Improper integral properties

Question

I have a question that I am not so sure about it.

It is not a H.W but looks like.

$f:[0,\infty) \rightarrow \mathbb{R}$ Continuous function and I know that

$\exists N \in \mathbb{N} \ \forall N\leq x \ f(x)>0$

and I know that $\ \int_{N}^{\infty} f(x) \, \mathrm{d}x $ is convergent

Can I conclude that $\ \int_{1}^{\infty} f(x) \, \mathrm{d}x $ is convergent ?

I have a feeling that I can but I do not know what.

Any help will be appreciate (I have a test in that subject next week)

Thanks in advanced !!

Answer 1

The integral $$\int_1^N f(x)\, dx < \infty$$ because $f$ is continuous on $[1,N]$. I can't understand your first assumption, which seems rather useless.

Answer 2

Consider $$I(B)=\int_1^B f(x)\,dx,$$ where $B\gt N$. Then $$I(B)=\int_1^N f(x)\,dx+\int_N^B f(x)\,dx.$$ Since $f$ is continuous in the closed interval $[1,B]$, all the integrals in the line above exist.

Now let $B\to\infty$. By the assumption that $\int_N^\infty$ exists, we know that $\lim_{B\to\infty} \int_N^B f(x)\,dx$ exists, and therefore $\lim_{B\to\infty}I(B)$ exists.

Remark: Informally, since our function is "well-behaved" on any finite interval, what happens "in the tail" determines whether the integral converges.