How do I determine the values of $r$ so $e^{t^r}$ has an antiderivative?

Question

It's fairly easy to show for individual values of $r$, just plug in the value and try it, but is there any way to find, in general, which values of $r$ work so that $\int{e^{t^r}dt}$ can be expressed in a way without integrals (using elementary functions)?

Answer 1

If you're willing to use the incomplete Gamma function, it can be done for all $r$.

Note btw that the substitution $\exp(t^r) = u$ gives you $$ \dfrac{1}{r} \int (\ln u)^{1/r - 1}\ du$$

and you can do this in elementary functions if $1/r$ is a positive integer.

Answer 2

Look up "integration in finite terms Liouville". This will give you a differential equation from $\int f(t)\, dt$ which has to have a certain type of solution for the integral to be integrable in finite terms.

You can then find out which values of $r$ enable the integration to be done.

Also, I think Chebychev worked on (and, maybe, solved) this particular problem.

Answer 3

As Robert Israel showed, if $1/r$ is a positive integer the antiderivative can be coomputed on the basis of elementary functions.

Suppose we write $1/r=n+1$, I give you below the expressions obtained for some values of $n$ (from $n=1$ to $n=7$).
$$2 e^{\sqrt{t}} \left(\sqrt{t}-1\right)$$ $$e^{\sqrt[3]{t}} \left(3 t^{2/3}-6 \sqrt[3]{t}+6\right)$$ $$e^{\sqrt[4]{t}} \left(4 t^{3/4}-12 \sqrt{t}+24 \sqrt[4]{t}-24\right)$$ $$e^{\sqrt[5]{t}} \left(5 t^{4/5}-20 t^{3/5}+60 t^{2/5}-120 \sqrt[5]{t}+120\right)$$ $$e^{\sqrt[6]{t}} \left(6 t^{5/6}-30 t^{2/3}+120 \sqrt{t}-360 \sqrt[3]{t}+720 \sqrt[6]{t}-720\right)$$ $$e^{\sqrt[7]{t}} \left(7 t^{6/7}-42 t^{5/7}+210 t^{4/7}-840 t^{3/7}+2520 t^{2/7}-5040 \sqrt[7]{t}+5040\right)$$

$$e^{\sqrt[8]{t}} \left(8 t^{7/8}-56 t^{3/4}+336 t^{5/8}+6720 t^{3/8}-1680 \sqrt{t}-20160 \sqrt[4]{t}+40320 \sqrt[8]{t}-40320\right)$$

For the most general case, the antiderivative is given by
$$-\frac{t E_{\frac{r-1}{r}}\left(-t^r\right)}{r}$$ where $E_a(x)$ stands for the exponential integral function.