# Are most integral formulas invariant to complex-valued parameters?

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.

1. Is it always the case that common integral formulas extend to the complex parameters (i.e. $$a,p \in \mathbb{C}$$)?
2. 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.

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.
• Thank you! Curiously enough, you mention that if $f(x) = g(x)$ for $x \in I \subset \mathbb{R}$ then that also extends to $f(z) = g(z)$ where $z$ is complex. Yet, since we had $x \in I$, how would $z$ be bounded? Mar 26, 2020 at 21:39
• I don't understand your question? We don't need that $z$ is bounded. Mar 27, 2020 at 0:13
• Hm, perhaps I mean something like: assume $f(x) = g(x)$ for $x\in[-1,1]$. Does that mean $f(z) = g(z)$ for all $z\in \mathbb{C}$? (Of course assuming $f$ and $g$ are analytic) Mar 27, 2020 at 0:44
• Yes it does. Even $f(x) = g(x)$ for $x \in \{1/n : n \in \mathbb Z_+\}$ implies $f(z) = g(z)$ for every $z \in \mathbb C$. Mar 27, 2020 at 4:16
• So if you want to prove $\sin^2(z) + \cos^2(z) = 1$ for every complex number $z$, you only need to show it for a sequence $z_n$ that has an accumulation point. Mar 27, 2020 at 4:17