Order of infinitesimal $\int_0^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr$

Question

How could I find the right order of infinitesimal of

$$\int_0^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr,\qquad c_1,c_2>0,$$

as $|x|$ tends to $+\infty$? Should it be trivial? For an upper bound, I tried something like

\begin{align} \int_0^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr&=\int_0^{|x|}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr+\int_{|x|}^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr\\ &\leq \frac{e^{-c_2|x|}}{c_1}(1-e^{-c_1|x|})+e^{-c_1|x|}\int_{|x|}^{|x|^2}e^{-c_2\frac{|x|^2}{r}}dr\\ &=\frac{e^{-c_2|x|}}{c_1}(1-e^{-c_1|x|})+|x|^2e^{-c_1|x|}\int_1^{|x|}\frac{e^{-c_2t}}{t^2}dt \end{align}

and the order of the upper bound should be $|x|^2e^{-c_1|x|}$ or $e^{-c_2|x|}$, depending on whose bigger between $c_1$ and $c_2$, and for a lower bound \begin{align} \int_0^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr&=\int_0^{|x|}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr+\int_{|x|}^{|x|^2}e^{-c_1 r-c_2\frac{|x|^2}{r}}dr\\ &\geq e^{-c_1|x|}\int_0^{|x|}e^{-c_2\frac{|x|^2}{r}}dr+e^{-c_2|x|}\int_{|x|}^{|x|^2}e^{-c_1r}dr\\ &=e^{-c_1|x|}\int_0^{|x|}e^{-c_2\frac{|x|^2}{r}}dr+\frac{e^{-c_2|x|}}{c_1}(e^{-c_1|x|}-e^{-c_1|x|^2})\\ &=|x|^2e^{-c_1|x|}\int_{|x|}^\infty \frac{e^{-c_2t}}{t^2}dt+\frac{e^{-c_2|x|}}{c_1}(e^{-c_1|x|}-e^{-c_1|x|^2})\\ \end{align}

where I don't exactly know how to handle $\int_{|x|}^\infty \frac{e^{-c_2t}}{t^2}dt$.

Thank you

Answer 1

One integration by parts gives $$\int\frac{e^{-ct}}{t^2}\,dt=-c\,\text{Ei}(-c t)-\frac{e^{-c t}}{t}$$ where appears the exponential integral function. $$\int_{|x|}^\infty \frac{e^{-ct}}{t^2}\,dt=c\, \text{Ei}(-c |x|)+\frac{e^{-c |x|}}{|x|}$$ Assuming $x>0$, the expansion would give $$\sim\frac{e^{-c |x|}}{c x^2}$$