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I am working on finding an upper bound of the function:

$$ \int_{0}^{\infty} \exp{\left[-(x + \frac{b x^a}{t})\right]} \, dx. $$ where $a, b > 0$ and $1> t > 0$.

I am attempting to show whether this can be upper bounded by a linear function of $t$, such as $\beta \cdot t$. My numerical experiments suggest that this function is increasing in $t$, converging to 1. As $t \rightarrow 0$, the value converges to $0$.

Is there a known solution to this integral, or could this be approximated in a simple expression? Can $\beta$ be expressed as a function of $a$ and $b$? Any insights or references would be greatly appreciated.

Thank you!

Dstarred
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mike
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1 Answers1

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I suggest an upper bound by using the AM-GM inequality and this answer:

$$\begin{align} \color{red}{\int_{0}^{+\infty}\exp\left(-\left(x + \frac{bx^a}{t} \right) \right)dx } &\le \int_{0}^{+\infty}\exp\left(- 2\sqrt{x \cdot \frac{bx^a}{t} } \right)dx\\&=\int_{0}^{+\infty}\exp\left(- 2\sqrt{ \frac{b}{t} } \cdot x^{\frac{a+1}{2}} \right)dx\\ &=\int_{0}^{+\infty} \left(2\sqrt{ \frac{b}{t}}\right)^{-\frac{2}{a+1}}\cdot\exp\left(- x^{\frac{a+1}{2}} \right)dx\\ &=\color{red}{\underbrace{\left(2\sqrt{b}\right)^{-\frac{2}{a+1}}\cdot \Gamma\left(1+ \frac{2}{a+1} \right)}_{=: \beta} \cdot t^{\frac{1}{a+1}}} \end{align}$$

And if you want a bound that is a linear function of $t$, it suffices to remind that $ t^{\frac{1}{a+1}}<t$ for $a>0$.

NN2
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