Convergence of $\int_{-\infty}^{\infty}f(x)dx$

Question

I posed a question to my calculus professor, asking how to evaluate the Riemann integral

$$\int_{-\infty}^\infty f(x) \, dx$$

I can simplify the above integral as

$$\int_{-\infty}^{\infty}f(x)\,dx = \int_{-\infty}^a f(x)\,dx + \int_a^\infty f(x)\,dx $$

$$\lim_{t\rightarrow -\infty} \int_t^c f(x) \, dx + \lim_{w\rightarrow \infty} \int_c^w f(x)\,dx$$

What happens if I get $ -\infty + \infty$ ?

My professor answered that we cre unable to know where the integral convergences. Is that really the case or not?

Answer 1

For improper Riemann integrals where any of the limits involved are $\infty$, we say that the integral does not converge, or diverges. So, for example, $\int_{-\infty}^\infty x\,dx$ does not converge since $\lim_{a \to \infty} \int_0^a x\,dx = \infty$ and since $\lim_{b \to -\infty} \int_b^0 x\,dx = -\infty$.

Note in this case that we may evaluate the limit $\lim_{a \to \infty}\int_{-a}^a x\,dx = 0$. Although the integral $\int_{-\infty}^\infty x\,dx$ diverges, we state that its Cauchy principal value is $0$.

Answer 2

With the integral $$ \int_{-\infty}^\infty \frac{x}{x^2+1}\,dx $$ you actually do get $-\infty+\infty$. And $$ \lim_{a\to\infty} \int_{-a}^a \frac{x}{x^2+1}\,dx =0. \tag 1 $$ And $$ \lim_{a\to\infty} \int_{-a}^{2a} \frac{x}{x^2+1}\,dx = \log_e 2. \tag 2 $$

The value in $(1)$ is the "Cauchy principal value" of the improper integral.

It is only when the positive and negative parts are both infinite that rearranging like this can change the value of the integral. Convergence like that in $(1)$, where the limit changes with a rearrangement because both the positive and negative parts are infinite, is called "conditional convergence". When the positive and negative parts are both finite, then one has "absolute convergnce" and rearrangement will not change the value.