Asymptotic Expansion of Logarithmic Integral with Fractional Powers $\int^\infty_0 e^{-tx} \sqrt{x^2-x\log(x)} \;dx$ 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!
 A: 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  {e^{ - tx} \sqrt {x^2  - x\log x} dx} & = \int_0^\varepsilon  {e^{ - tx} \sqrt { - x\log x} dx}  + \int_0^\varepsilon  {e^{ - tx} \mathcal{O}(x^{3/2} )dx} \\ & = \int_0^\varepsilon  {e^{ - tx} \sqrt { - x\log x} dx}  + \mathcal{O}\!\left( {\frac{1}{{t^{5/2} }}} \right)
\end{align*}
as $t\to +\infty$. According to Theorem 2 in Chapter II, $\S$2 of R. Wong's Asymptotic Approximations of Integrals,
$$
\int_0^\varepsilon  {e^{ - tx} \sqrt { - x\log x} dx}  = \frac{{\sqrt \pi  }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right)
$$
as $t\to +\infty$. Consequently,
$$
\int_0^\varepsilon  {e^{ - tx} \sqrt {x^2  - x\log x} dx}  = \frac{{\sqrt \pi  }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }} + \mathcal{O}\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right) = \frac{{\sqrt \pi  }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log t}}} \right)} \right)
$$
as $t\to +\infty$.  We also have
\begin{align*}
\int_\varepsilon ^{ + \infty } {e^{ - tx} \sqrt {x^2  - x\log x} dx} & = \int_\varepsilon ^{ + \infty } {e^{ - tx} \mathcal{O}(x)dx}  = e^{ - t\varepsilon } \int_0^{ + \infty } {e^{ - tx} \mathcal{O}(x + \varepsilon )dx} \\ & = \mathcal{O}\!\left( {\frac{{e^{ - t\varepsilon } }}{t}} \right) = o\!\left( {\frac{1}{{t^{3/2} \sqrt {\log t} }}} \right)
\end{align*}
as $t\to +\infty$. Hence, finally,
$$
\int_0^{ + \infty } {e^{ - tx} \sqrt {x^2  - x\log x} dx}  = \frac{{\sqrt \pi  }}{2}\frac{{\sqrt {\log t} }}{{t^{3/2} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log t}}} \right)} \right)
$$
as $t\to +\infty$.
A: You may not be able to use the Taylor expansion of the messy function $\sqrt{x^2 - x\ln x}$ around $x =0$, but you can use the Taylor expansion of $\sqrt{1-x}$:
$$\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \dots$$
and then use the fact that
$$\sqrt{x^2 - x\ln x} = \sqrt{-x\ln x}\cdot \sqrt{1 - \frac{x}{\ln x}}$$
Since $x/\ln x \to 0$ as $x\to 0$, we have that
$$\sqrt{x^2 - x\ln x} \sim \sqrt{- x\ln x}\cdot \Bigg(1 - \frac{x}{2\ln x} - \frac{x^2}{8\ln^2 x} - \dots\Bigg)$$
You were right: the leading order term is $\sqrt{- x \ln x}$.
Now, as for finding the asymptotically dominant term of $\int_0^\infty e^{-tx}\sqrt{-x\ln x}dx$... I’m still trying to figure that out myself.
