I am interested in the leading order behavior of the following integral when $t \to \infty$:
$$\int^\infty_0 e^{-tx} \sqrt{x^2-x\log(x)} \;dx$$
Here $x^2-x\log(x)$ is always positive, so the square root is well defined everywhere even at $x=0$. However, I can't use the usual tricks I use (e.g. Watson's lemma) to find the leading order behavior, because well, it is not continuously differentiable at 0, so I cannot derive any useful Taylor expansion.
I expect the order to be something like $O(\sqrt{\frac{\log(t)}{t^2}})$ because heuristically speaking when $x$ is close to 0, we have $|x\log x|\gg x^2$, so the integral should operate like $ \int^\infty_0 e^{-tx} \sqrt{-x\log(x)}\;dx$. Now I don't know how to solve that either but I do know that the leading order of:
$$\int^\infty_0 e^{-tx} x\log(x) \;dx$$
is $O(\frac{\log(t)}{t^2})$ (this can be seen easily through change of variables or integration by parts), so intuitively speaking I'm expecting the square root to carry over, but I am 1. not sure that my reasoning is right, and 2. don't know how to actually get there.
Any help would be greatly appreciated!
Best Answer
Fix $0<\varepsilon \ll 1$. For $0<x<\varepsilon$, \begin{align*} \sqrt {x^2 - x\log x} = \sqrt { - x\log x} \sqrt {1 - \frac{x}{{\log x}}} & = \sqrt { - x\log x} + \mathcal{O}\!\left( {\frac{{x^{3/2} }}{{\sqrt { - \log x} }}} \right) \\ & = \sqrt { - x\log x} + \mathcal{O}(x^{3/2} ). \end{align*} Therefore, \begin{align*} \int_0^\varepsilon {\mathrm{e}^{ - tx} \sqrt {x^2 - x\log x} \,\mathrm{d}x} & = \int_0^\varepsilon {\mathrm{e}^{ - tx} \sqrt { - x\log x} \,\mathrm{d}x} + \int_0^\varepsilon {\mathrm{e}^{ - tx} \mathcal{O}(x^{3/2} )\,\mathrm{d}x} \\ & = \int_0^\varepsilon {\mathrm{e}^{ - tx} \sqrt { - x\log x} \,\mathrm{d}x} + \mathcal{O}\!\left( {\frac{1}{{t^{5/2} }}} \right) \end{align*} as $t\to +\infty$. According to Theorem $2$ in Chapter II, $\S2$ of R. Wong's book Asymptotic Approximations of Integrals, $$ \int_0^\varepsilon {\mathrm{e}^{ - tx} \sqrt { - x\log x} \,\mathrm{d}x} = \frac{{\sqrt \pi }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right) $$ as $t\to +\infty$. Consequently, \begin{align*} \int_0^\varepsilon {\mathrm{e}^{ - tx} \sqrt {x^2 - x\log x} \,\mathrm{d}x} & = \frac{{\sqrt \pi }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right) \\ &= \frac{{\sqrt \pi }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log t}}} \right)} \right) \end{align*} as $t\to +\infty$. We also have \begin{align*} \int_\varepsilon ^{ + \infty } {\mathrm{e}^{ - tx} \sqrt {x^2 - x\log x} \,\mathrm{d}x} & = \int_\varepsilon ^{ + \infty } {\mathrm{e}^{ - tx} \mathcal{O}(x)\,\mathrm{d}x} = \mathrm{e}^{ - t\varepsilon } \int_0^{ + \infty } {\mathrm{e}^{ - tx} \mathcal{O}(x + \varepsilon )\,\mathrm{d}x} \\ & = \mathcal{O}\!\left( {\frac{{\mathrm{e}^{ - t\varepsilon } }}{t}} \right) = o\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right) \end{align*} as $t\to +\infty$. Hence, finally, $$ \int_0^{ + \infty } {\mathrm{e}^{ - tx} \sqrt {x^2 - x\log x} \,\mathrm{d}x} = \frac{{\sqrt \pi }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log t}}} \right)} \right) $$ as $t\to +\infty$.