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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 !?

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  • $\begingroup$ Try substituting $u=ax+x^b$. This will create a "bad" term, namely $a+b(x^{-1}(u))^{b-1}$, but it will make the exponential "good". Approximations of this bad term as a linear combination of $x^{c_n}$ terms result in a linear combination of Gamma functions for the integral itself. Presumably the difficulty then is how to get an asymptotic approximation of $x^{-1}(u)$. $\endgroup$ – Ian Aug 20 '18 at 17:26
  • $\begingroup$ (Also, I assume $b \geq 0$ here, if not then things are different...) $\endgroup$ – Ian Aug 20 '18 at 17:30
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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:

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

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