Consider integral formulas derived for common functions, such as
$$
\int(x+a)^{p} \mathrm{d} x=\frac{(x+a)^{p+1}}{p+1}, p \neq-1
$$
Or even something less trivial, such as,
$$
\int x^{p} e^{a x} \mathrm{d} x=\frac{x^{p} e^{a x}}{a}-\frac{p}{a} \int x^{p-1} e^{a x} \mathrm{d} x
$$
I have been wondering recently, why do these formulas still work even when $a,p \in \mathbb{C}$?
Initially I thought these formulas were derived for $a,p \in \mathbb{R}$ but they still worked fine after I tested with complex numbers.
- Is it always the case that common integral formulas extend to the complex parameters (i.e. $a,p \in \mathbb{C}$)?
- What about complex variables (i.e. $x \in \mathbb{C}$)?
Of course, when a formula has a restriction that $p$ is an integer, or when $a>0$ is restricted, that is self-explanatory. I am referring to when there are no restrictions that force the parameter to be non-complex.
Thank you for your insight.
Suppose you have two analytic functions $f(z)$ and $g(z)$ defined for $z \in \mathbb C$. Then if $f(x) = g(x)$ for all $x \in \mathbb R$, then necessarily $f(z) = g(z)$ for all $z \in \mathbb C$.
This is true in a lot more generality than I stated here, for example, knowing $f(x) = g(x)$ for $x$ in any small sub interval of the real numbers would be enough.
Look up the principle of isolated zeros. (Also if you haven't taken a class on complex variables, you will also need to look up the definition of analytic function.)