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I am reading the Florida Mu Alpha Theta Sponsors Guide. Page 43 is a list of clarifications and disputes commonly made, and their resolutions. One of their clarifications is this:

A function which is not integrable on an interval A is not integrable on any interval B, where B contains A. I.e. no “the negative signs cancel” arguments.

What is this argument they're referring to?

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Consider the function $f(x)=\frac{1}{x}$ for $x \neq 0$ and $f(0)=0$. You can show that $f$ is not integrable on $[0,1]$. The statement in your question tells us $f$ is not integrable on $[-1,1]$. One might argue that because $f$ is an odd function and $[-1,1]$ is centered at the origin, the positive and negative area "cancel out" and $\int_{-1}^1 f(x)dx=0$ (similarly to how $\int_{-\pi}^\pi \sin(x)dx=0$), but this would be incorrect.

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For example, $\int_{-\infty}^\infty x \; dx$ is not $0$ (except in the Cauchy principal value sense).

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The rule that $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx $ should hold, and its generalization to all partitions of an interval into a finite number of closed sub-intervals. This means $f$ should have an integral on all closed subintervals, if it is to be integrable.

For the Riemann integral and its generalizations (Lebesgue, Stieltjes etc) the partition property is a consequence of $\int_a^b f $ existing. Thus the word "should" can be replaced by "must" in the first paragraph.

Cancellation of singularities happens in some regularization schemes for divergent or improper integrals, which is a different matter that does not affect integrability.

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I would guess this means that, for example, if $f$ is a real valued function on $[-1,1]$ which is odd (meaning $f(-x)=-f(x)$ for all $x$) and which is not integrable on $[0,1]$, then it would be fallacious to say that the function is integrable on $[-1,1]$ and the integral evaluates to zero".

Using the symmetry of the function to arrive at a value for a definite integral without consideration for integrability of the function would certainly be a mistake...

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