Can integrals approximate sums or series over more complicated fields than $\mathbb R$?

Question

For the real numbers, it is well known that many series can be approximated with known integrals due to them being Riemann sums.

$$\int_{a}^{a+\epsilon}f(x)dx \approx {\epsilon}f(x_0)\\ x_0\in[a,a+\epsilon]$$

And then a sum or series of such terms on the right over same $\epsilon$-width windows.

For other functions than real valued functions, can this idea be generalized?

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For example $$\sum_{k=0}^\infty\left(\frac i {2^{1/4}}\right)^k$$

can be readily split into real and imaginary parts and calculated with geometric series. We can also do a geometric series for complex numbers (why?), but would it for example make sense to do a spiral integral over the complex numbers that passes through the numbers?

Answer 1

If I understand correctly, you would like to know if

$$\sum_{k=0}^\infty\left(\frac i {2^{1/4}}\right)^k=\int_0^{\infty}\left(\frac i {2^{1/4}}\right)^x dx$$

Both of these expressions can be expressed in closed form and it is apparent that they are not the same, to wit,

$$\sum_{k=0}^\infty\left(\frac i {2^{1/4}}\right)^k=\frac{2+i^{3/4}}{2+\sqrt{2}}\\ \int_0^{\infty}\left(\frac i {2^{1/4}}\right)^x dx=\frac{4}{\log2-2i\pi} $$

The figure below shows the respective summand (square spiral) and integrand (round spiral). Even if the $dx$ is forced to unity, so that the integrand is indeed a square spiral, the results would still not be the same. Just think of the trapezoidal integration... the first term is cut in half. In fact, if $dx=1$ we find that $$\sum_{k=0}^\infty\left(\frac i {2^{1/4}}\right)^k=\int_0^{\infty}\left(\frac i {2^{1/4}}\right)^x dx+\frac{1}{2}$$

Answer 2

About "We can also do a geometric series for complex numbers (why?)"

For finite geometric sums we have:

$$\sum_{k=0}^n a^k =\frac{1 - a^{n+1}}{1-a}\qquad(a\ne 1).$$ And $$\sum_{k=0}^\infty a^k = \lim_{n\to\infty}\sum_{k=0}^n a^k$$ exists iff $\displaystyle\lim_{n\to\infty}a^{n+1}$ exists iff $|a|<1$.