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Let $x$ be a real number and $z \neq 0$ a complex number, the following can be proved using the Taylor series of a power.

$$\int x^z = \frac{1}zx^{z-1}$$

Do you know any more elementary solution?

Nisba
  • 777

2 Answers2

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Let $x^z=\exp(z\ln(x))$. It is then not too hard to see that

$$\frac d{dx}\exp(z\ln(x))=\frac zx\exp(z\ln(x))=zx^{z-1}$$

which is the power rule. In reverse, we get the antiderivative:

$$\int x^z\ dx=\frac{x^{z+1}}{z+1}+c$$

For the case when $z=-1$, the antiderivative may be handled using this answer of mine, which follows through power rule.

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This is no more complicated than just defining the complex derivative as $$\lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$$
Where both $h$ and $z$ are in $\mathbb{C}$. Now we just note that $$\frac{d}{dz} \frac{x^{z+1}}{z+1} = x^z$$
and we define the complex antiderivative as the operator that inverts complex differentiation.