What is the correct Substitution for this integral

Question

$$\int_0^{1} ab^ac^{-a-1} dc$$ What would be the substitution so that this problem could be done by integration by parts would it be $~c = a~$? Also $~c ≥ b~$.

Answer 1

As pointed out, this is really straight forward (if say, $a\lt0$): $\int_0^1 ab^ac^{-a-1}\operatorname dc=-b^a[c^{-a}]_0^1=-b^a$.

If $a\gt0$, it appears to not converge.

Answer 2

As $a$ and $b$ are constant with respect to $c$, you can rewrite the integral as (provided $a \neq 0$)

$$ab^a \int_0^1 c^{-a-1}dc=ab^a \Big[\frac{c^{-a}}{-a}\Big]_0^1=-b^a\Big[{1}^{-a}-{0}^{-a}\Big]$$

which can then be analyzed by cases.

Case 1: Suppose that $a=0$. Then $$ab^a \int_0^1 c^{-a-1}dc=0$$

Case 2: Suppose that $a<0$. Then

$$-b^a \Big[1^{-a}-0^{-a}\Big]=-b^a\Big[1\Big]=-b^a$$

Case 3: Suppose that $a>0$. Then

$$-b^a \Big[1^{-a} -0^{-a}\Big]$$

is undefined as it is necessary to divide by zero in the calculation of $0^{-a}$.