Residue integral calculation, Cauchy Principal Value, etc.

First: the Cauchy Principal Value. Remember when you were in calculus and did \int^{\infty}_{-\infty} \frac{dx}{1+x^2} ? If you said something like lim_{b \rightarrow \infty} \int^{b}_{-b} \frac{1}{1+x^2} dx you lost some credit and said that the integral converged if and only if both lim_{a \rightarrow \infty} \int^{a}_{0} \frac{1}{1+x^2} dx AND lim_{b \rightarrow \infty} \int^{0}_{-b} \frac{1}{1+x^2} dx BOTH converged..the limits had to be independent of one another.

That is correct, of course, but IF you knew that both integrals converged then, in fact, lim_{b \rightarrow \infty} \int^{b}_{-b} \frac{1}{1+x^2} dx gave the correct answer.

As for why you need convergence: lim_{b \rightarrow \infty} \int^{b}_{-b} x dx = lim_{b \rightarrow \infty}|^b_{-b} \frac{x^2}{2} =0 so while the integral diverges, this particular limit is zero. So this limit has a name: it is called the Cauchy Principal Value of the integral and is equal to the value of the improper integral PROVIDED the integral converges.

We will use this concept in some of our calculations.

One type of integral: Let f(z) be a function with at most a finite number of isolated singularities, none of which lie on the real axis. Suppose for R large enough, |f(Re^{it})| \leq \frac{|M|}{R^p} for p > 1 . Let H denote the upper half plane (Im(z) > 0 ) Then \int^{\infty}_{-\infty} f(z) dz = 2 \pi i \sum \{ \text{residues in } H\}

The idea: let C represent the contour shown above: upper half of the circle |z| = R followed by the diameter, taken once around in the standard direction. Of course \int_C f(z) dz = 2 \pi i \sum \{ \text{residues enclosed by } C \}  and if R is large enough, the contour encloses all of the singularities in the upper half plane.

But the integral is also equal to the integral along the real axis followed by the integral along the upper half circle. But as far as the integral along the upper half circle (call it \Gamma )

|\int_{\Gamma} f(z) dz | \leq \frac{|M|}{R^p} \pi R = \frac{|M|}{R^{p-1}} \rightarrow 0 as R \rightarrow \infty because p-1 > 0 . The bound comes from the “maximum of the function being integrated times the arc length of the integral”.

So the only non-zero part left is lim_{R \rightarrow \infty} \int^{R}_{-R} f(z) dz which is the Cauchy Principal value, which, IF the integral is convergent, is equal to 2 \pi i \sum \{ \text{residues in } H\} .

Application: \int^{\infty}_{-\infty} \frac{x^2}{x^4+1} dx Note that if f(z) = \frac{z^2}{z^4+1} then |f(Re^it)| \leq \frac{R^2}{R^4-1} which meets our criteria. So the upper half plane singularities are at z_0 = e^{i\frac{\pi}{4}}, z_1 = e^{i\frac{3\pi}{4}} ; both are simple poles. So we can plug these values into \frac{z^2}{4z^3} = \frac{1}{4}z^{-1} (the \frac{h}{g'} formula for simple poles)

So to finish the integral up, the value is 2 \pi i \frac{1}{4}((e^{i\frac{\pi}{4}})^{-1} +(e^{i\frac{3\pi}{4}})^{-1} ) = \frac{1}{2} \pi i (-2i\frac{1}{\sqrt{2}}) = \frac{\pi}{\sqrt{2}}

Now for a more delicate example
This example will feature many concepts, including the Cauchy Principal Value.

We would like to calculate \int^{\infty}_0 \frac{sin(x)}{x} dx with the understanding that lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1 so we assume that \frac{sin(0)}{0} \text{``="} 1 .

First, we should show that \int^{\infty}_0 \frac{sin(x)}{x} dx converges. We can do this via the alternating series test:

\int^{\infty}_0 \frac{sin(x)}{x} dx = \sum^{\infty}_{k=0} \int_{\pi k}^{\pi (k+1)} \frac{sin(x)}{x} dx which forms an alternating series of terms whose magnitudes are decreasing to 0 (we are adding the signed areas bounded by the graph of \frac{sin(x)}{x} and the x axis and note that the signed areas alternate between positive and negative and the absolute values of the areas decrease to zero as |\frac{1}{x}| \geq |\frac{sin(x)}{x} | for x > 0

Above: we have the graphs of Si(x) = \int^x_0 \frac{sin(t)}{t} dt \text{ and } \frac{sin(x)}{x} .

Aside: note that \int^{\infty}_0 |\frac{sin(x)}{x}| dx diverges.
Reason: |  \int^{(k+1) \pi}_{k \pi} |\frac{sin(x)}{x}| dx > |\frac{1}{(k+1) \pi}| \int^{(k+1) \pi}_{k \pi}|sin(x)| dx =|\frac{2}{(k+1) \pi}| and these form the terms of a divergent series.

So we know that our integral converges. So what do we do? Our basic examples assume no poles on the axis. And we do have the trig function.

So here is what turns out to work: use f(z) = \frac{e^{iz}}{z} . Why? For one, along the real axis we have \frac{e^{ix-y}}{x} = \frac{e^{ix}}{x} =\frac{cos(x)}{x} + i \frac{sin(x)}{x} and so what we want will be the imaginary part of our expression..once we figure out how to handle zero AND the real part. Now what about our contour?

Via (someone else’s) cleverness, we use:

We have a small half circle around the origin (upper half plane), the line segment to the larger half circle, the larger half circle in the upper half plane, the segment from the larger half circle to the smaller one. If we call the complete contour \gamma then by Cauchy’s theorem \int_{\gamma} \frac{e^{iz}}{z} dz = 0 no matter what r \text{ and } R are provided 0 < r < R . So the idea will be to let r \rightarrow 0, R \rightarrow \infty and to see what happens.

Let L stand for the larger half circle and we hope that as R \rightarrow \infty we have \int_L \frac{e^{iz}}{z} dz \rightarrow 0

This estimate is a bit trickier than the other estimates.

Reason: if we try |\frac{e^{iRe^{it}}}{Re^{it}}| = |\frac{e^{iRcos(t)}e^{-Rsin(t)}}{Re^{it}} = |\frac{e^{-Rsin(t)}}{R}| which could equal |\frac{1}{R}| when t = 0, t = \pi . Then multiplying this by the length of the curve (\pi R ) we simply get \pi which does NOT go to zero.

So we turn to integration by parts: \int_L \frac{e^{iz}}{z} dz =  and use u = \frac{1}{z}, dv = e^{iz}, du = -\frac{1}{z^2}, v = -ie^{iz} \rightarrow \int_L \frac{e^{iz}}{z} dz= -ie^{iz}\frac{1}{z}|_L - i \int_L \frac{e^{iz}}{z^2} dz

Now as R \rightarrow \infty we have |-ie^{iz}\frac{1}{z}|_L | \leq |\frac{1}{R}| which does go to zero.

And the integral: we now have a z^2 in the denominator instead of merely z hence

|\int_L \frac{e^{iz}}{z^2} dz | \leq |\frac{1}{R^2}| \pi R = \frac{\pi} {R} which also goes to zero.

So we have what we want on the upper semicircle.

The lower semicircle We will NOT get zero as we integrate along the lower semicircle and let the radius shrink to zero. Let’s denote this semicircle by l (and note that we are going in the opposite of the standard direction) so we are really calculating $latex -\int_l \frac{e^{iz}}{z} dz  . This is not an elementary integral. But what we can do is expand \frac{e^{iz}}{z} into its Laurent series centered at zero and obtain:
\frac{e^{iz}}{z} = \frac{1}{z} (1 + (iz) + (iz)^2 \frac{1}{2!} + (iz)^3\frac{1}{3!} + ...) = (\frac{1}{z} + i -z \frac{1}{2} +-z^2 \frac{1}{3!} + ...) = \frac{1}{z} + g(z) where g(z) represents the regular part..the analytic part.

So now -\int_l \frac{e^{iz}}{z} dz = -\int_l \frac{1}{z} dz  -\int_l g(z) dz . Now the second integral goes to zero as r \rightarrow 0 because it is the integral of an analytic function over a smooth curve whose length is shrinking to zero. (or alternately: whose endpoints are getting closer together; we’d have G(-r) - G(r) where G is a primitive of g and -r \rightarrow r .)

The part of the integral that survives is -\int_l \frac{1}{z} dz = -\pi i for ANY r (check this: use z(t) = re^{it}, t \in [0, \pi] .

So adding up the integrals (I am suppressing the integrand for brevity)

\int_L + \int^{-r}_{-R} -\int_l + \int^{R}_r = 0 \rightarrow \int^{-R}_{r} + \int^{R}_r = \int_l = \pi i which is true for r \rightarrow 0, R \rightarrow \infty .

So we have lim_{R \rightarrow \infty, r \rightarrow 0 } \int^{-r}_{-R} \frac{cos(x)}{x} + i \frac{sin(x)}{x} dx +  \int^{R}_{r} \frac{cos(x)}{x} + i \frac{sin(x)}{x} dx =\pi i

Now for the cosine integral: yes, as r \rightarrow 0 this integral diverges. BUT note that for all finite R, r > 0 we have:\int^{-r}_{-R} \frac{cos(x)}{x}  +  \int^{R}_{r} \frac{cos(x)}{x}  dx = 0 as \frac{cos(x)}{x} is an odd function. So the Cauchy Principal Value of \int^{\infty}_{-\infty} \frac{cos(x)}{x} dx = 0

We can now use the imaginary parts and note that the integrals converge (as previously noted) and so:

\int^{\infty}_{-\infty} i\frac{sin(x)}{x} dx = i \pi \rightarrow \int^{\infty}_0 \frac{sin(x)}{x}dx = \frac{\pi}{2}