# [SOLVED] Line integral, Cauchy's integral formula

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• Sep 15th 2009, 02:13 PM
arbolis
[SOLVED] Line integral, Cauchy's integral formula
The problem is the same as in this thread : http://www.mathhelpforum.com/math-he...l-formula.html, with the exception of the integrand : $\frac{\cosh z}{z^4}$.
My attempt : I see the singularity at $z=0$.
So $\int _C \frac{\cosh z}{z^4}dz=2\pi i f(0)$ with $f(z)=\cosh z$.
As $\cosh 0=1$, the integral is worth $2 \pi i$.

Edit : If the integrand is $\frac{z}{2z+1}$, I get the result $-\pi i$. Am I right?
• Sep 15th 2009, 02:42 PM
shawsend
Quote:

Originally Posted by arbolis
T
So $\int _C \frac{\cosh z}{z^4}dz=2\pi i f(0)$ with $f(z)=\cosh z$.
As $\cosh 0=1$, the integral is worth $2 \pi i$.

No because that's not in the form $\oint \frac{f(z)}{z-a}$. However it is in the form of $\oint \frac{f(z)}{(z-a)^k}$ and there is an extention of Cauchy's Integral Formula which covers this case.

Quote:

If the integrand is $\frac{z}{2z+1}$, I get the result $-\pi i$. Am I right?
That's not in the form $\oint \frac{f(z)}{z-a}dz$ but you can write it as:

$\oint \frac{\frac{z}{2}}{z+1/2}dz$
• Sep 15th 2009, 03:47 PM
arbolis
Quote:

Originally Posted by shawsend
No because that's not in the form $\oint \frac{f(z)}{z-a}$. However it is in the form of $\oint \frac{f(z)}{(z-a)^k}$ and there is an extention of Cauchy's Integral Formula which covers this case.

That's not in the form $\oint \frac{f(z)}{z-a}dz$ but you can write it as:

$\oint \frac{\frac{z}{2}}{z+1/2}dz$

Ah thanks a lot. It was very useful information.

EDIT: I'm searching for the extension of the CIF but without success. If you have a link, feel free to share.
Thanks again.
• Sep 15th 2009, 04:35 PM
shawsend
Quote:

Originally Posted by arbolis
Ah thanks a lot. It was very useful information.

EDIT: I'm searching for the extension of the CIF but without success. If you have a link, feel free to share.
Thanks again.

Arbolis, that's the next section after the introduction of Cauchy's Integral Formula. It's called "Differentiability of Cauchy-Type Integrals". It's in any text on Complex Analysis. My favorite is "Basic Complex Analysis" by Marsden and Hoffman.
• Sep 16th 2009, 09:09 AM
arbolis
Another attempt!
Quote:

Originally Posted by shawsend
Arbolis, that's the next section after the introduction of Cauchy's Integral Formula. It's called "Differentiability of Cauchy-Type Integrals". It's in any text on Complex Analysis. My favorite is "Basic Complex Analysis" by Marsden and Hoffman.

Ok thanks for sharing, I think I now know the formula.

For $\oint \frac{\frac{z}{2}}{z+1/2}dz$, I get that it equals $-\frac{ i \pi}{2}$.

Now comes $\int _C \frac{\cosh z}{z^4}dz$.
The formula states that $f^{n}(z)=\frac{n!}{2\pi i} \int _C \frac{f(\xi)}{(\xi -z)^{n+1}} d \xi$. I notice that in the exercise $z=0$ and $n=3$.
Thus I have $f^{(3)}(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi = \frac{3}{\pi i} \int _C \frac {\cosh z}{z^4} dz = \cosh (0) ^{(3)}= \sinh 0 =0$... I don't think it's possible.
What did I do wrong?
• Sep 16th 2009, 11:26 AM
shawsend
Quote:

Originally Posted by arbolis
Thus I have $f^{(3)}(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi = \frac{3}{\pi i} \int _C \frac {\cosh z}{z^4} dz = \cosh (0) ^{(3)}= \sinh 0 =0$... I don't think it's possible.
What did I do wrong?

That's ill-formed: $\cosh(0)^{(3)}$ is poor syntax. Rather, if $f(z)=\cosh(z)$, then $f^{(3)}(z)=\sinh(z)$ and therefore:

$\sinh(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi$ and that implies the integral is zero.
• Sep 16th 2009, 11:37 AM
arbolis
Quote:

Originally Posted by shawsend
That's ill-formed: $\cosh(0)^{(3)}$ is poor syntax. Rather, if $f(z)=\cosh(z)$, then $f^{(3)}(z)=\sinh(z)$ and therefore:

$\sinh(0)=\frac{3!}{2 \pi i} \int _C \frac{f(\xi)}{\xi ^4} d\xi$ and that implies the integral is zero.

Thanks for the reply and sorry for the syntax. So my result was right?
• Sep 16th 2009, 11:50 AM
shawsend
Quote:

Originally Posted by arbolis
Thanks for the reply and sorry for the syntax. So my result was right?

Yes. It's very easy to integrate it numerically in Mathematica:

Code:

NIntegrate[(Cosh[z]/z^4)*I*Exp[I*t] /.   z -> Exp[I*t], {t, 0, 2*Pi}]
And this gives zero.