Residues: integrals, calculation, etc.

Here is the idea: we’d like (relatively) easy ways to integrate functions around simple closed curves. For the purposes of this particular note, we assume the usual about the closed curve (piecewise smooth, once around in the standard direction) and that the function in question does not have singularities along the said closed curve and only a finite number of isolated singularities in the simply connected region enclosed by the curve. If we have no singularities in the said region, then our integral is zero and there is nothing more to do.

Now by the theory of cuts, we can assume that the integral around the given simple closed curve is equal to the sum of the integrals of the function around small circles surrounding each singular point.

So, integrals of this type just boil down to calculating the integrals of the function around these isolated singularities and then just adding.

So lets examine a typical singularity: say z_0 . Let R done the circle of radius R surrounding z_0 and assume that R is small enough that we don’t enclose any other singularity.

Now on a punctured disk (excluding z_0 ) bounded by this circle, we have

f(z) = \sum^{\infty}_{k=1} b_k \frac{1}{(z-z_0)^k} + \sum^{\infty}_{k=0} c_k(z-z_0)^k . Now we have

\int_R f(z)dz = \int_R \sum^{\infty}_{k=1} b_k \frac{1}{(z-z_0)^k} + \sum^{\infty}_{k=0} c_k(z-z_0)^k

= \int_R \sum^{\infty}_{k=1} b_k \frac{1}{(z-z_0)^k} dz + \int_R \sum^{\infty}_{k=0} c_k(z-z_0)^k dz

Now the second integral is the integral of a power series (the regular part of the Laurent series) and therefore the integral is zero since we are integrating along a simple closed curve. So only the integral of the principal part matters.

Now we can interchange integration and summation by the work we did to develop the Laurent series to begin with.

Doing that we obtain:

\int_R f(z)dz =  \sum^{\infty}_{k=1} \int_R b_k \frac{1}{(z-z_0)^k} dz and keep in mind that the b_k are constants. So we have:

\int_R f(z)dz =  \sum^{\infty}_{k=1} b_k \int_R  \frac{1}{(z-z_0)^k} dz and so we need only evaluate

\int_R \frac{1}{(z-z_0)^k} dz to evaluate our integral.

Now for k \in \{2, 3, 4, ... \} we note that \frac{1}{(z-z_0)^k} has a primitive in the punctured disk. OR..if you don’t like that, you can do the substitution z(t) = Re^{it} +z_0, dz = Rie^{it}dt, t \in [0, 2 \pi] and find that these integrals all evaluate to zero.

So only the integral \int_R \frac{1}{z-z_0} dz matters and the aforementioned substitution shows that this integral evaluates to 2 \pi i

So all of this work shows that \int_R f(z) dz = 2 \pi i b_1 ; only the coefficient of the \frac{1}{z-z_0} term matters.

b_1 is called the residue of f at z_0 and sometimes denoted Res(f,z_0)

So now we see that the integral around our original closed curve is just 2 \pi i \sum_k Res(f, z_k) , that is, 2 \pi i times the sum of the residues of f which are enclosed by our closed curve.

This means that it would be good to have “easy” ways to calculate residues of a function.

Now of course, we could always just calculate the Laurent series of the function evaluated at each singularity in question. SOMETIMES, this is easy. And if the singularity in question is an essential one, it is what we have to do.

But sometimes the calculation of the series is tedious and it would be great to have another method.

The idea
Ok, suppose we have a simple pole at z_0 (we’ll build upon this idea). This means that f(z) = b_1\frac{1}{z-z_0} + \sum^{\infty}_{k=0} c_k (z-z_0)^k Now note that (z-z_0)f(z) = b_1 +  \sum^{\infty}_{k=0} c_k (z-z_0)^{k+1} so we obtain:

b_1 = lim_{z \rightarrow z_0} (z-z_0)f(z)

Example: calculate the residue of f(z) = \frac{1}{sin(z)} at z = 0 .

lim_{z \rightarrow 0}z\frac{1}{sin(z)} =lim_{z \rightarrow 0}\frac{z}{sin(z)} =1 .

Now sometimes this is straight forward, and sometimes it isn’t.

Now if f(z) = \frac{1}{g(z)} where g(z) has an isolated zero at z = z_0 (as is the case when we have a pole), we can then write (z-z_0) f(z) = \frac{z-z_0}{g(z)} where both the numerator and denominator functions are going to zero.

This should remind you of L’Hopital’s rule from calculus. So we should derive a version for our use.

Now if both g, h have isolated zeros of finite order at z_0 then

\frac{h(z_0+h)}{g(z_0+h)} =  \frac{h(z_0+h) +h(z_0)}{g(z_0+h) + g(z_0)} (because h(z_0)=g(z_0)=0 )

= \frac{\frac{h(z_0+h) +h(z_0)}{h}}{\frac{g(z_0+h) + g(z_0)}{h}} Now take a limit as h \rightarrow 0 to obtain \frac{h'(z_0)}{g'(z_0)} which can be easier to calculate.

If we apply this technique to \frac{1}{sin(z)} we get \frac{(z)'}{sin(z)'} = \frac{1}{cos(z)} which is 1 at z = 0

Ok, my flight is boarding so I’ll write part II later.

Advertisements

Fresnel Integrals

In this case, we want to calculate \int^{\infty}_0 cos(x^2) dx and \int^{\infty}_0 sin(x^2) dx . Like the previous example, we will integrate a specific function along a made up curve. Also, we will want one of the integrals to be zero. But unlike the previous example: we will be integrating an analytic function and so the integral along the simple closed curve will be zero. We want one leg of the integral to be zero though.

The function we integrate is f(z) = e^{iz^2} . Now along the real axis, y = 0 and so e^{i(x+iy)^2} =e^{ix^2} = cos(x^2) + isin(x^2) and dz = dx So the integral along the positive real axis will be the integrals we want with \int^{\infty}_0 cos(x^2) dx being the real part and \int^{\infty}_0 sin(x^2) dx being the imaginary part.

So here is the contour

Now look at the top wedge: z = te^{i \frac{\pi}{4}}, t \in [0,R] (taken in the “negative direction”)

So z^2 = t^2 e^{i \frac{\pi}{2}} =t^2 (0 + isin(\frac{\pi}{2}) = it^2 \rightarrow e^{iz^2} = e^{i^2t^2} = e^{-t^2}

We still need dz = cos(\frac{\pi}{4}) + isin(\frac{\pi}{4}) dt = \frac{\sqrt{2}}{2}(1+i)dt

So the integral along this line becomes - \frac{\sqrt{2}}{2}(1+i)\int^{R}_0 e^{-t^2} dt = -\frac{\sqrt{\pi}}{2}\frac{\sqrt{2}}{2}(1+i) as R \rightarrow \infty . Now IF we can show that, as R \rightarrow \infty the integral along the circular arc goes to zero, we have:

\frac{\sqrt{\pi}}{2}\frac{\sqrt{2}}{2}(1+i) = \int^{\infty}_0 cos(x^2) dx + i  \int^{\infty}_0 sin(x^2) dx . Now equate real and imaginary parts to obtain:

\int^{\infty}_0 cos(x^2) dx = \frac{\sqrt{\pi}}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{2 \pi}}{4} and \int^{\infty}_0 sin(x^2) dx =\frac{\sqrt{\pi}}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{2 \pi}}{4}

So let’s set out to do just that: here z = Re^{it}, t \in [0, \frac{\pi}{4}] so e^{iz^2} = e^{iR^2e^{2it}}e^{iR^2(cos(2t) + isin(2t)} = e^{iR^2(cos(2t)}e^{-R^2sin(2t)} . We now have dz = iRe^{it} dt so now |\int^{\frac{\pi}{4}}_0 e^{iR^2(cos(2t)}e^{-R^2sin(2t)}iRe^{it} dt| \leq \int^{\frac{\pi}{4}}_0|e^{iR^2(cos(2t)} || e^{-R^2sin(2t)} |iRe^{it}| dt =

\int^{\frac{\pi}{4}}_0| e^{-R^2sin(2t)} R dt

Now note: for t \in [0, \frac{\pi}{4}] we have sin(2t) \geq \frac{2}{\pi}t

\rightarrow e^{-R^2 sin(2t)} \leq e^{-R^2\frac{2}{\pi}t} hence

\int^{\frac{\pi}{4}}_0 e^{-R^2sin(2t)} R dt \leq \int^{\frac{\pi}{4}}_0 e^{-R^2frac{2}{\pi}t} R dt = -\frac{1}{R}\frac{\pi}{2}(1-e^{-R^2\frac{1}{2}}) and this goes to zero as R goes to infinity.

Some real integral calculation

Note to other readers: if you know what a “residue integral” is, this post is too elementary for you.

Recall Cauchy’s Theorem (which we proved in class): if f is analytic on a simply connected open set A and \gamma is some piecewise smooth simple closed curve in A and z_0 is in the region enclose by \gamma then f(z_0) =\frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{w-z_0} dw

This is somewhat startling that the integral a related function along the boundary curve of a region determines the value of the function in that said region.

And, this fact, plus this fact: |\int_{\gamma} f(w)dw | \leq M l(\gamma) where l(\gamma) is the length of the curve \gamma and M = max \{|f(w)|, w \in \gamma \} can lead to the solution to integrals of real variable functions.

Here is one collection of examples: let’s try to calculate \int^{\infty}_{-\infty} \frac{dx}{x^{2a} + 1} a \in \{1,2,3,... \}

Now consider the curve \gamma which runs from -R to R along the real axis and then from R to -R along the “top semicircle” of |z| = R (positive imaginary part). Denote that curve by C_r See the following figure:

So if we attempt to integrate \frac{1}{z^{2a} + 1} along this contour we get \int_{-R}^{R} \frac{1}{x^{2a} + 1}dx + \int_{C_r} \frac{1}{z^{2a} + 1} dz

Now as we take a limit as R \rightarrow \infty the first integral becomes the integral we wish to evaluate. The second integral: remember that |z| constant along that curve and we know that |z^{2a} + 1| \geq |z^{2a}| - 1 =R^{2a} -1 along this curve, hence \frac{1}{z^{2a} + 1} \leq \frac{1}{R^{2a}-1} along C_r (assuming R > 1 )

So |\int_{C_r} \frac{1}{z^{2a} + 1} dz| \leq \frac{1}{R^{2a} -1} R \pi because l(C_r) = \pi R (think: magnitude of the integrand times the arc length).

Now lim_{R \rightarrow \infty} \frac{1}{R^{2a} -1} R \pi =0 (provided, of course, that a \in \{1, 2,...\} . So as R goes to infinity, the integral around the entire curve becomes the integral along the real axis, which is the integral that we are attempting to calculate. Note that because 2a is even, \frac{1}{x^{2a} + 1} is continuous on the whole real line.

This, of course, does not tell us what \int^{\infty}_{-\infty} \frac{1}{x^{2a} + 1} dx is but we can use Cauchy’s theorem to calculate the integral around the whole curve, which is equal to the integral along the entire real axis.

So, in order to calculate the integral along the curve, we have to deal with where \frac{1}{z^{2a} + 1} is NOT analytic. This means finding the roots of the denominator: z^{2a} + 1 that lie in the upper half plane (and are therefore contained within the curve when R is large enough). There will be a of these in the upper half plane.

Label these roots w_1, w_2,...w_a . Now draw small circles C_1, C_2, ..C_a around each of these ..the circles are small enough to contain exactly ONE w_k . Within each of the circles, \frac{1}{z^{2a}+1 } is analytic EXCEPT at that said root.

Now here is the key: for each root w_k , write z^{2a} + 1 = (z-w_k)(p_k(z) ) where p_k(z) = \frac{z^{2a} + 1}{z-w_k} Then for all k , \int_{C_k} \frac{1}{z^{2a} + 1} dz = \int_{C_k} \frac{\frac{1}{p_k(z)}}{z-w_k }dz = 2 \pi i \frac{1}{p_k(w_k)} by Cauchy’s Theorem (\frac{1}{p_k(z)} is analytic within C_k as we divided out the root within that region).

Now by using the method of cuts, the integral around the large curve \gamma is just the sum of the integrals along the smaller circles around the roots. This figure is the one for \frac{1}{z^4 + 1} .

So, putting it all together:

\int^{\infty}_{-\infty} \frac{1}{x^{2a} + 1} dx = (2 \pi i)(\frac{1}{p_1(w_1)} + \frac{1}{p_2(w_2)} +...+ \frac{1}{p_a(w_a)})

And YES, the i always cancels out so we do get a real valued answer.

I admit that calculation of p_k(w_k) can get a bit tedious but conceptually it isn’t hard.

Let’s do this for 2a= 2 and again for 2a = 4.

For \int^{\infty}_{-\infty} \frac{1}{x^2+1} dx  note that p_1(z) = \frac{z^2+1}{z-i} = z+i and so the integral is 2 \pi i (\frac{1}{i+i}) = 2 \pi \frac{1}{2} = \pi as expected (you can do this one with calculus techniques; use \int \frac{1}{x^2+1} dx = arctan(x) + C )

Now for \int^{\infty}_{-\infty} \frac{1}{x^4+1} dx

Label the roots w_1 = \frac{\sqrt{2}}{2}(1+i), w_2 = \frac{\sqrt{2}}{2}(-1+i), w_3 = \frac{\sqrt{2}}{2}(-1-i), w_4 = \frac{\sqrt{2}}{2}(1-i)

So z^4+1 = (z-w_1)(z-w_2)(z-w_3)(z-w_4) \rightarrow

p_1(w_1) = (w_1-w_2)(w_1-w_3)(w_1-w_4), p_2(w_2) = (w_2-w_1)(w_2-w_3)(w_2-w_4)

So \int^{\infty}_{-\infty} \frac{1}{x^4+1} dx = 2 \pi i(\frac{1}{p_1(w_1)} + \frac{1}{p_2(w_2)}) =

2 \pi i (\frac{1}{(\sqrt{2})^3})(\frac{1}{(1)(1+i)(i)} + \frac{1}{(-1)(i)(-1+i)} = \frac{\pi}{\sqrt{2}} i \frac{1}{i}(\frac{1}{1+i} - \frac{1}{-1+i})

= \frac{\pi}{\sqrt{2}}\frac{-1+i -( 1+i)}{(1+i)(-1+i)} = \frac{\pi}{\sqrt{2}}\frac{-2}{-2} = \frac{\pi}{\sqrt{2}}