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Why does $\int_{-\infty}^{\infty}\ x\ dx\;$ diverge?

I'm aware that when you find the principal value that it equals $0$ because you create a limit, but I'm still unsure as to why this would diverge (as is implied by my textbook).

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marked as duplicate by leonbloy, user99914, Hans Lundmark, Claude Leibovici integration Dec 13 '17 at 7:23

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The cauchy principal value equals zero. But for the integral to exist (and have value $I$) it must be the case that $$ \int_{a_n}^{b_n}\ x\ dx\to I. $$ as $n\to\infty$ for all sequences where $a_n\to-\infty$ and $b_n\to\infty$. But $$ \int_{-n}^{n}\ x\ dx\to 0 $$ while $$ \int_{-n}^{n^2}\ x\ dx=\frac{1}{2}(n^4-n^2)\to \infty $$ as $n\to\infty$

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    $\begingroup$ In short, the sequences may tend to infinity in different manners/speeds. $\endgroup$ – user441558 Dec 13 '17 at 3:08
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You probably assume that since $-\infty$ and $+\infty$ are additive inverses and the integrand is an odd function, the integral must be zero. But infinity does not work like that; $-\infty$ and $+\infty$ are not numbers but limiting conditions. To get the given integral to converge all combinations of limiting cases leading to a lower limit of $-\infty$ and an upper limit of $+\infty$ must agree. Certainly if we select lower and upper limits $a=-M, b=+M, M\rightarrow +\infty$ then indeed the integral is zero for that limiting case. But for another limiting case, let us say $a=-M, b=+2M$, the integral is nowhere near converging to zero. So, no dice on convergence.

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While $\int_{-n}^{n} x \ dx = 0$ for all n.

but $\int_{-n}^{n} x \ dx = \int_{-n}^{0} x \ dx + \int_{0}^{n} x \ dx$

If we are going to talk about the integral converging or

$\lim_\limits {n\to\infty} \int_{-n}^{n} f(x) \ dx $ existing then

$\lim_\limits{n\to\infty} \int_{-n}^{a} f(x) \ dx + \lim_\limits{n\to\infty} \int_{a}^{n} f(x) \ dx$

Must exist as well.

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