I know that the same question for divergent sums is false, but cannot find much on divergent integrals.
For example, consider the following divergent integral for positive reals $a,b$:
$\displaystyle \int_0^\infty \frac{e^{-ax} - e^{-bx}}{x^2} \, \mathrm{d}x$ "=" $(a-b)(\gamma -1)$
The regularised value is obtained by replacing $2$ with $q$ and evaluating the corresponding generalised integral, then using uniqueness of analytic continuation and setting $q=2$, and taking $\Gamma(-1)$ to be the regularised value of $\gamma-1$.
However, there could be some other more convoluted means of generalising the integral which does not yield the same value, and I am wondering if this is the case.
All divergent sums have unique natural regularization. You can call anything "regularization" but there is a set of mutually-compatible methods that produce the same result.
The same holds for the integrals. Yet, the method based on analytic continuation does not belong here. Analytic continuation can produce very different results depending on what expression you take to analytically continue.