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

REPLY [2 votes]: 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 $h(x_i)=f(x_i),i=0,1.$
{"set_name": "stack_exchange", "score": 0, "question_id": 2868552}