Upper bound for $\int_{c_0}^{+\infty} e^{-(x + c_1 x^{-c_2})} dx$

Question

I'm trying to place an upper bound on $$\int_{c_0}^{+\infty} e^{-\left(x + \frac{c_1}{x^{c_2}}\right)} dx, \ \ \ c_i > 0$$ in terms of $c_0, c_1, c_2$ which is at least asymptotically correct. A factor isn't significant. Laplace's method is applicable with $M = c_1, h(x) = e^{-x}, g(x) = \frac{c_1}{x^{c_2}}$, but I was hoping to avoid the horrendous calculation for the relative error.

Answer 1

Write the integral in the form $$ \exp \left( { - \left( {c_0 + \frac{{c_1 }}{{c_0^{c_2 } }}} \right)} \right)\int_{c_0 }^{ + \infty } {\exp \left[ { - \left( {\left( {x + \frac{{c_1 }}{{x^{c_2 } }}} \right) - \left( {c_0 + \frac{{c_1 }}{{c_0^{c_2 } }}} \right)} \right)} \right]dx} . $$ By the mean value theorem, $$ \left( x + \frac{{c_1 }}{{x^{c_2 } }} \right) - \left( {c_0 + \frac{{c_1 }}{{c_0^{c_2 } }}} \right) > \left( {1 - \frac{{c_1 c_2 }}{{c_0^{c_2 + 1} }}} \right)(x - c_0 ) $$ for all $x>c_0$. Thus, an upper bound for your integral is \begin{align*} &\exp \left( { - \left( {c_0 + \frac{{c_1 }}{{c_0^{c_2 } }}} \right)} \right)\int_{c_0 }^{ + \infty } {\exp \left[ { - \left( {1 - \frac{{c_1 c_2 }}{{c_0^{c_2 + 1} }}} \right)(x - c_0 )} \right]dx} \\ & = \exp \left( { - \left( {c_0 + \frac{{c_1 }}{{c_0^{c_2 } }}} \right)} \right)\frac{1}{{1 - \frac{{c_1 c_2 }}{{c_0^{c_2 + 1} }}}}, \end{align*} provided that $c_1 c_2 < c_0^{c_2 + 1}$. For large positive $c_0$, this is asymptotically sharp (it agrees with the leading order term coming from Laplace's method).

Answer 2

$$\int_{c_0}^\infty\exp \left(-(x+c_1x^{-c_2})\right)\,dx=\int_{c_0}^\infty e^{-x}e^{-\frac{c_1}{x^{c_2}}}\,dx$$ now since: $$1\le e^{-\frac{c_1}{x^{c_2}}}\le e^{-\frac{c_1}{c_0^{c_2}}}$$ we can use the integral of $e^{-x}$ multiplied by these bounds