Recently I had an argument with Luboš Motl on Quora, where he had argued that $0^0$ should be left undefined in computer algebra systems, because $x^y$ has no limit at $(0,0)$ and $0^x=0$ at all $x>0$.
So, I decided to represent the function $0^x$ in integral form. Once heaving a universal representation of the function one would be able to see what its value at $x=0$ is (as well as around that point).
For that purpose, I armed myself with divergent integrals, Laplace transforms, etc, etc.
So, assuming that
$$0^{-n}=\frac{W^{n+1}-w^{n+1}}{(n+1)!}$$
where
$$w^n=B_n(0)+n\int_0^\infty B_{n-1}(x)dx$$
and
$$W^n=B_n(1)+n\int_0^\infty B_{n-1}(x+1)dx$$
I obtained the following formula:
$$0^x=\frac{B_{1-x}(1)-B_{1-x}+\int_0^{\infty } (x-1) x t^{-x-1} \, dt}{\Gamma (2-x)}=$$ $$\frac{(x-1) \zeta (x,1)-(x-1) \zeta (x,0)+\int_0^{\infty } (x-1) x t^{-x-1} \, dt}{\Gamma (2-x)}$$
The formula with Bernoulli polynomials works well in Mathematica at positive and negative integer $x$, as well as at zero, returning expected divergent integrals or finite values, but the formula with Hurwitz Zeta fails at positive integers, returning expression with the symbol ComplexInfinity.
The both formulas fail at non-integer $x$. So, I wonder, whether it is possible to re-write these formulas in such a way that I would always obtain a convergent or divergent integral?
I tried to apply the integral representations of Hurwitz Zeta function from Wikipedia, but those return wrong results outside of their range of validity (unlike the built-in functions).