Calculate definite integral $\int^b_a e^{-x^a}dx $

Question

Is there any closed form expression for $\int^b_a e^{-x^k}dx$ ?. With $z=x^k$, we have $\int^{b^k}_{a^k} 1/k e^{-z} z^{1/k-1} dz$. Then I have no idea what to do next. It appears I have a gamma function. Any suggestion will be appreciated!

Answer 1

There is no closed form solution except for a special case, as has been articulated in a comment. In case you are interested here it is: \begin{align} \int_0^{\infty} e^{-x^\alpha}\,dx=\Gamma\left(1+\frac{1}{\alpha}\right) \end{align} EDIT: Yes, this integral can be expressed in terms of the incomplete Gamma functions $\Gamma(n,x)$ and $\gamma(n,x)$. \begin{align} I&=\int_a^b e^{-x^k}\,dx \qquad u=x^k\\ I&=\frac{1}{k}\int_{a^k}^{b^k} u^{\frac{1}{k}-1} e^{-u}\,du\\ I&=\Gamma\left(1+\frac{1}{k}\right)-\frac{1}{k}\,\gamma\left(\frac{1}{k},a^k\right)-\frac{1}{k}\,\Gamma\left(\frac{1}{k},b^k\right) \end{align} This can even be simplified: \begin{align} \int_a^b e^{-x^k}\,dx=\frac{1}{k}\,\Gamma\left(\frac{1}{k},a^k\right)-\frac{1}{k}\,\Gamma\left(\frac{1}{k},b^k\right) \end{align}