# finding the limit $\lim_{a\rightarrow \infty} \frac{a^N}{\log a} \int_{0}^\infty \frac{e^{-x}}{(1+ag(x))^N}dx = c$

I am realy stuck in solving the following limit problem. Can you find any function $g(x)$ by which $$\lim_{a\rightarrow \infty} \frac{a^N}{\log a} \int_{0}^\infty \frac{e^{-x}}{(1+ag(x))^N}dx = c$$ where $c$ is a nonzero constant.

The solution to this problem may contain some general properties on $g(x)$. But I can't even find a specific $g(x)$ for a specific $N$, say $N=2$.

• Also posted here. – Lucian Apr 28 '14 at 9:58
• I guess you need/want $g(x)$ to be independent of $a$, right? (otherwise trivially pick $g(x)=cst=1/(\log a)^N$). If so, this is impossible unless $g$ oscillates or vanishes (for finite or infinite points): for example if $0<c_1\leq g(x)\leq c_2$ for two fixed constants then the integral is of order $\mathcal{O}(1/(a^N))$, so the log kills everything. – leo monsaingeon Apr 28 '14 at 10:24

I propose you take $$g(x):=\frac{N}{c}\exp\frac{-x}{N}.$$ I too doubt that you can find an answer independent on $N$.
The method I used to determine $g$ might lead to an answer which is more intreseting, so I provide it here. Define $$\phi(a):=\int_0^\infty\frac{\exp(-x)}{(1+ag(x))^N} dx$$ so that $$\phi'(a)=\int_0^\infty u'(x)v(x) dx$$ with $$u(x)=\frac{1}{a(1+ag(x))^N} \\ v(x)=\frac{g(x)\exp(-x)}{g'(x)}.$$Integrating by parts gives $$a\phi'(a)=\alpha\phi(a)-\frac{g(0)}{g'(0)(1+ag(0))^N}$$ if we have the relation $$v'(x)=-\alpha\exp(-x) , ~~~~\alpha\in\mathbb C$$ which holds whenever $$g(g'+g'')=(\alpha+1)(g')^2 .$$ Assume for now that this equation is satisfied. Then $$\phi(a)=-a^\alpha \int_0^a \frac{g(0)}{g'(0)(1+tg(0))^Nt^{1+\alpha}} dt$$ which is of the required growth-type only if $\alpha=-N$, in which case $$\phi(a)\sim_\infty -\frac{\ln a}{a^Ng'(0)},$$ so that $$c=-\frac{1}{g'(0)}.$$
Now we solve for $g$, which satisfies the equation $$1+\frac{g''}{g'}=(1+\alpha)\frac{g'}{g}$$. This equation is integrable and there exists two constants $\delta, \gamma$ such that $$g(x)=(\gamma+\delta\exp(-x))^{-1/\alpha}.$$ The constant $\gamma$ is useless, while $\alpha$ and $\delta$ are completely determined.
As has been pointed out, if $g(x)>\epsilon$ for some $\epsilon>0$, the limit will be zero. On the other hand, if $g(x)=0$ on a set of positive measure, the limit is $\infty$, so it all depends on how fast $g$ tends to zero. The function $g(x)=x^{1/N}$ actually is a solution with limit $N$, as one sees by substituting $y=a^nx$, using de l'Hospital and substituting back. However, this solution depends on $N$. But the initial thought implies that, if there is a solution that depends on $N$, there probably is no solution independent of $N$.