Asymptotics of $F(\varepsilon)=\int^b_0\frac{dx}{\varepsilon +\phi(x)}$ as $\varepsilon\rightarrow0$ where $\phi(x)\sim Ax$ as $x\rightarrow0$.

Question

This is one part of an old qualifier analysis problem that I am trying to complete.

Suppose $\phi$ is a continuous function on a finite interval $[0,b]$ and that $\phi(x)>0$ for all $x\in[0,b]$. Assume that $\phi(x)\sim A x^r$ for some constant $A>0$ and $r\geq0$. Define $$F(\varepsilon)=\int^b_0 \frac{dx}{\varepsilon + \phi(x)}$$

1. If $0\leq r<1$, then $\lim_{\varepsilon\rightarrow0}=\int^b_0\frac{dx}{\phi(x)}<\infty$.
2. If $r>1$, then $$F(\varepsilon)\sim \frac{\pi}{r A^{1/r}\sin(\pi/r)} \varepsilon^{-(1-\frac1r)}\quad\text{as}\quad\varepsilon\rightarrow0$$
3. If $r=1$, then $$F(\varepsilon)\sim\frac{1}{A}\log(\varepsilon^{-1})\quad\text{as}\quad\varepsilon\rightarrow0$$

Part (1) is a simple application of monotone convergence.

When $r\geq1$, $\lim_{\varepsilon0}F(\varepsilon)=\infty$ by monotone convergence.

Part (2) I have done by first applying the of variable $u=\varepsilon^{-1}x^r$, followed by application of dominated convergence. All that gives $$\lim_{\varepsilon\rightarrow0}\varepsilon^{1-\frac1r}F(\varepsilon)=\int^\infty_0\frac{u^{\frac1r-1}}{1+Au}\,du$$ The later is a integral of the Mellin transform type and the expression in the problem can be obtained by complex contour integration.

Part (3) is where I am having difficulties. The asymptotic seems given in the problem seems correct. By assuming that $\phi(t)=At$, then a direct computation gives $$F(\varepsilon)=\frac{1}{A}\log\Big(\frac1A+\frac{b}{\varepsilon}\Big)\sim\frac{1}{A}\log(\varepsilon^{-1})$$ I thing dominated convergence still useful but a trick may be needed. For example, the function $x\mapsto x^{-1}\phi(x)$ can be extended to $[0,b]$ as a strictly positive continuous function on $[0,b]$ and so, there is $M>0$ such that $\phi(x)\geq Bx$ for $x\in[0,b]$. Hence $$\frac{1}{\varepsilon +\phi(x)}\leq\frac{1}{\varepsilon+Bx}$$

Any hints (not necessarily a complete solution) are appreciated. Thank you!

Answer 1

Hint:

Given any $0 < t < A$ there's a $\delta > 0$ such that $(A - t)x^r < \phi(x) < (A + t)x^r$ when $x < \delta$. Insert this into the $x < \delta$ portion of the integral and you'll get lower and upper bounds for this part of the integral which will be close to the asymptotics you need as $\epsilon \rightarrow 0$. The $x > \delta$ integral will be an error term as $\epsilon$ goes to zero. Letting $t \rightarrow 0$ gives the desired asymptotics.

Answer 2

If assuming that $\phi(t) = At$ directly yields the desired asymptotics, then one natural strategy would be to control the error of this assumption. Namely, we consider $$ \frac{1}{\epsilon + \phi(t)} = \frac{1}{\epsilon + At} + \frac{At - \phi(t)}{(\epsilon + \phi(t))(\epsilon + At)}. $$ Then it remains to show that $$ \int_0^b \frac{At - \phi(t)}{(\epsilon + \phi(t))(\epsilon + At)}dt $$ is asymptotically subdominant to $\frac{1}{A}\log(\epsilon^{-1})$. Now, it's not entirely clear to me what definition of the notation $\phi(t) \sim At$ is being used here, but I would presume you could use it to control the numerator here, and consequently the integral overall.

Answer 3

Just to explain two different methods that I have now to solved this problem. The first is based on Zarrax's solution above.

I have already established in my posting that there is $B>0$ scuch that $\phi(x)\geq Bx$ or all $x\in[0,b]$.

Method 1: For any $0<t<<1$ chose $\delta_t>0$ such that $(A-t)x<\phi(x)<(A+t)x$ for all $0<t<\delta$. Then $$\frac{1}{\varepsilon+(A+t)x}<\frac{1}{\varepsilon+\phi(x)}<\frac{1}{\varepsilon+(A-t)x}\quad\text{for}\quad 0<x<\delta$$ Splitting the the interval of integration gives $$\frac{1}{\log(\varepsilon^{-1})}\int^b_0\frac{dx}{\varepsilon+\phi(x)}=\frac{1}{\log(\varepsilon^{-1})}\int^\delta_0\frac{dx}{\varepsilon+\phi(x)} +\frac{1}{\log(\varepsilon^{-1})}\int^b_\delta\frac{dx}{\varepsilon+\phi(x)}$$

The second term on the right-hand side is small, that is \begin{align*} \frac{1}{\log(\varepsilon^{-1})}\int^b_\delta\frac{dx}{\varepsilon +\phi(x)}&\leq \frac{1}{\log(\varepsilon^{-1})}\int^b_\delta\frac{dx}{\varepsilon+Bx}\\ &=\frac{1}{B\log(\varepsilon^{-1})}\log\Big(\frac{\varepsilon+Bb}{\varepsilon+B\delta}\Big)\xrightarrow{\varepsilon\rightarrow0}0 \end{align*}

The first terms has bounds $$\frac{1}{(A+t)\log(\varepsilon^{-1})}\log\Big(\frac{\varepsilon+(A+t)\delta}{\varepsilon}\Big)\leq \frac{1}{\log(\varepsilon^{-1})}\int^\delta_0\frac{dx}{\varepsilon +\phi(x)}\leq \frac{1}{(A-t)\log(\varepsilon^{-1})}\log\Big(\frac{\varepsilon+(A-t)\delta}{\varepsilon}\Big) $$ Letting $\varepsilon\rightarrow\infty$ gives

$$ \frac{1}{A+t}\leq\liminf_{\varepsilon\rightarrow0}\frac{1}{\log(\varepsilon^{-1})}\int^b_0\frac{dx}{\varepsilon+\phi(x)}\leq\limsup_{\varepsilon\rightarrow0}\frac{1}{\log(\varepsilon^{-1})}\int^b_0\frac{dx}{\varepsilon+\phi(x)}\leq\frac{1}{A-t} $$ for all $0<t\ll 1$. Ir follows that $$ \lim_{\varepsilon\rightarrow0}\frac{1}{\log(\varepsilon^{-1})}\int^b_0\frac{dx}{\varepsilon +\phi(x)}=\frac{1}{A} $$

Method 2: This I found from a hint given by one of may study mates. Split the interval of integration as follows $$\int^b_0\frac{dx}{\varepsilon+\phi(x)}=\int^{\varepsilon b}_0+\int^b_{\varepsilon b}\frac{dx}{\varepsilon+\phi(x)}$$ The first itnegral can be bounded as \begin{align} \int^{\varepsilon b}_0\frac{dx}{\varepsilon+\phi(x)}&\leq\int^{\varepsilon b}_0\frac{dx}{\varepsilon + Bx}\\ &=\frac{1}{B}\log(1+Bb)<\infty \end{align} Since $F(0+)=\infty$, $$F(\varepsilon)\sim \int^b_{\varepsilon b}\frac{dx}{\varepsilon + \phi(x)}$$ The change of variables $x(t)=b\varepsilon^t$ (equivalently $t=\frac{\log(b^{-1}x)}{\log \varepsilon}$) gives \begin{align*} \int^b_{\varepsilon b}\frac{dx}{\varepsilon + \phi(x)}&=\log(\varepsilon^{-1})\int^1_0\frac{b\varepsilon^t}{\varepsilon + \phi(b\varepsilon^t)}\,dt \end{align*}

Since $\frac{b\varepsilon^t}{\varepsilon + \phi(b\varepsilon^t)}\leq \frac{b\varepsilon^t}{\varepsilon + Bb\varepsilon^t}=\frac{b}{\varepsilon^{1-t}+Bb}\xrightarrow{\varepsilon\rightarrow0}\frac{1}{B}$, by dominated convergence \begin{align*} \lim_{\varepsilon\rightarrow0}\int^1_0\frac{b\varepsilon^t}{\varepsilon + \phi(b\varepsilon^t)}\,dt=\frac1{A} \end{align*} In conclussion, $F(\varepsilon)\sim \frac{1}{\log(\varepsilon^{-1})}$.