Meaning of imaginary part of $\int_0^6 e^{x^3} dx$

Question

My question is the title itself. How can it be possible that integral of real numbers can have an imaginary part?

Answer 1

If $x \in [0,6]$, then $x$ is real, so $x^3$ is real, so $e^{x^3}$ is (positive) real. So: $$\int_0^6 e^{x^3}dx \in \mathbb{R}_{>0}.$$

So the way I see it, there's basically two possibilities.

1. Wolfram alpha is just wrong. This is entirely possible, since its not based on a rigorous formalization of mathematics, more like a haphazard mess of ideas that seem to work most of the time.

2. Alternatively, perhaps the problem is that the symbol $\approx$ is fundamentally meaningless. What's meaningful is $a \approx b : r$, which can be defined to mean $d(a,b) \leq r$ in any metric space. Btw, this is basically how you get a uniform structure from a metric. Anyway, the point is that mathematical software that fails to compute a rigorous upper bound on its error terms is fundamentally pretty useless.

In motto form: "The Error term is God."