residue theorem

**cribby** · May 16th, 2010, 10:58 PM

Find an expression for $\int_0^{\infty} \frac{1}{p(x)} \, dx$ where $p(x)$ is a real polynomial having no nonnegative real zeroes.

Ok. So I think it's clear that I'm supposed to set this up for an application of the residue theorem by considering $\int_C \frac{\log z}{p(z)}dz$ where $C$ is the cut-annulus contour (an annulus missing a band about the real line). Then the residue theorem says that this complex line integral has the value $2 \pi i \cdot \sum_{j=1}^n \left( Res _{1/p(x)}(z_j) \cdot Ind _{C}(z_j) \right)$ , where $n$ is the number of zeroes of $p(z)$ and each $z_j$ is one of these zeroes. We can ignore the index for each zero given the contour because for any zero (which produces an isolated singularity) the index will be one.

But now what? All I've really done is write down the residue theorem. I don't know the zeroes of the polynomial, so no further calculation can be done with the above. Am I supposed to break the integral up now and look at the integral over each piece? If I let $C1$ be the staight-line piece just above the real axis, $C2$ be the outer ring, $C3$ be the straight-line piece just below the real axis, and $C4$ be the inner ring, would that go something like:

$\int_{C1}\frac{\log z}{p(z)}dz + \int_{C2}\frac{\log z}{p(z)}dz + \int_{C3}\frac{\log z}{p(z)}dz + \int_{C4}\frac{\log z}{p(z)}dz$ ?

But then I have to fix the inner radius, the outer radius, and the width of the deleted band of the annulus, and then take limits as the width and inner radius go to zero and the outer radius goes to infinity? I think what happens is that the integrals over the rings vanish, but the other 2 integrals do not...but I have no information really about the polynomial so I'm lost. How should I proceed (especially formally)?

**shawsend** · May 17th, 2010, 04:17 AM

If no zeros are on the real axis or at the origin, then the integral over the leg above the real axis is:

$\int_0^{\infty}\frac{\log(x)}{p(x)}dx$

and over the lower leg is:

$\int_{\infty}^0\frac{\log(x)+2\pi i}{p(x)}dx$

What happens when you then add them?

**cribby** · May 17th, 2010, 08:19 AM

So I change limits on the second integral before I can add them, then $\log x \, - \log x = \log (\frac{x}{x}) =0$ , so the sum is equal to $2\pi i \cdot \int_0^{\infty} \frac{1}{p(x)} \, dx$ .

So, $\int_0^{\infty} \frac{1}{p(x)} \, dx$ is just the sum of the residues?

**shawsend** · May 17th, 2010, 09:41 AM

No. It's the negative and I don't think it converges unless the polynomial has a degree 2 or more. Try and slowly go over each minuscule step. Draw a nice key-hole contour with the branch cut along the positive real axis, label all the "legs", prove the integral over the circular parts go to zero and then if use the branch $\log(z)=\ln(r)+i\theta,\quad 0\leq \theta<2\pi$ , then over the branch-cut contours, we get:

$\int_0^{\infty}\frac{\log(x)}{p(x)}dx+\int_{\infty }^0 \frac{\log(x)+2\pi i}{p(x)}dx=2\pi i\sum_{z_i}\mathop \text{Res}\limits_{z=z_i}\left\{\frac{log(z)}{p(z) }\right\}$

The log parts cancel when you switch the limits on the second one and a $\frac{-2\pi i}{p(x)}$ for the remainder.

**cribby** · May 17th, 2010, 11:59 AM

Right, I missed that negative.

I don't really know what I'm doing with branch cuts, just automatically applying a method I saw in an example. What's the intuition of that application here? Sorry if this seems like a ridiculous question but I've looked it up in my text where there is a cursory note on it, however the examples lack any helpful explanation in application.

**shawsend** · May 17th, 2010, 12:21 PM

Originally Posted by cribby

Right, I missed that negative.

I don't really know what I'm doing with branch cuts, just automatically applying a method I saw in an example. What's the intuition of that application here? Sorry if this seems like a ridiculous question but I've looked it up in my text where there is a cursory note on it, however the examples lack any helpful explanation in application.

The intuition is the underlying Riemann surfaces that make up multi-valued functions. If you were to draw the real and imaginary components of log(z), you would notice the imaginary component is in the form of a helix in which the height difference between the surface at each revolution is 2pi. log(z)/p(x) on the lower contour is 2pi i difference from the contour above the real axis because p(x) is analytic and the 2pi comes from going only 2pi around the log surface. If you went around twice, it would be 4pi and so forth. The key to understanding branch-cut integrals is understanding this geometry.

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