Cauchy residue theorem tells us that for a function $$f(z) = \sum_{k \in \mathbb{Z}} a(k) z^k,$$ the coefficient $a(k)$ can be extracted by an integral formula $$a(k) = \frac{1}{2\pi i}\oint f(z) z^{-k-1},$$ with a contour around zero. Now, there is nothing to prevent us from thinking of $a\colon \mathbb{Z} \to \mathbb{C}$ as a function over the complex domain, defined by the above integral. In this way, we have naturally "continued" the function $a(k)$ over the integers to one over the complex numbers. Is there any meaning to this beyond just something amusing?

For example, consider $f(z)=\exp(z)$. In this case, $a(k)=1/k!$ which one would think would be continued to $1/\Gamma(1+k)$. But at least if you attempt to plot what happens above, this is not the case. We get a function with Bessel-like oscillations (blue) which nevertheless coincides with $1/\Gamma(1+k)$ (orange) on integral points!

One can experiment with other choices of $f(z)$. When $f(z)=(z+1)^n$, it seems like the continuation actually coincides with what one would naturally think (the Beta function). Which is actually quite strange. For example for $n=5$ and $f(z) = 1 + 5 z + 10 z^2 + 10 z^3 + 5 z^4 + z^5$, how does this computation "learn" that the sequence $1,5,10,10,5,1$ actually corresponds to the binomial coefficients and thus should be continued to the Beta function? If we perturb one of the coefficients, say change $5 z^4$ to $2 z^4$, the result would look like a "perturbed" Beta function (blue below vs the Beta function in orange).

Any explanation (or references in the literature) for what is going on here?

notactually uniquely defined. Because of the branch cuts in $z^{-k-1}$ for non-integer $k$, the value of the integral depends precisely on the shape of the contour (well, more precisely on the location of your chosen branch cut and where it intersects the contour). I noticed that you didn't specify the contour in your post. So, as you change the contour, your blue curve will change, except for its intersections with the orange curve. $\endgroup$ – Igor Khavkine Feb 22 at 6:55