A continuous function $f$ verifies : $\int_{0}^{+\infty} f(x).dx= +\infty$. $\int_{-\infty}^{0} f(x).dx= -\infty$.

Question

In general,for a continuous function $f$ verifies : $\int_{0}^{+\infty} f(x).dx= +\infty$. $\int_{-\infty}^{0} f(x).dx= -\infty$. What can we say about $\int_{-\infty}^{+\infty} f(x).dx$. I took the example of $f(x)=2x$ it verifies the hypothesis, and I've found that : $\int_{-\infty}^{+\infty} 2xdx$ doesn't exit ! I think there are examples where the integral may exist ! (like a finite number, or infinity).

Answer 1

If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $\mathbb{R}$ is 0. So, $$\lim_{a\rightarrow \infty} \int_{-a}^{a} f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.

To get the PV to equal some $b\in \mathbb{R}$ you can take an odd function $f$ and manipulate it to a function $$ g(x)= \begin{cases}\ h(x) \quad\text{for}\quad x\in [x_0,x_1],\\ f(x) \quad \text{else}, \end{cases} $$ where $$\int_{x_0}^{x_1} h(x) dx=\int_{x_0}^{x_1} f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$