Integral over antisymmetric function diverges I was wondering why the integral
$$ S = \int_{-\infty}^{\infty} \frac{x}{1+x^2} \, \mathrm{d}x $$
does not converge. Since the function
$$f(x) = \frac{x}{1+x^2}$$
is antisymmetric, I could calculate the integral as follows
$$S \enspace = \enspace \int_{-\infty}^{\infty} f(x) \, \mathrm{d}x \enspace = \enspace \int_0^{\infty} f(x) \, \mathrm{d}x + \int_{-\infty}^0 f(x) \, \mathrm{d}x$$
Now, I substitute $x \rightarrow (-y)$ in the 2nd integral and then use the antisymmetry of $f(x)$:
$$S \enspace = \enspace \int_0^{\infty} f(x) \, \mathrm{d}x + \int_{\infty}^0 f(-y) \, \mathrm{d}(-y) \enspace = \enspace \int_0^{\infty} f(x) \, \mathrm{d}x + \int_{\infty}^0 f(y) \, \mathrm{d}(y) \enspace = \enspace \int_0^{\infty} f(x) \, \mathrm{d}x - \int_{0}^{\infty} f(y) \, \mathrm{d}(y) \enspace = \enspace 0$$
What is the problem with this kind of reasoning?
 A: By definition $$\int_{0}^{\infty} f(x) dx=\displaystyle\lim_{n \to \infty}\int_{0}^{n} f(x) dx$$
$$\int_{-\infty}^{0} f(x) dx=\displaystyle\lim_{m \to \infty}\int_{-m}^{0} f(x) dx$$.
Thus $$\int_{-\infty}^{\infty} f(x) dx=\int_{-\infty}^{0} f(x) dx+\int_{0}^{\infty} f(x) dx\\=\displaystyle\lim_{n \to \infty}\int_{0}^{n} f(x) dx+\displaystyle\lim_{m \to \infty}\int_{-m}^{0} f(x) dx$$
Taking $x=-y$ in the second integral and assuming $f$ to be odd we get,
$$ \int_{-\infty}^{\infty} f(x) dx=\displaystyle\lim_{n \to \infty}\int_{0}^{n} f(x) dx-\displaystyle\lim_{m \to \infty}\int_{0}^{m} f(y) dy$$.
Since we can't claim that $n=m$, thus $$ \int_{-\infty}^{\infty} f(x) dx \neq 0$$ unless $f(x)=0 \ \forall x$ obviously.
But Cauchy principal value of $ \int_{-\infty}^{\infty} f(x) dx$ is defined to be $\displaystyle\lim_{n \to \infty}\int_{-n}^{n} f(x) dx$.
Thus in this case you can say principal value of the integral is $0$.
A: If we have an integral:
$$I(n)=\int_1^\infty\frac{1}{x^n}dx$$
This will only converge for $n>1$, notice how this does not include $n=1$. If we look at the function you are integrating:
$$\frac x{x^2+1}=\frac 1{x+\frac1x}\approx\frac 1x$$ and so your integral will not converge.
If we look at the entire domain of your integral however, as you meantioned it is assymetric but since parts of this domain are divergent we consider integrals like this in terms of their cauchy principle value, in other terms it is best to write your integral as:
$$I(A)=\int_{-A}^0\frac x{x^2+1}dx+\int_0^A\frac x{x^2+1}dx$$ and then take the limit as $A\to\infty$ to show the value for which this integral converges towards. Basically, for an integral to be considered truely divergent you should be able to split up the domain of the integral and each part be also convergent. Hope this helps :)
A: The problem is simply that, by definition, this improper integral converges if & only the improper integrals $\int_0^\infty\frac x{1+x^2}\,\mathrm dx$ and $\int_{-\infty}^0\frac x{1+x^2}\,\mathrm dx$ both converge independently. Now
$$\lim_{A\to\infty}\int_0^A\frac x{1+x^2}\,\mathrm dx= \lim_{A\to\infty}\tfrac12\ln(1+A^2)\to\infty=\infty,$$
and similarly for the other integral.
