I have quite an interesting question to ask.
How does one evaluate the following indefinite integral?
$$I = \int {e^{x^n}}dx$$
Note: $n$ is a natural number ($n = 1, 2, 3, 4, 5, ...$).
It would be best if the incomplete gamma function is used:
$$\Gamma (s, x) = \int^{\infty}_x {t^{s - 1}e^{-t}}dt$$
Here is what I have done:
$$I = \int {e^{x^n}}dx$$
$$\text {Let } y = - x^n\text {.}$$
$$\frac {dy}{dx} = - n x^{n - 1}$$
$$dx = \cfrac {dy}{-nx^{n - 1}}$$
$$\text {Since } y = - x^n\text {,}$$
$$x = (-y)^{\frac {1}{n}}$$
$$\text {Therefore,}$$
$$dx = \frac {dy}{-n(-y)^{{\frac {1}{n}}{(n - 1)}}}$$
$$\text {Thus,}$$
$$I = \int {e^{-y}\cdot {\frac {dy}{-n(-y)^{{\frac {1}{n}}{(n - 1)}}}}}$$
$$I = -\frac {1}{n}\int {e^{-y}\cdot {(-y)^{{-\frac {1}{n}}{(n - 1)}}} \; dy}$$
$$I = -\frac {1}{n}\int {(-y)^{\frac {1}{n} - 1}}\cdot {e^{-y} \; dy}$$
How does one proceed to incorporate the incomplete gamma function?
Looking forward to a response! Also, I'm still new to MSE, so I would really appreciate any and all feedback. Thank you!