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).


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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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 $n\to\infty$

  • 1
    $\begingroup$ In short, the sequences may tend to infinity in different manners/speeds. $\endgroup$ – user441558 Dec 13 '17 at 3:08

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.


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.


Not the answer you're looking for? Browse other questions tagged or ask your own question.