Variation of residue theorem? Residue theorem can be stated informally as
$$\oint_C f(z)dz=2\pi i\sum a_{-1}$$
A contour integral sums up all the $-1$ coefficients inside.
Then, one would naturally ask: 

Is there something like
  $$\text{something of }f(z)=\sum a_{-2}$$ where $\text{something}$ is a sort of operator?

I encounter difficulties as the residue theorem makes use of the fact that $\frac1{z-c}$ has no antiderivative, but $\frac1{(z-c)^2}$ does have an antiderivative, so the same trick can’t be used.
Any idea?
EDIT:
@David C. Ullrich's answer does rule out some possibilities. However, I am not desperate: I still have the hope that there might exist some operators $\operatorname{P}$ that:


*

*only requires information of $f(z)$ along a contour $\gamma$

*$$\operatorname{P}_\gamma[f(z)]=\sum_{\text{all poles included}} a_{-2}$$ where $a_{-2}$ is the coefficient of $z^{-2}$ of the Laurent expansion of $f(z)$ around the pole.


($f(z)$ can be assumed to be meromorphic on $\mathbb C$.)
Therefore, I started a bounty to draw more attention.
 A: Various positive suggestions in comments depend on for example knowing where all the poles are. This is not a residue-theoremish thing; an analog of RT would give, for the disk, a complex measure $\mu$ on $|z|=1$ such that $\int_{|z|=1}f\,d\mu$ equals the sum of the $a_{-2}$ at all the poles. With no information except the values of $f$ on the boundary, is the point.
There is no such measure. Taking $f(z)=z^n$ for $n\in\Bbb Z$ tells you what the Fourier coefficients of $\mu$ would be, and it follows that the only $\mu$ that could possibly work is the one defined by $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz.$$
But that gives the wrong answer for other $f$, for example $f(z)=1/(z-1/2)$.
In more detail: First, it's clear that $$\frac1{2\pi i}\int_{|z|=1}z\cdot z^n\,dz=\begin{cases}1,&(n=-2),
\\0,&(n\ne-2).\end{cases}$$
Now assume that $\mu$ "works". Then we have $\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz$ if $f(z)=z^n$. Since trigonometric  polynomials are uniformly dense in the continuous functions on the circle it follows that $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz$$for every $f$ continuous on the unit circle.
But now let $f(z)=1/(z-1/2)$. It follows that $$\int_{|z|=1}f(z)\,d\mu=\frac1{2\pi i}\int_{|z|=1}zf(z)\,dz\ne0,$$by the actual Residue Theorem. But $\sum a_{-2}=0$, so $\mu$ doesn't work for this $f$.
