Convergence of doubly infinite improper integral for odd functions.

Question

I was working on this integral:

$$\int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}$$

Calculations shows that the limits DNE, and therefore the integral diverge. I used Mathematica and found the same result.

But, the integrand is an odd functions, therefore:

$$\forall c \in \Bbb R : \int_{-c}^{+c} \frac{x \, dx}{1+x^2} = 0 $$

So why don't we just say that: $$\int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{c\to\infty} \int_{-c}^{+c} \frac{x \, dx}{1+x^2}=0$$ And the same for any other odd functions?

Answer 1

I would say that, if the integral $$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2} \tag1 $$ does exist, then we have $$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{c\to\infty} \int_{-c}^{+c} \frac{x \, dx}{1+x^2}. \tag2 $$ You have to first prove that the integral in $(1)$ exists to deduce $(2)$.

Think about the following analog situation, you can not assert that $$ (-1)^{\infty}=\lim_{n \to \infty}(-1)^{2n}=1. \tag3 $$ One may recall that

$$ \int_{-\infty}^{+\infty} \frac{x \, dx}{1+x^2}=\lim_{a \to -\infty}\int_a^c \frac{x \, dx}{1+x^2}+\lim_{b \to +\infty}\int_c^b \frac{x \, dx}{1+x^2},\quad \text{for }\color{red}{\text{any }}c \in \mathbb{R}. $$

Answer 2

$$ \int_{-c}^{2c} \frac{x\,dx}{1+x^2} = \frac 1 2 \log\frac{1+4c^2}{1+c^2} \to \frac 1 2 \log 4 \ne 0 \text{ as }c\to\infty. $$

As always with conditionally convergent things, the limit depends on how the bounds approach $\infty$.

Answer 3

The definition of the Improper Integral is

$$\begin{align} \int_{-\infty}^{\infty}\frac{x}{1+x^2}dx&\equiv\lim_{L^{-}\to -\infty}\,\,\lim_{L^{+}\to \infty}\int_{L^{-}}^{L^{+}}\frac{x}{1+x^2}dx\\\\ &=\lim_{L^{-}\to -\infty}\,\,\lim_{L^{+}\to \infty} \frac12 \log\left(\frac{(L^{+})^2+1}{(L^{-})^2+1}\right) \end{align}$$

where the integral is defined by taking two separate limits. Inasmuch as this limit does not exist, the integral is undefined.

However, if we interpret the integral as a Cauchy Principal Value, then the upper and lower limits are identical and we have

$$\begin{align} \text{P.V.}\int_{-\infty}^{\infty}\frac{x}{1+x^2}dx&\equiv\lim_{L\to \infty}\int_{-L}^{L}\frac{x}{1+x^2}dx\\\\ &=\lim_{L\to \infty}\frac12 \log\left(\frac{L^2+1}{L^2+1}\right)\\\\ &=0 \end{align}$$