Antiderivative of $e^{x^n}$

Question

How do you prove the following identity ? $$\int \exp(x^n)dx = - \frac{\Gamma(1/n, -x^n)}{n \cdot i^{2/n}} + C$$

where $\Gamma$ is the Incomplete gamma function and $i$ the imaginary unit (initially, $i^{2/n}$ was written $(-1)^{1/n}$).

Answer 1

Define $$ \Gamma(a,z) = \int_z^\infty e^{-t} t^{a-1}\;dt,\qquad a > 0. $$ So $$ \Gamma\left(\frac{1}{n},-x^n\right) = \int_{-x^n}^\infty e^{-t} t^{1/n-1}\;dt $$ Differentiate, using the fundamental theorem of calculus and the chain rule: \begin{align} \frac{d}{dx} \Gamma\left(\frac{1}{n},-x^n\right) &= - e^{-(-x^n)}\;(-x^n)^{1/n-1}\;(-n x^{n-1}) = e^{x^n} (-1)^{1/n-1} (x^n)^{1/n-1} (-n) x^{n-1} \\ &= n e^{x^n} (-1)^{1/n} x^{1-n} x^{n-1} = n e^{x^n} i^{2/n} \end{align}

Therefore, $$ \int e^{x^2}\;dx = \frac{1}{n i^{2/n}} \Gamma\left(\frac{1}{n},-x^n\right) +C$$

Answer 2

\begin{align*} \frac{d}{dx}\left(-\frac{\Gamma(1/n,-x^n)}{n\cdot (-1)^{1/n}}+C\right)&=-\frac{1}{n(-1)^{1/n}}\frac{d}{dx}\int_{-x^n}^\infty t^{1/n-1}e^{-t}dt\\ &= -\frac{1}{n(-1)^{1/n}}\left((-x^n)^{1/n-1}\cdot e^{x^n}\right)\frac{d}{dx}(x^n)\\ &= -\frac{1}{n(-1)^{1/n}} \left((-1)^{1/n-1}\cdot x^{1-n}\cdot e^{x^n}\right)(nx^{n-1})\\ &= \frac{1}{n(-1)^{1/n}} \left((-1)^{1/n}\cdot x^{-(n-1)}\cdot e^{x^n}\right)(nx^{n-1})\\ &=\exp(x^n) \end{align*}