# 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?

• Have you tried consulting R. Wong's 'Asymptotic Approximation of Integrals' (SIAM, 2001) already? Aug 26 '13 at 23:31
• Not without more information about $g$. As stated now, there are counterexamples. Aug 28 '13 at 14:34
• @fedja, would you mind posting your counterexample? I'm interested to see it. Aug 29 '13 at 18:22
• @Antonio Vargas It is very simple: take a sequence $a_k$ that grows fast enough and put $g_0(x)=(1+x^2)^{-1}\sum_k k^{-2} \sin (a_k x)$. Note that when $\lambda=a_k^{-1}$, there is one resonance term, and its integral over, say, $(2ka_k,(2k+1)a_k)$ is of order $(\log a_k)/k^2\to\infty$ as $k\to\infty$. However, this part depends only on the values of $g_0$ on $(2k,2k+1)$, and for every precision sequence $\varepsilon_k$, we can find an entire function $g$ such that $|g-g_0|<\varepsilon_k$ on $(2k,2k+1)$. In other words, the purely qualitative condition of real analyticity is useless. Aug 31 '13 at 14:19
• Any reasonable "quantitative" version of real analyticity will yield the desired result, of course, but you should state it explicitly as an assumption. Most likely, the case is not that what you need is wrong but that what you state is not what you need. Just revise the post, and I'll revise the answer. :) Aug 31 '13 at 14:24

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.