Indefinite integrals of $x^x$ sin(x)/x and relation to elementary integrals

Question

I was just doing some math when I got these questions, and I haven't been able to find a clear answer online. My first question is: are elementary integrals the same as indefinite integrals? I understand an elementary integral is an integral expressed as the finite sum of exponents, algebraic operations, etc, but how does this differ from an indefinite integral? Secondly, are the integrals of $x ^ x$ and sin(x)/x elementary integrals, and can their indefinite integrals be found? Finally, by taking ln(y) = xln(x), why would it not be possible to integrate $x^x$? Thanks a lot for the help!

Answer 1

Are elementary integrals the same as indefinite integrals?

How does this differ from an indefinite integral?

No, they are not the same in general. All elementary integrals are indefinite integrals but the converse is not always true.

Integrals may be indefinite integrals (i.e. antiderivatives), where we don't specify the bounds of the integral, so the answer is a function. Or, integrals could be definite integrals, where we do specify the bounds, and our answer is a number. I'll use the term antiderivative from here, to avoid confusion.
Conversely, elementary integrals are a subtype of antiderivative. Elementary functions are the common, neat functions we like to deal with (e.g. $\sin$, $\log$, $\sqrt{\cdot}$). When you integrate an elementary function to get its antiderivative, one of two things can happen. We can either get an antiderivative that is also an elementary function (this is an elementary integral) or we can get one that isn't elementary (this is a non-elementary integral).

For example, in (a), we have specified the bounds of the integral, so it is a definite integral. In (b) and (c), we haven't specified the upper bound, so they are both indefinite integrals (antiderivatives). However, when (b) is evaluated, we get an elementary function, but this isn't true for (c). So (b) is an elementary integral, while (c) is a non-elementary integral.

$$\begin{aligned} \int_0^\pi \sin(x) \,\mathrm{d}x &= 2 &\text{(a) : definite} \\ \int_0^z \sin(x) \,\mathrm{d}x &= -\cos(z) &\text{(b) : indefinite and elementary} \\ \int_0^z \frac{\sin(x)}{x} \,\mathrm{d}x &= \,? &\text{(c) : indefinite and non-elementary} \\ \end{aligned}$$

Lastly, for non-elementary integrals, since there's no neat way to express them in terms of elementary functions, we just give them new names (if necessary). So in fact, (c) is referred to as $\mathrm{Si}(z).$

Are the integrals of $x^x$ and $\frac{\sin(x)}x$ elementary integrals, and can their indefinite integrals be found?

No, their antiderivatives are both non-elementary integrals. Whether they can be "found" is a tricky one, because it depends what you mean by "found". We know they exist, and we can evaluate them but we can't express them in terms of elementary functions. But for all intents and purposes, we have "found" them as much as we need to.

By taking $\ln(y) = x\ln(x)$, why would it not be possible to integrate x^x?

You're welcome to try it, but it's been proven to be impossible. You might find that you keep ending up in a loop between certain integrals.

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Addendum

@Zombaka asked me in a comment whether there are functions with no antiderivative (be it elementary or not). This the case; not all functions have an antiderivative that exists. A function only has an antiderivative (in the "normal" sense of integration…) if it is only discontinuous at a very small (measure zero) set of points. So functions like $\mathrm{sgn}(x)$ are fine to integrate, since they're only discontinuous at tiny sets of points. But functions like the Dirichlet function, which are discontinous at loads of places can't be integrated (in the "normal" sense), which is to say they have no antiderivative.
But why am I making this case about the "normal" (Riemann) sense of integration? Well it turns out that you can extend the typical way of integrating functions. And functions that might have no Riemann integral (i.e. no antiderivative via Riemann integration) might still have a Lebesgue integral. In fact this is the case for the Dirichlet function.

$$\begin{aligned} \int_0^z \mathrm{sgn}(x) \,\mathrm{d}x &\quad\quad\text{(d) : discontinuous but integrable} \\ \int_0^z I_\mathbb{Q}(x) \,\mathrm{d}x &\quad\quad\text{(e) : not Riemann integrable (although Lebesgue integrable)} \\ \end{aligned}$$

Where $I_\mathbb{Q}(x)$ is the Dirichlet function. All Riemann integrable functions are Lebesgue integrable. Now the obvious next question is whether there are functions who have no antiderivative by Lebesgue integration - I'm not actually sure about this. But there are definitely some definite integrals that can't be found by Lebesgue integration, such as in (f).

$$\int_0^{+\infty} \frac1x \,\mathrm{d}x \quad\quad\text{(f) : neither Riemann integrable nor Lebesgue integrable} \\$$

This text is quite good, if you're interested in reading some more about integration.