# Cauchys integral formula.

1. Sep 24, 2010

### Dissonance in E

If i have a closed pathintegral of the form:
(sin(z)+3cos(z) + 3e^z)/((z-(pi/2)^2)
How is cauchys integral formula applicable? If I split the integral into partial fractions wont i still get A/(z-pi/2) + B/(z-pi/2)^2, the B part of which wont be applicable to cauchys formula since the denominator is still squared.

In short, how do i apply cauchys formula to integrals with a denominator with a power higher than 1.

2. Sep 24, 2010

### HallsofIvy

Apparently, by "Cauchy's formula" you are referring to
$$f(a)= \frac{1}{2\pi i}\cint_\gamma \frac{f(z)}{z- a} dz$$

However, it can be easily generalized to
$$f^{(n)}(a)= \frac{n!}{2\pi i}\cint_\gamma \frac{f(z)}{(z- a)^{n+1}}dz$$
where the left sided is the nth derivative of f evaluated at z= a.

In your case, that integral is equal to the derivative of $f(z)= sin(z)+ 3cos(z)+ 3e^z$ evaluated at $z= \pi/2$, multiplied by $2\pi i$.

Last edited by a moderator: Sep 25, 2010
3. Sep 25, 2010

### snipez90

...multiplied by $2\pi i$.

4. Sep 25, 2010

### HallsofIvy

Right, thanks. I will now edit my post so I can pretend I didn't make that foolish mistake!