Solving Complex Integrals with Cauchy's & Residue Theorem

[skrat]

Homework Statement

Calculate following integrals:

a) $\int _{|z|=1}\frac{e^z}{z^3}dz$

b) $\int _{|z|=1}\frac{sin^6(z)dz}{(z-\pi /6)^3}$

Homework Equations

The Attempt at a Solution

I am really confused, so before writing my solutions I would need somebody to please tell me:
- What is the difference between Cauchy's integral formula and Residue theorem? I am guessing there is a difference because I just don't get the same results.
- Which do I use here? or in other words: When do I use Residue theorem and when is is wiser to go with the Cauchy's formula?

a)
$\frac{e^z}{z^3}$ has a pol of order 3 in $z=0$. Therefore the residue is $1/2$ which brings me to

$\int _{|z|=1}\frac{e^z}{z^3}dz=\pi i$

b)
$\frac{sin^6(z)dz}{(z-\pi /6)^3}$ also pole of order 3, this time in $z=\pi /6$ and therefore the residue $-\frac{30}{2}(\frac{1}{2})^4=\frac{15}{16}$ and finally

$\int _{|z|=1}\frac{sin^6(z)dz}{(z-\pi /6)^3}=2\pi i\frac{15}{16}$That's IF using Residue theorem is ok...

 

[vanhees71]

It's the theorem of residues here. Of course you can also interpret it as an application of the Cauchy-integral formula at 0 argument. To find the residue you need the coefficient $a_{-1}$ of the integrand's Laurent expansion. Your first example is right, because
$$\frac{\exp z}{z^3}=\cdots+\frac{1}{2z}+\cdots$$
The only residue inside the unit circle is $a_{-1}=1/2$ and the integral is thus $\mathrm{i} \pi$ provided you have to integrate counter-clockwise (mathematically positive direction).

If I've not made a mistake (using Mathematica ;-)), you should better check your 2nd solution.

 

[skrat]

Let's go step by step..

By definition $Res(f,a)=lim_{z->a}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(f(z)(z-a)^n)$.

Therefore $Res(f,\pi /6)=lim_{z->\pi /6}\frac{1}{2!}\frac{d^2}{dz^2}(\frac{sin^6(z)}{(z-\pi /6)^3}(z-\pi /6)^3)=lim_{z->\pi /6}\frac{1}{2!}\frac{d^2}{dz^2}(sin^6(z))$.

Now $\frac{d^2}{dz^2}(sin^6(z))=\frac{d}{dz}(6cos^5(z))=-30sin^4(z)$.

$Res(f,\pi /6)=lim_{z->\pi /6}\frac{1}{2!}(-30sin^4(z))=-\frac{1}{2}30(\frac{1}{2})^4=-\frac{15}{16}$

And finally the integral $\int _{|z|=1}\frac{sin^6(z)dz}{(z-\pi /6)^3}=-2\pi i\frac{15}{16}$.

If sign error is all you had in mind than ok, if not, i give up!

 

[CAF123]

Hi skrat,

What is the difference between Cauchy's integral formula and Residue theorem?

Cauchy's integral formula is simply a special case of the Residue theorem where the function you are integrating over the closed contour, $\gamma$, is analytic everywhere in $\text{Int} \gamma$. So this means the RHS of the Residue theorem collapses to zero.

Now $\frac{d^2}{dz^2}(sin^6(z))=\frac{d}{dz}(6cos^5(z))=-30sin^4(z)$.

There is an error in your differentiation there.

 

[skrat]

Hi skrat,
Cauchy's integral formula is simply a special case of the Residue theorem where the function you are integrating over the closed contour, $\gamma$, is analytic everywhere in $\text{Int} \gamma$. So this means the RHS of the Residue theorem collapses to zero.

-> So one could say that the Residue Theorem is more general?

-> Now if f has singularity points (therefore not analytic in those points) in $\text{Int} \gamma$, than the Residue Theorem says that the integral is equal to $2\pi i \sum Res(f,a)$. Right?

-> Is there anything we can say, IF the singularity point happens to be right on the edge?

AND

HAHAHAHA for my awesome derivation!

$\frac{d^2}{dz^2}(sin^6(z))=\frac{d}{dz}(6sin^5(z)cos(z))$ ...

 

[CAF123]

-> So one could say that the Residue Theorem is more general?

-> Now if f has singularity points (therefore not analytic in those points) in $\text{Int} \gamma$, than the Residue Theorem says that the integral is equal to $2\pi i \sum Res(f,a)$. Right?

Right.

-> Is there anything we can say, IF the singularity point happens to be right on the edge?

So you mean if the singularity happens to coincide with $\gamma$? In this case, you will have to distort the contour in some way that is suitable to the problem at hand and then consider the limit when this distortion tends to zero. In your other thread, when you were computing the integral of sinz/z from 0 to ∞, to allow you to apply the Residue theorem, you had to find a way to close the contour. Since $\sin z$ is not bounded on $\mathbb{C}$, an equivalent and more helpful way to formulate the question is to consider $$\text{Im} \int_0^\infty \frac{e^{iz}}{z}\,\text{d}z$$ and notice that $e^{iz}$ is bounded on a circle... The integrand has a pole at the origin, that lies on the created closed contour.