Why is this integral diverging? $\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$ $$\int\limits^{\infty}_{-\infty} \,\frac{x}{x^2+1} dx$$
I can easily prove that this integral is diverging by taking the limit over the following proper integral,
$$\lim_{R_1,R_2\to \infty} \int\limits^{R_2}_{-R_1} \,\frac{x}{x^2+1} dx$$
Mathematically, this makes sense to me, but intuitively I am not able to absorb this.
If we observe the function $y=\frac{x}{x^2+1}$, it is clearly an odd function. And since, integrals return signed areas, an integral to an odd function having limits that are the negatives of each other, should evaluate to zero.
As clearly seen by the following plot.

Can someone please clarify this?
Thank you.
 A: This integral is not absolutely convergent and thus you cannot "easily prove that this integral is diverging"; such expression is usually regarded as ill-defined.
This is similar to $\int_{-\infty}^{\infty}xdx$, the "meaning" of which depends on how you interpret the symbol $\int_{-\infty}^\infty$.
If you take it as "a Cauchy principal value"
$$
\lim_{N\to\infty}\int_{-N}^N xdx
$$
then it is zero because the function is odd.
However, if you interpret it as an iterated limit:
$$
\lim_{a\to-\infty}\left(\lim_{b\to\infty}\int_{a}^b xdx\right)
$$
you already have a problem with the inner limit.

On the other hand, if a function is absolutely convergent,  then all the following "interpretations" of $\int_{-\infty}^\infty$ are the same:

*

*$\int_{-\infty}^\infty:=\lim_{a\to\infty}\int_{-a}^a$;

*$\int_{-\infty}^\infty:=\lim_{a\to-\infty}\lim_{b\to\infty}\int_a^b$;

*$\int_{-\infty}^\infty:=\lim_{b\to\infty}\lim_{a\to-\infty}\int_a^b$;

*$\int_{-\infty}^\infty:=\int_0^\infty+\int_{-\infty}^0$
A: You have to look more carefully at the definition of $\int_{- \infty}^{\infty}.$  Your intuition is treating it as though it means $\lim_{x \to \infty} \int_{-x}^x$, but that's not correct.  It's actually $\lim_{x \to - \infty}(\lim_{y \to \infty} \int_x^y)$ (assuming the integrand is in fact defined everywhere), and for this function the inner limit does not exist for any $y$.
A: Let $ f $ be locally integrable at $ (-\infty,+\infty)$
By définition, we say that the integral $$\int_{-\infty}^{+\infty}f(x)dx$$
is convergent if only if there exists $ c\in \Bbb R $ such that, Both integrals
$$\int_{-\infty}^cf(x)dx\text{ and } \int_c^{+\infty}f(x)dx$$
Are convergent.
If ine of them is divergent, the integral will be divergent.
In your case, the integral $$\int_0^{+\infty}\frac{xdx}{1+x^2}$$ is divergent.
A: $$\lim_{l\to\infty}\lim_{u\to\infty}\int_l^u f(t)\,dt\ne\lim_{u\to\infty}\int_{-u}^u f(t)\,dt.$$
A: so:
$$\frac{x}{x^2+1}=\frac{1}{x+1/x}$$
which means that for large $x$,
$$\frac{x}{x^2+1}\approx\frac1x$$
which the integral diverges for. what you have is the integral of the odd function so in other words,
$$\int_{-a}^af(x)dx=0\,a\to\infty$$
Normally the condition for convergence is considered to be that the interval can be split into parts and still converge, this cannot but we can take the cauchy principle value which does go to zero
A: From the graph of $\frac{x}{1+x^2}$;
For $x>0$, $I_1=\lim_{R_1\rightarrow\infty}\int_0^{R_1}\frac{x}{1+x^2}dx=\lim_{R_1\rightarrow\infty}\frac{1}{2}In(1+R_1^2)$
For $x<0$, $I_2=\lim_{R_2\rightarrow\infty}\int_0^{R_2}\frac{x}{1+x^2}dx=\lim_{R_2\rightarrow\infty}-\frac{1}{2}In(1+R_2^2)$
Signed area$=I_1+I_2=\lim_{R_1\rightarrow\infty}\frac{1}{2}In(1+R_1^2)-\lim_{R_2\rightarrow\infty}\frac{1}{2}In(1+R_2^2)$

The problem is that $RHS=\infty-\infty\ne 0$

