Help with a cauchy integral, where the denominator of f(z) is zero I have an integral to solve and my tutor suggested doing it as a cauchy integral. The integral is 
$$\int_C \frac{\sin(z)}{(z+1)^7} dz $$
With C being the circle of radius 5, centered at z=2 in the complex plane.
I tried turning it into a cauchy integral with $f(x)=\frac{\sin(z)}{(z+1)^6}$ With the only root at $z=-1$ so using the cauchy integral formula, I had
$$2\pi i \frac{\sin(z)}{(z+1)^6}=\int_C \frac{\frac {\sin(z)}{(z+1)^6}}{z+1} dz$$
Substituting the $ z=-1$ I then ended up with the left hand side denominator being zero, so it hasn't gotten me anywhere...
I've also tried using Liouville, but the root $ z=-1$ is inside the circle so I don't know it that works..
Thanks so much!!
 A: It seems you are trying to apply a form of Cauchy's Theorem which requires that the function you're denoting by $f(z)$ to be holomorphic. But the function
$$\frac{\sin z}{(z + 1)^6}$$
is not holomorphic.
It is more natural to think of the entire integrand
$$f(z) = \frac{\sin z}{(z + 1)^7}$$
and to consider Cauchy's Residue Theorem (which has several names. Many simply call it Cauchy's Theorem). Then you're looking for the residue, which is attainable from the (not entirely pleasant) Taylor expansion of $\sin z$ at $z = -1$.
But we can extend Cauchy's Integral Formula to cases with higher powers in the denominator.

Theorem:
  If $f(z)$ is holomorphic in a simply connected region containing $z_0$, then
  $$f'(z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^2}dz. \tag{1}$$
  More generally,
  $$ f^{(n)} (z_0) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - z_0)^{n + 1}}dz.$$

Let's prove $(1)$. We can write the derivative as
$$f'(z_0) = \frac{f(z_0 + h) - f(z_0)}{h}$$
as $h \to 0$. We apply the ordinary Cauchy's Integral formula to this expression. So
$$ \begin{align}
f'(z_0) &= \frac{f(z_0 + h) - f(z_0)}{h} \\
&= \frac{1}{2\pi i h} \left( \oint \frac{f(z)}{z - z_0 - h} + \frac{f(z)}{z - z_0} dz \right) \\
&= \frac{1}{2\pi i h} \oint \frac{f(z)h}{(z - z_0 - h)(z - z_0)}dz \\
&= \frac{1}{2 \pi i} \oint \frac{f(z)}{(z - z_0)^2}dz,
\end{align}$$
where the last equality came after taking $h \to 0$. Now one can proceed in many ways to prove the general case, perhaps inductively by iterating this process. $\diamondsuit$
With this, you can recognize your question as the sixth derivative of the sine function at $-1$ (with a factorial factor). And if you know the residue theorem, you'll immediately recognize that this is precisely the residue as given by Taylor's theorem.
A: What you want here is the more general Cauchy integral formula, obtained by differentiating the one we all know and love: suppose $C$ is a simple positively-oriented curve enclosing $a$. Then
$$ f^{(n)}(a) = \frac{n!}{2\pi i}\int_{C} \frac{f(z)}{(z-a)^{n+1}} \, dz. $$
You can apply this to your question with $f(z)=\sin{z}$, $n=6$, $a=-1$.
