The convergence rate of the integrals $\int_0^M\log(1+A\delta^{-1}\exp(-B\delta^{-2}x^{2}))dx$

Question

Let $A$, $B$, $M$ be fixed, I am going to find the convergence rate of the integrals $$\int_0^M\log(1+A\delta^{-1}\exp(-B\delta^{-2}x^{2}))dx$$ as $\delta \to 0$. I have tried to find the derivative of it and to apply L'Hôpital's rule, but the resulting integral is also hard to compute.

Answer 1

I'll show my analysis for the case $M=\infty$. I use $f\ll g$ to mean $f=O(g)$ as $x\to 0$. By rescaling, we can take $A=1, B=1$.You can probably use similar ideas for the case $M=1$. First, substitute $x\to\delta x$ to find that your integral is $$\mathcal{A}(\delta)=\delta\int_0^\infty\log(1+\delta^{-1}e^{-x^2})\,dx:=\delta\int_0^\infty f_\delta(x)\,dx.$$ Let $0< \epsilon<1$ be fixed and $0<\delta<1$ arbitrary. For $0<x<c_\delta:=\log^{1/2}(1/\delta^{1-\epsilon})$, $f_\delta(x)> \log(1+\delta^{-\epsilon}).$ Thus, $$\log^{3/2}(1/\delta) \ll\int_0^{c_\delta}f_\delta\,dx\ll \log^{3/2}(1/\delta).$$ Let $d_\delta=\log^{1/2}(1/\delta)$. Then for all $x>d_\delta$, $f_\delta(x)<\log(2)$ and $0<T_\delta:=\int_{d_\delta}^\infty f_\delta(x)\,dx<\infty$. Hence $T_\delta=O(1)$. Finally, consider. $$N_\delta=\int_{c_\delta}^{d_\delta}f_\delta(x)\,dx.$$ We have that $$ \log^{1/2}(1/\delta) \ll N_\delta\ll \log^{3/2}(1/\delta).$$ Thus, we conclude that $$\delta\log^{3/2}(1/\delta)\ll\mathcal{A}\ll \delta\log^{3/2}(1/\delta).$$ Any errors pointed out are appreciated.

Answer 2

Using the fact that,

$$ue^{-\alpha u^2} \le \sqrt{\frac{e}{2\alpha}}$$

\begin{align} h_\delta(x) = \log \left(1+A\delta^{-1}\exp \left(-B \delta^{-2} x^2\right)\right)&\le \log \left(1+A\frac{1}{\sqrt{2Bx^2}}\sqrt{e}\right)\\ &= \log\left(A\sqrt e + x\sqrt{2B}\right) - \log\left(x\right) - \frac12\log(2B) \end{align}

since $x\mapsto \log\left(A\sqrt e + x\sqrt {2B}\right) - \log\left(x\right) - \frac12\log(2B)$ is integrable on $[0,M]$ and $\lim\limits_{\delta \to 0} h_\delta(x) = 0$ for all $x\in [0,M]$ then $$\lim\limits_{\delta \to 0} \int_0^M h_\delta(x)\mathrm d x = 0$$