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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.

Thank you for your insight.

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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.)

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  • $\begingroup$ 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? $\endgroup$
    – ex.nihil
    Mar 26, 2020 at 21:39
  • $\begingroup$ I don't understand your question? We don't need that $z$ is bounded. $\endgroup$ Mar 27, 2020 at 0:13
  • $\begingroup$ 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) $\endgroup$
    – ex.nihil
    Mar 27, 2020 at 0:44
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    $\begingroup$ Yes it does. Even $f(x) = g(x)$ for $x \in \{1/n : n \in \mathbb Z_+\}$ implies $f(z) = g(z)$ for every $z \in \mathbb C$. $\endgroup$ Mar 27, 2020 at 4:16
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    $\begingroup$ 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. $\endgroup$ Mar 27, 2020 at 4:17

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