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 $x\rightarrow\infty$. I think the following integral should be bounded as $\lambda\rightarrow0$, but would like to have a proof:
$$
J(\lambda) = \int\_0^{\infty}dy \sin\left(y\right)\log(y)g(\lambda y)
$$


I know how to deal with similar asymptotic expansions when there is no log term (integration by parts, method of steepest descent), but these seem to fail here. Can somebody advise how to approach this?



Boundedness of an Oscillating Integral
Yes it is bounded. Assume throughout that $\lambda<1$ and split
$$J(\lambda)=\int\_0 ^1 \sin(y) \log(y) g(\lambda y)dy + \int\_1 ^\infty \sin(y) \log(y) g(\lambda y)dy=:J\_0(\lambda)+J\_1(\lambda) .$$
In $[0,1]$ we have that $0<\lambda y <\lambda <1$ so that $|g(\lambda y)|\lesssim\_g 1$ and this is best possible since $g(0)\neq 0$. We estimate $J\_0$ by brute force
$$|J\_0(\lambda)|\lesssim\_g \int\_0 ^1 |\sin(y)\log(y)|dy\leq\int\_0 ^1 y \log(1/y)dy\lesssim 1.$$
For $J\_1$ we first integrate by parts once using that $\lim\_{y\to\infty}g(y)=0$:
$$J\_1(\lambda)=-\cos(y)\log(y)g(\lambda y)\bigg|\_{1} ^\infty +\int\_1 ^\infty \cos(y)\frac{d}{dy}\big(\log (y) g(\lambda y)\big)dy$$
$$ = \int\_1 ^\infty \cos(y)\frac{g(\lambda y)}{y}dy +\lambda \int\_1 ^\infty \cos(y)\log(y) g'(\lambda y)dy=:A+B.$$
For $A,B$ it is convenient to split the interval of integration to $[1,1/\lambda]$ and $[1/\lambda,\infty]$ where different bounds for $g(\lambda \cdot)$ apply and integrate by parts as many times as necessary.
$$A\_1:=\int\_1 ^\frac{1}{\lambda} \cos(y)\frac{g(\lambda y)}{y}dy= \sin(y) \frac{g(\lambda y)}{y}\bigg|\_{1} ^{1/\lambda}-\int\_1 ^\frac{1}{\lambda} \sin(y) \big(\frac{\lambda g'(\lambda y)}{y}-\frac{g(\lambda y)}{y^2} \big)dy$$
$$=\lambda\sin(1/\lambda)g(1)-\sin(1)g(\lambda)-\lambda\int\_1 ^\frac{1}{\lambda}\sin(y)\frac{g'(\lambda y)}{y}dy +\int\_1 ^\frac{1}{\lambda}\sin(y)\frac{g(\lambda y)}{y^2}dy.$$
Since $|\lambda y|< 1$ and $g$ is real analytic we have $|g'(\lambda y)|+|g(\lambda y)|\lesssim\_g 1$ for $y\in[1,1/\lambda]$. Using these estimates we get $|A\_1|\lesssim\_g 1 $ (one can be more careful here in order to get the main term). 


Similarly
$$A\_2:=\sin(y)\frac{g(\lambda y)}{y}\bigg|\_{\frac{1}{\lambda}}^\infty - \int \_ {\frac{1}{\lambda}} ^\infty \sin(y)\frac{d}{dy}\big( \frac{g(\lambda y)}{y}\big)dy $$
so that $|A\_2|\lesssim \lambda |g(1)|\lesssim\_g \lambda.$ Now for $B$ we split again
$$B\_1:= \lambda \sin(y) \log(y) g'(\lambda y) \bigg|\_{1} ^{1/\lambda}-\lambda\int\_1 ^\frac{1}{\lambda}\sin(y)\frac{d}{dy}\big(\log(y)g'(\lambda y)\big)dy$$
which shows that $|B\_1|\lesssim \lambda \sin(1/\lambda)\log(1/\lambda)|g'(1)|\lesssim\_g \lambda \log(1/\lambda).$ 


Finally
$$B\_2:= \lambda \sin(y)\log(y)g'(\lambda y)\big|\_\frac{1}{\lambda} ^\infty-\lambda \int \_\ frac{1}{\lambda} ^\infty \sin(y)\bigg(\frac{g'(\lambda y)}{y}+\log(y)\lambda g''(\lambda y)\bigg)dy.$$


Using the asmyptotic decay for the derivatives of $g(y)$, $g^{(n)}(x)=O(x^{-2-n})$ for large $y$, and the boundedness for small $y$ we can estimate
$$|B\_2|\leq |g'(1)|\lambda \log(1/\lambda)+\lambda \int\_{\frac{1}{\lambda}} ^\infty \frac{1}{\lambda^3 y^4}dy+\lambda^2 \int\_{\frac{1}{\lambda}} ^\infty \frac{\log(y)}{\lambda^4 y^4}dy $$
$$\lesssim\_g \lambda \log(1/\lambda)+\lambda+\lambda\log(1/\lambda).$$
So $|A|\lesssim\_g 1$ and $|B|\lesssim\_g \lambda \log(1/\lambda)$ as $\lambda \to 0^+$ which implies that $|J(\lambda)|\lesssim\_g 1$. 


If the function $g$ is something like $g(x)=1/(1+x^2)$ (which satisfies all the hypotheses), I am pretty sure that you also have $|J(\lambda)|\gtrsim 1$ but I don't have the patience to check the details.


