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?

  • $\begingroup$ Do you have similar decay information on the derivatives of $g$? Then I think it's OK. $\endgroup$ Apr 28, 2013 at 19:08
  • $\begingroup$ Yes, you may assume $g^{(n)}(x)=O(x^{-2-n})$. $\endgroup$ Apr 29, 2013 at 17:17
  • $\begingroup$ On a second thought (after quite a bit more time of thinking) I'm not at all sure whether your $J(\lambda)$ is bounded in $\lambda$. I keep getting an upper bound of the type $\log(1/\lambda)$ though I cannot prove yet that this is also a lower bound. $\endgroup$ Apr 30, 2013 at 12:19

1 Answer 1


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.

  • $\begingroup$ Great! Numerical experiments suggest that for $g(x)=1/(1+x^2)$, the integral tends to a nonzero constant, so the limit is optimal. Thanks very much. $\endgroup$ May 2, 2013 at 1:00

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