Asymptotic Expansion of an Oscillating Integral Let $g(x):\mathbb{R}_{\geq0}\rightarrow\mathbb{R}$ be real analytic s.t. $g(0)\neq 0$ and $g(x)=O(x^{-2})$ as $x\rightarrow\infty$.
What is the leading order in $\lambda$ as $\lambda\rightarrow 0$ of the following integral?
$$
I(\lambda) = \int_0^{\infty}dx \cos\left(\frac{x}{\lambda}\right)x\log(x)g(x)
$$
I think it should be $I(\lambda)\sim -\lambda^2(\log\lambda) g(0)$ based on integration by parts, but I haven't got a complete argument.
Edit: Integrating by parts and throwing away subleading terms easily shows that it suffices to prove that
$$
J(\lambda) = \int_0^{\infty}dy \sin\left(y\right)\log(y)g(\lambda y)
$$
is bounded as $\lambda\rightarrow 0$ under the above conditions on $g$. Numerical experiments suggest this to be the case. Can somebody prove it, please?
 A: This answer is a work in progress:
Following Asymptotic Expansions of Integrals by Bleistein and Handelsman, section 6.3, we can use the method of Mellin transforms.
Notice that $$I(\epsilon)=\int_0^\infty g(x) \ x \log({x}) \cos{\big(\epsilon x\big)} \ dx = Re\bigg[ \int_0^\infty g(x) \ x \log({x}) e^{i \epsilon x} \ dx \bigg]$$ with $\epsilon = \frac{1}{\lambda}$ and we will evaluate as $\epsilon \to \infty$.
As $\epsilon$ becomes large, the oscillations of the integrand become more and more rapid. These rapid oscillations tend to induce cancellation of the integral in a neighborhood of every point, except those where the argument of the exponential goes to zero, or the derivative of the argument goes to zero. Since $x$ is monotonically increasing on $(0,\infty]$, the only such point is $x=0$. Thus, the primary contribution to the integral will come from a small region near the origin.
For now, I will simply say that we can write this integral as an inverse Mellin Transform:
$$I(\epsilon) = Re \bigg[ \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} \epsilon^{-z} M[e^{i\epsilon x};z] M[F_a;1-z] dz \bigg] $$ where $M[...]$ denotes the Mellin Transform of the expression and $$F_a(x) = \left\{ \begin{array}{lr}  
f_a & 0 \le x \le \gamma \\
0 &  x > \gamma \\
\end{array} \right.$$ with $f_a \equiv g(x) \ x \log({x})$ in a positive half-neighborhood of $0$ and goes to $0$ smoothly as $ t \to \gamma$ such that $f_a \in C^\infty(0,\infty)$
We can then close the contour for this expression of $I(\epsilon)$ in the right half-plane and by considering the residues find an asymptotic expansion. I will quote the result with explanation to follow at a later time that since $$e^{i \epsilon x} \sim 1, \ x \to 0$$ and $$g(x) \ x \log({x}) \sim g(0) \ x \log({x}), \ x \to 0$$ we have $$I(\lambda) \sim \lambda^2 g(0) \big( \gamma_c - 1 - \log{\lambda}\big)  $$ where $\gamma_c$ is the Euler-Mascheroni constant. I have verified this approximation by comparing to some numerical results for simple $g(x)$ generated by Mathematica. For using the integration by parts technique, I am used to the restriction that the functions must be infinitely differentiable over the closed interval of integration. I think it is interesting that this method produces the correct leading term despite the problem at the origin.
