[SOLVED] Line integral, Cauchy's integral formula

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September 15th 2009, 12:39 PM
arbolis

[SOLVED] Line integral, Cauchy's integral formula

Let $C$ be the frontier of the square whose sides are the lines $x=\pm 2$ and $y= \pm 2$ . (orientation is positive)
Calculate the following integral using Cauchy's integral formula for a function or its derivative.

$\int _C \frac{e^{-z}dz}{z-\frac{i\pi}{2}}$ .

My attempt : I find confusing the "using Cauchy's integral formula for a function or its derivative.", I don't really know what they mean by that.

So Cauchy's formula is $f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi$ .

Looking at the integrand of the integral, I see that it is not defined for $z=i \frac{\pi}{2}$ . Does that mean that $f$ is not analytic in this point?
I believe the integral should equals $2 \pi i$ , I don't really know why.

I need help, I find all this new theory somewhat hard to grasp due to the amount of it.
(Bow)
September 15th 2009, 01:03 PM
shawsend

Cauchy's Integral Formula states that if $f(z)$ is analytic in $C$ and the point $z$ is not on C, then:

$f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi<br />$

Ok, $f(z)=e^{-z}$ and that's analytic in C. Don't confuse the ENTIRE integrand with the expression for $f(z)$ when using this formula.

So plugging in $z=\pi i/2$ , then the integral is just:

$2\pi i f(\pi i/2)$ right?

Try and get good at calculating these in Mathematica so that you can check your results. For example (where I used an equivalent circle of radius 2 via deformation of path theorem):

Code:

In[111]:= NIntegrate[(Exp[-z]/(z - Pi*(I/2)))*2*I* Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}] N[2*Pi*I*Exp[(-Pi)*(I/2)]] Out[111]= 6.283185307179801 - 2.708944180085382*^-14*I Out[112]= 6.283185307179586
September 15th 2009, 01:18 PM
arbolis

Thanks for the reply, however I'm extremely confused.

Quote:

Originally Posted by shawsend

Cauchy's Integral Formula states that if $f(z)$ is analytic in $C$ and the point $z$ is not on C, then:

$f(z)=\frac{1}{2\pi i} \int _C \frac{f(\xi)}{\xi -z}d \xi<br />$

but $z$ has to be in the interior of $C$ , right?
If so, then in the exercise I can consider $z$ as $i \frac{\pi}{2}$ and I notice that indeed $z$ is inside the square $C$ .
Also z is a pole I think. I don't know if it's important.

Quote:

Ok, $f(z)=e^{-z}$ and that's analytic in C. Don't confuse the ENTIRE integrand with the expression for $f(z)$ when using this formula.

So plugging in $z=\pi i/2$ , then the integral is just:

$2\pi i f(\pi i/2)$ right?

Here I don't understand why you worked with the numerator of the integrand only. I know that it is analytic in $C$ . However I think the denominator is not analytic at $z$ , inside $C$ , the square.

Quote:

Try and get good at calculating these in Mathematica so that you can check your results. For example (where I used an equivalent circle of radius 2 via deformation of path theorem):

Code:

In[111]:= NIntegrate[(Exp[-z]/(z - Pi*(I/2)))*2*I* Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}] N[2*Pi*I*Exp[(-Pi)*(I/2)]] Out[111]= 6.283185307179801 - 2.708944180085382*^-14*I Out[112]= 6.283185307179586

I never heard of such a theorem... looks interesting though.
Before computing I'd like to understand what I'm doing. I know much less than you think.
September 15th 2009, 01:28 PM
arbolis

Oh... I think I just understood what you meant :
Thanks to CFI we have $2 \pi i f(\frac{i \pi}{2})= \int _C \frac{e^{-z}dz}{z- \frac{i \pi}{2}}$ .
But $f(\frac{i\pi}{2})=e^{-\frac{i \pi}{2}}\Rightarrow \int _C \frac{e^{-z}dz}{z- \frac{i \pi}{2}}=2\pi i \cdot (-i)= 2\pi$ .
Fantastic, I get your result!