Indefinite integral of $e^{x^n}$

Question

I came across this integral $\int e^{x^n} dx $ and was just wondering how you go about solving it. In this Quora question there approach is to let $y = -x^n$, then $$ dy = -nx^{n-1} dx = -n(-y)^{\frac{n-1}{n}} \ dx \ \Rightarrow \ dx = -\frac{1}{n}(-y)^{\frac{1-n}{n}} \ dy, $$ then sub into the integral, we have $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n}\int_{-\infty}^\infty y^{\frac{1}{n}-1} e^{-y} \ dy, $$ then they referenced the primary definition of incomplete gamma function, $$ \Gamma(a,y) = \int_y^\infty t^{a-1} e^{-t} \ dt, $$ and claim that we can write our integral in terms of $\Gamma$ function, and the end result is $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, y). $$ Can anyone fill in the missing step for me? Our integral does look very similar to the gamma function, but what happened to the lower bound of the integral? In our integral the lower bound is $-\infty$; but to equate it to the gamma function we need the lower bound to be $y$.

Anyway substitute back the definition of $y$ we have the final result $$ \int e^{x^n} dx = \frac{(-1)^{\frac{1}{n}-1}}{n} \Gamma(\frac{1}{n}, -x^n). $$ I̶'̶m̶ ̶g̶u̶e̶s̶s̶i̶n̶g̶ ̶t̶h̶e̶i̶r̶ ̶a̶n̶s̶w̶e̶r̶ ̶i̶s̶ ̶c̶o̶r̶r̶e̶c̶t̶ ̶b̶e̶c̶a̶u̶s̶e̶ ̶M̶a̶t̶h̶e̶m̶a̶t̶i̶c̶a̶ ̶a̶l̶s̶o̶ ̶g̶i̶v̶e̶s̶ ̶t̶h̶e̶ ̶s̶a̶m̶e̶ ̶r̶e̶s̶u̶l̶t̶.̶ ̶B̶u̶t̶ ̶a̶n̶o̶t̶h̶e̶r̶ ̶t̶h̶i̶n̶g̶ ̶I̶ ̶d̶o̶n̶'̶t̶ ̶u̶n̶d̶e̶r̶s̶t̶a̶n̶d̶ ̶i̶s̶ ̶w̶h̶y̶ ̶d̶o̶e̶s̶ ̶t̶h̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶l̶ ̶d̶e̶p̶e̶n̶d̶s̶ ̶o̶n̶ ̶$̶x̶$̶?̶ ̶Y̶o̶u̶'̶d̶ ̶t̶h̶i̶n̶k̶ ̶$̶x̶$̶ ̶s̶h̶o̶u̶l̶d̶ ̶b̶e̶ ̶i̶n̶t̶e̶g̶r̶a̶t̶e̶d̶ ̶o̶v̶e̶r̶.̶ Any help is appreciated, thanks.

Answer 1

$\newcommand{\d}{\mathrm{d}}$ $\newcommand{\e}{\mathrm{e}}$

We can evade confusion by working exclusively with parametrized definite integrals and using different symbols for bound and free variables. Consider

$$\begin{align}I(x)&=\int_0^x\e^{u^{\nu}}\d u\text{,}& \nu &>0\text{.} \end{align}$$ Set $$\begin{align} t&=-u^{\nu}& \d t &= -\nu u^{\nu-1}\d u\text{.} \end{align}$$ Then $$\begin{split}I(x)&=\frac{(-1)^{(1/\nu) - 1}}{\nu}\left(\int_0^{-x^{\nu}}t^{(1/\nu) - 1}\e^{-t}\d t\right)\\ &=\frac{(-1)^{(1/\nu) - 1}}{\nu}\left( \Gamma(\tfrac{1}{\nu},-x^{\nu})-\Gamma(\tfrac{1}{\nu})\right) \end{split}$$

Answer 2

Also notice that: $$e^y=\sum_{i=0}^\infty\frac{y^i}{i!}$$ And so: $$e^{x^n}=\sum_{i=0}^\infty\frac{x^{ni}}{i!}$$ And so: $$\int e^{x^n}dx=\int\sum_{i=0}^\infty\frac{x^{ni}}{i!}dx=\sum_{i=0}^\infty\int\frac{x^{ni}}{i!}dx=\sum_{i=0}^\infty\frac{x^{ni+1}}{i!(ni+1)}$$