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Suppose $f(x)$ is a function defined on $\mathbb{R}\setminus\{c\}$, where $c$ is a scalar. Consider the integral

$$\int_a^bf(x)dx,$$

where $a$ and $b$ are such that $a<c<b$. All Calculus books I checked define the integral above to be convergent if and only if $\int_a^cf(x)dx$ and $\int_c^bf(x)dx$ are both convergent, in which case we have:

$$\int_a^bf(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx$$

It is only a definition, but does it make sense? For example, I can certainly have two series $x_t$ and $y_t$ that are both divergent, but their sum is convergent (the most obvious example is the one where $y_t=-x_t$). The same could occur with integrals. For instance, if I try to integrate $1/x$ from $-1$ to $1$ using the Riemann Sum I would get

$$\int_{-1}^1 \frac{1}{x}dx= \int_{-1}^0\frac{1}{x}dx+\int_0^1\frac{1}{x}dx=\lim_{n\rightarrow \infty}\left[\sum_{j=1}^n\left(\frac{1}{n}\right)\left(\frac{1}{-\frac{1}{n}j}\right)+\sum_{j=1}^n\left(\frac{1}{n}\right)\left(\frac{1}{\frac{1}{n}j}\right)\right]=\lim_{n\rightarrow\infty}\sum_{j=1}^n0=0$$

And therefore, I would conclude this integral is convergent. But the definition tells me otherwise (since $\int_{-1}^0\frac{1}{x}dx$ is divergent). Any thoughts on why it is defined that way?

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    $\begingroup$ In the equality where you substitute the integrals by series, you have changed the order of the terms of two series which are divergent, the first to $-\infty$ and the second to $+\infty$, and that sum can be any number with the appropiate arrangement by Riemann's theorem. Thus, that limit doesn't exist. I think that the problem is you are using an equality that doesn't hold: $\lim_{x\rightarrow x_0}(f(x)+g(x))=\lim_{x\rightarrow x_0}f(x)+\lim_{x\rightarrow x_0}g(x).$ $\endgroup$ – Fernando May 10 '16 at 23:55
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    $\begingroup$ You might be interested in the Cauchy principal value en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$ – fqq May 11 '16 at 0:06
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    $\begingroup$ You cannot just use the same $n$ in the limit of the two integrals. They should be independent of each other. See here where a similar question is asked. $\endgroup$ – Eff May 11 '16 at 0:32
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The theorem you stated is a property of the definite integral. If we go back to the first fundamental theorem of calculus seen here, it says that the functions need to be continuous. If we consider the proof the theorem you stated, you can see it uses the first fundamental theorem of calculus. Thus, the theorem you talk about assumes that f(x) is continuous for [a,b]. For the integral $$\int_{-1}^1 \frac{1}{x}dx$$, this is an improper integral so different rules would apply.

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