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When I use Simplify[] or FullSimplify[], specifying domain reals, I get terms with $0^n$ as the coefficient. Isn't that always zero, and thus be simplified out? A example of input/output is below, with the expressions truncated/replaced since they are extremely long.

Input:
Assuming[Reals && c > 5 , FullSimplify[expression]]
Output: \int_0^1 i x... +0^{-1+c} \text{Hypergeometric2F1Regularized}\left[\frac{-1+c}{c},1-c i,2-\frac{1}{c},0^c\right]\ F'[x] \, dx

Note, I added the assumption that $c >5$ as overkill to avoid any possible issue with raising 0 to a negative number (or zero), but it didn't help. (Also, note, the $i$ above is a variable, not the imaginary number). Does anyone know why this happens?

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    @J.M. oops, I missed this comment somehow. I have forgotten the expression -- I think is was a series of order distributions of a function. I had a lot of them... If I find one, I'll repost.2011-09-18

1 Answers 1

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Usually, especially in the case of binomial identities, $ 0^n=\left\{\begin{array}{c}1\text{ if }n=0\\0\text{ if }n>0\end{array}\right. $ is used.

In your particular example, the exponent of $0$ is assumed to be greater than $0$, so I don't see the need for $0^{-1+c}$ or $0^c$. The answer is still correct, but unnecessarily complicated. If your assumption was $c\ge 1$, then $0^{-1+c}$ would be needed, but not $0^c$.

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    @JandR: $0^0$ is one of the classic "indeterminate forms" of beginning calculus. For example $\displaystyle\lim_{x\to 0^+}0^x=0$ and $\displaystyle\lim_{x\to 0}x^0=1$. From a set-theoretic point of view, $0^0$ represents the functions from the empty set to the empty set, which is just the empty function, so from that point of view $0^0=1$. See the [sci.math FAQ about $0^0$](http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/).2011-08-22