Why does not $\int_{-\infty}^\infty x\,\mathrm{d}x$ converge?

Question

It seems natural that it should converge, because for any $A\in\mathbb{R}$, $$\int_{-A}^A x\,\mathrm{d}x=\int_{-A}^0 x\,\mathrm{d}x+\int_0^A x\,\mathrm{d}x=\left[\frac{x^2}{2}\right]_{-A}^0+\left[\frac{x^2}{2}\right]_0^A=-\frac{(-A)^2}{2}+\frac{A^2}{2}=0.$$

So the integral is zero for every such value $A$, yet it doesn't converge, according to wolfram alpha. How come?

It's desirable that $\int_{-a}^b f(x),dx$ be close to the limit whenever both $a>0$ and $b>0$ are large. — David Mitra, Jan 21 '15 at 17:58
As others have pointed out, $\int_{-\infty}^\infty xdx$ does not converge; however the calculation in the OP is still useful and is called the principal value. — vadim123, Jan 21 '15 at 17:59

score 12 · Answer 1 · answered Jan 21 '15 at 17:51

All that is true, but the definition of convergence of integrals of this type is

The integral $\int_{-\infty}^\infty f(x)\,dx$ converges if for each $a$ the integrals $\int_a^{\infty}f(x)\,dx$ and $\int_{-\infty}^a f(x)\,dx$ converge. In this case $$ \int_{-\infty}^\infty f(x)\,dx=\int_{-\infty}^a f(x)\,dx+\int_a^{\infty}f(x)\,dx $$

Since neither $\int_0^\infty x\,dx$ or $\int_{-\infty}^0x\,dx$ converge, the two-sided improper integral does not converge.

score 3 · Answer 2 · answered Jan 21 '15 at 20:19

The integral $\displaystyle\int_{-\infty}^\infty x~dx$ diverges for the same reason that $\displaystyle\sum_{n=-\infty}^\infty n$ also diverges.
You write that $\displaystyle\int_{-A}^Ax~dx=0$ for all real A. True enough! But $\displaystyle\int_{-A}^{A+B}x~dx\neq0$, for $B\neq0$.

$~\quad~$ Now, let $A\to\infty$ in the latter.

Why does not $\int_{-\infty}^\infty x\,\mathrm{d}x$ converge?

2 Answers2

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