Confusing closed form for $\int_{-\infty}^{0}e^{(x^{2n+1})}dx$ for $n\in\Bbb N$

Question

Given that $n\in\Bbb N$, $$f(n)=\int_{-\infty}^{0}e^{(x^{2n+1})}dx$$
How can one get to a closed form of $f(n)$? According to the online integral calculator, $$\int e^{x^{2n+1}}dx=\frac{-\Gamma(\frac{1}{2n+1},-x^{2n+1})}{(2n+1)(-1)^{\frac{1}{2n+1}}}+C$$ Which I have no idea how to derive.

Answer 1

Let $x^{2n+1}=-u$ or $x=-u^{\frac{1}{2n+1}}$ then $$dx=-\frac{1}{2n+1}u^{\frac{1}{2n+1}-1}du$$ therefore $$f(n)=\int_{-\infty}^{0}e^{(x^{2n+1})}dx=\int_0^\infty e^{-u}\frac{1}{2n+1}u^{\frac{1}{2n+1}-1}du=\frac{1}{2n+1}\Gamma\left(\frac{1}{2n+1}\right)=\Gamma\left(\frac{2n+2}{2n+1}\right)$$