How to prove $\int^\infty_{-\infty}x dx$ is divergent? Well, intuitively, I thought $\int^\infty_{-\infty}x dx$ is $0$
 ,but the answer is divergent.
How to prove $\int^\infty_{-\infty}x dx$ is divergent?
 A: This is very closely related to the question on whether $\int_{-\infty}^\infty \sin x\,\mathrm d x$ converges, where I responded.
The answer is the following:
No, the integral does not converge (It is divergent). If it did converge to some limit $L$, then for every pair of sequences $a_n\to-\infty$ and $b_n\to\infty$, we would have that
$$\lim\limits_{n\to\infty}\int_{a_n}^{b_n}x \,\mathrm{d} x = L$$
But this is not the case. This is easily proven, try for example the case with $a_n = -n$ and $b_n = 2n$. In this case $a_n\to -\infty$ and $b_n\to+\infty$ as $n\to+\infty$ but the value of the integral does not go to zero.
The important take-home message is that when we consider limits, they must converge to the same limit no matter the path. No matter how we approach the limit. Otherwise it is not considered convergent. In this instance it should not matter how the lower bound goes to minus infinity and how the upper limit goes to plus infinity, but it does matter. We conclude that the integral is divergent.
The Cauchy Principal Value is different to usual convergence (And is therefore not what is being asked in the problem) and this value does exist, it is
$$\lim\limits_{a\to\infty}\int_{-a}^a x \,\mathrm{d} x = 0.$$
A: The improper integral $\int\limits_{-\infty}^{\infty}f$ exists, if both $\int\limits_{-\infty}^{0}f$ and $\int\limits_{0}^{\infty}f$ exists and finite. And then $\int\limits_{-\infty}^{\infty}f=\lim\limits_{a \to \infty} \int\limits_{-a}^{a}f$.
In the case of $f(x)=x$, the improper integrals are not finite. And in the case of $f(x)=\sin(x)$, they don't exists, because the $\lim\limits_{x \to \infty} \cos(x)$ does not exist.
A: The integral $\int^\infty_{-\infty}x dx$ does not converge .
We can get two different answers by taking two different limits $$\lim_{n\to\infty} \int_{-n}^{n}x \,dx=\lim_{n\to\infty}\left[\frac{n^2}{2}-\frac{(-n)^2}{2}\right]=0$$ $$\lim_{n\to\infty} \int_{-n}^{2n}x \,dx=\lim_{n\to\infty}\left[\frac{(2n)^2}{2}-\frac{(-n)^2}{2}\right]=\lim_{n\to\infty}\frac{3n^2}{2}=\infty$$
Two different answers are not possible so the integral does not exist.
The correct way to evaluate $\int^\infty_{-\infty}x dx$ is: $$ \lim_{a \rightarrow - \infty , b \rightarrow \infty} \int_a ^ b f(x) \, dx $$ the integral exists if and only if the limit exists.
However, note that the Cauchy principal value exists, $$PV\int^\infty_{-\infty}x dx=0$$
