How do we express “$\int_a^\infty f(x) \, dx$ is convergent” using predicate logic?

How do we express $$\displaystyle \int_a^\infty f(x) \, dx \text{ is convergent}$$ using predicate logic?

I know how to express that a simple function is convergent: $$f(x) \text{ is convergent}$$ means $$\lim_{x \to \infty} f(x) = L$$ which in turn can be expanded to $$\exists L \in \mathbb{R}, \forall \epsilon > 0, \exists M \in \mathbb{R}, \forall x \in \mathbb{R}, x \geq M \Longrightarrow | L - f(x) | < \epsilon.$$

Based on that, here is my attempt.

I already know that $$\displaystyle \int_a^\infty f(x) \, dx \text{ is convergent}$$ means $$\lim_{b \to \infty} \int_a^b f(x) \, dx = L.$$

To expand it further, maybe all I have to do is use a similar expression as above, but change $$x$$ to $$b$$, and ensure $$b > a$$, as follows:

$$\exists L \in \mathbb{R}, \forall \epsilon > 0, \exists M \in \mathbb{R}, \forall b \in \mathbb{R}, ( b \geq M \wedge b > a ) \Longrightarrow \left|L - \int_a^b f(x)\, dx \right| < \epsilon$$

Is this correct?

• Yes it's correct – Jakobian Aug 13 at 17:44
• You can omit $b>a$, because you can always pick an $M$ greater than $a$ if you wish. – Naj Kamp Aug 13 at 19:11

Your answer is correct. Let $$\forall x \in \mathbb{R}, \, g(x) = \int_a^x f(t) dt$$ As you noticed \begin{align} \displaystyle \int_a^\infty f(t) \, dt \text{ is convergent} &\iff (\exists L,\lim_{x \rightarrow + \infty} \int_a^x f(t) \, dt =L)\\ &\iff (\exists L,\lim_{x \rightarrow + \infty} g(x) = L) \end{align} Notice that $$g$$ is just function. Thus, by definition $$\exists L \in \mathbb{R}, \forall \epsilon > 0, \exists M \in \mathbb{R}, \forall x \in \mathbb{R}, x \geq M \Longrightarrow | L - g(x) | < \epsilon.$$
• Note that in a general case $g$ is not an antiderivative of $f$, because it's not necessarily differentiable. For example, take $|x|=\int_0^x \text{sgn}(t)\mathrm{d}t$ – Botond Aug 15 at 17:49