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I have the following function $$f(\textbf{x})=\frac{Q(\textbf{x})}{P(\textbf{x})}$$ where $P(\textbf{x})$ is such that $P(\textbf{x})=0\iff \textbf{x}=0$ and at the same time $Q(\textbf{x})=0\iff\textbf{x}=0$. The functions $P(\textbf{x})$ and $Q(\textbf{x})$ are such that: $\lim \limits _{\textbf{x}\rightarrow0}f(\textbf{x})=0$ and $\int \limits _{\Omega}f(\textbf{x})\ \mathrm{d}\textbf{x}=m$.

Suppose now that $Q(\textbf{x})$ is a polynomial function and that I need to algebrically expand it. What one would get is: $$f(\textbf{x})=\frac{Q_1(\textbf{x})+Q_2(\textbf{x})+\cdots+Q_n(\textbf{x})}{P(\textbf{x})}=\sum \limits _{i=1}^n\frac{Q_i(\textbf{x})}{P(\textbf{x})}$$

Unfortunately, the functions are such that $$\lim \limits _{\textbf{x}\rightarrow0} f(\textbf{x})\neq \sum \limits _{i=1}^n\lim \limits _{\textbf{x}\rightarrow0} \frac{Q_i(\textbf{x})}{P(\textbf{x})}$$ and also $$\int \limits _{\Omega}f(\textbf{x})\ \mathrm{d}\textbf{x}\neq \sum \limits _{i=1}^n\int \limits _{\Omega} \frac{Q_i(\textbf{x})}{P(\textbf{x})} \mathrm{d}\textbf{x}$$

What I would like to know is, when one has to work with the "expanded" version of the function (meaning that one has to evaluate the integral or the limit forcibly in this form), how can he/she adapt the inequalities in order to be sure that he retrieves the results of the "compact" formulation?

I have read When can a sum and integral be interchanged?, but still find this situation obscure

A very simple example would be:

$$f(x)=\frac{x-2}{x-2}$$ and $$g(x)=\frac{x}{x-2},\ h(x)=-\frac{2}{x-2}$$ integrated over (for example) $I=[0,3]$

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  • $\begingroup$ I think it has to come down to the limits $Q_i(x)/P(x)$ for $x\to 0$ not existing, otherwise those equalities would have to hold. $\endgroup$ – SvanN Jan 18 '19 at 12:21
  • $\begingroup$ Yes I get some of the $Q_i(x)/P(x)$ tending to $\infty $ for $x\rightarrow 0$ $\endgroup$ – Riccardo Jan 18 '19 at 12:26
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I'm not sure why you'd want to split it up if you had it in such a nice form, as your example shows you'll run into problems. Let me address some of those problems though; hopefully they show why you wanna avoid breaking things up into 'bad-behaved' parts.

Whenever two functions are integrable on some interval, then we may reverse the order of the integral and the sum. But, the integrability condition is crucial here. Your example at the end is just such a case where this doesn't apply: $x/(x-2)$ has integral $-\infty$ over $[0,3]$, while $-2/(x-2)$ has integral $+\infty$ over $[0,3]$. (If you happen to know some measure theory, this is exactly the type of problems one wishes to avoid when defining the Lebesgue integral for a function.)

The other issue is similar: the limit $Q_i(x)/P(x)$ as $x\to 0$ need not exist, so that we cannot even talk about the sum of these limits (let alone decide if it's equal to the limit of the sum).

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