\(\displaystyle \int_{x=0}^{x=1}{{ax}^a dx}\)

\(\displaystyle =a\int_{x=0}^{x=1}{x^a dx}\)

\(\displaystyle =a \left[\frac{x^{a+1}}{a+1}\right]^{x=1}_{x=0} \quad \text{with } a>-1\)

\(\displaystyle =\frac{a}{a+1}\left(\left(1^{a+1}-0^{a+1}\right)\right)\)

\(\displaystyle =\frac{a}{a+1}, \quad a>-1\)

1. You seem to have been substituting \(\displaystyle a\) in for \(\displaystyle x\). That's why I explicitly put in \(\displaystyle x=1\) and \(\displaystyle x=0\).

2. Re. your statement: "and anything divided by zero is zero, so the value of the lower bound is neutral". The first part is not true. You cannot divide by 0. (There is no answer, 0 or otherwise). The second half of the statement I don't understand.

3. As you see from my answer, this integral only exists for \(\displaystyle a>-1\). To get an idea of what's happening you could graph \(\displaystyle x^a\) for some negative values of \(\displaystyle a\). (Even though, this doesn't tell the whole story).