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

1. ## [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?

2. 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.

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:

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

3. 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.

4. 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.

5. ## Another attempt!

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?

6. 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.

7. 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?

8. 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.