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Assume that I have a real-valued positive function $f(x)$, and I need to find the following limit:

$\lim\limits_{y\rightarrow\infty} \int\limits_{0}^{\infty} y x e^{-y f(x)}dx$. What approaches can I follow to find the limit?

My exact function is $f(x)=a_{1}x^3+a_{2}x^2+a_{3}x$, where $a_3,a_2,a_1$ are positive constants. But I'm also interested in learning about general approaches to solve this problem.

My approach:

The given integration is upper-bounded as follows:

$\int\limits_{0}^{\infty} y x e^{-y (a_{1}x^3+a_{2}x^2+a_{3}x)}dx \leq \int\limits_{0}^{\infty} y x e^{-y (a_{2}x^2)}dx=\frac{1}{2a_2}$

And we also know that it is positive, so the limit is bounded between $0$ and $\frac{1}{2a_2}$ and it is finite. However, I can't exchange the order of the limit and the integration since this will result in $\int\limits_{0}^{\infty} 0 dx$. I know that I should check the dominated convergence theorem to be able to do so, but even if the conditions of it are satisfied, changing the order will not solve the problem.

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For the case $f(x)=a_1x^3+a_2x^2+a_3x$, with $a_i>0$, we have

$$xye^{-yf(x)}\le \frac{x}{ef(x)}=\frac{1}{e(a_1x^2+a_2x+a_3)}\in L^1$$

Hence, the Dominated Convergence Theorem guarantees that

$$\lim_{y\to \infty}\int_0^\infty xye^{-yf(x)}\,dx=\int_0^\infty \lim_{y\to \infty}\left(xye^{-yf(x)}\right)\,dx=0$$

And we are done!

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    Can you please explain to me how did you get the upper-bound? – 2017-02-21
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    Sure. For $f(x)=xe^{-ax}$, $f'(x)=(1-ax)e^{-ax}=0$ at $x=1/a$, where $f$ is a maximum. – 2017-02-22