demanded that $\arg{z} \in [0,2 \pi)$. Thus,

$$\log{z_+} = \frac12 \log{12} + i \left (\pi- \arctan{\sqrt{11}}\right )$$ $$\log{z_-} = \frac12 \log{12} +i \left (\pi+ \arctan{\sqrt{11}}\right )$$

Putting this altogether, we find, after some arithmetic, that

$$\int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} = \frac{1}{22 \sqrt{11}} \arctan{\sqrt{11}} - \frac1{264} $$

ADDENDUM

Another way of evaluating this integral is to observe that

$$\int_0^{\infty} \frac{dx}{(x^2+2 x+b)^2} = -\frac{\partial}{\partial b} \int_0^{\infty} \frac{dx}{x^2+2 x +b} $$

then consider

$$\oint_C dz \frac{\log{z}}{z^2+2 z+b} $$

The analysis is the same as above; the result is that

$$-\int_0^{\infty} \frac{dx}{x^2+2 x +b} = \frac{\log{z_+}-\log{z_-}}{z_+-z_-} = -\frac{\arctan{\sqrt{b-1}}}{\sqrt{b-1}}$$

Differentiating, we get

$$\int_0^{\infty} \frac{dx}{(x^2+2 x+b)^2} = \frac{\arctan{\sqrt{b-1}}}{2 (b-1)^{3/2}}-\frac{1}{2 b (b-1)}$$

Plugging in $b=12$ produces the above result.

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  • $\begingroup$ (+1) I'm now happy I did not write some half-crappy solution to this :) $\endgroup$
    – mickep
    Commented Jan 11, 2015 at 19:11
  • $\begingroup$ Wonderful, thank you! $\endgroup$
    – Minethlos
    Commented Jan 11, 2015 at 19:39

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