Tight upper bound for an improper integral

Question

I need to find a more tight upper bound for the improper integral $\int_0^{\infty}e^{-ax-x^b}dx$ rather than the following approximation based on Jensen's inequality $\mathbb{E}(f(X))<f(\mathbb{E}(X))$:

$\int_0^{\infty}e^{-ax-x^b}dx<1-e^{-a\Gamma(1+\frac{1}{b})}$ for $a$ and $b$ are two positive reals, $\Gamma(.)$ is the complete gamma function

Have you any ideas !?

Answer 1

This is not an answer, but an extended comment.

I think this observation may be of interest, even though it's not related to the OP's question.

Consider:

$$f(a,b)=\int_0^{\infty}e^{-ax-x^b}dx$$

Now, substitute:

$$y=(ax)^b$$

$$x=y^{1/b}/a$$

$$f(a,b)=\frac{1}{ab}\int_0^{\infty}e^{-y^{1/b}-y/a^b} y^{1/b-1}dy$$

Using integration by parts, it's easy to see that:

$$f(a,b)=\frac{1}{a}-\frac{1}{a^{b+1}}\int_0^{\infty}e^{-y^{1/b}-y/a^b} dy$$

So we obtain a functional equation:

$$a f(a,b)+\frac{1}{a^b} f \left( \frac{1}{a^b}, \frac{1}{b} \right)=1$$