Difficult integral of exponential function

Question

Is there a closed-form solution for integrals of the form $$ V(a,b,c) := \int_a^\infty \exp( - b \, x^c) \, \mathrm{d}x \quad a,b,c > 0 . $$

Only for the special case $c=1$ I can figure out $V(a,b,1) = \exp(-b a) / b$. I would guess that it holds something like $V(a,b,c) \approx \exp(-b a^c)$, but is there an exact formula?

Thanks very much!

Answer 1

For $a,b,c>0$, the following changes of variables do the trick: \begin{align*} \int_{a}^{\infty}e^{-bx^{c}}dx & =a\int_{1}^{\infty}e^{-b\left(ax\right)^{c}}dx & \text{(} x & \rightarrow ax \text{)}\\ & =a\int_{1}^{\infty}e^{-a^{c}bx^{c}}dx\\ & =\frac{a}{c}\int_{1}^{\infty}e^{-a^{c}bx}x^{\frac{c}{c-1}}dx & \text{(} x & \rightarrow x^{1/c} \text{)}\\ & =\frac{a}{c}\int_{1}^{\infty}e^{-a^{c}bx}/x^{\frac{c-1}{c}}dx\\ & =\frac{a}{c}E_n(a^{c}b) & \text{ where } n & ={\frac{c-1}{c}}. \end{align*}

The exponential integral $E_n$ is defined here.

Answer 2

No there isn't, except for some special cases. Consider $\int \exp(-x^2)dx$ (as part of the normal distribution).

Answer 3

According to Section 3.381 in Gradshteyn and Ryzhik: Table of Integrals, Series, and Products the answer is $$ V(a,b,c) = \frac{1}{c b^{1/c}} \Gamma(1/c, b a^c), $$ where $\Gamma$ denotes the upper incomplete gamma function.