Divergent integrals $\int_{a}^{\infty}\frac{dx}{f(x)}$ and $\int_{a}^{\infty}\frac{dx}{f(x)+b}$ for positive $f$ and $b$ Assume that $f\colon [a,\infty)\to(0,\infty)$ is continuous ($a\in\mathbb{R}$) and such that
$$\int_{a}^{\infty}\frac{dx}{f(x)}=+\infty$$
and let $b>0$. Can we claim that
$$\int_{a}^{\infty}\frac{dx}{f(x)+b}=+\infty\quad ?$$
If $f$ is nondecreasing, then it is true. And in general? What else makes this implication true?
Proof for nondecreasing $f$: we have
$$\frac{1}{f(x)}=\frac{1}{f(x)+b}\left(1+\frac{b}{f(x)}\right)\leq\frac{1}{f(x)+b}\left(1+\frac{b}{f(a)}\right)$$
and we compare the integrals.

This is false in general. First, we may as well take $a=0$. If $f(x)$ is not positive for $[0,a)$, modify the function and consider $f(x-a)$ instead of $f(x)$. Second, note that it is sufficient to consider a positive function that is discontinuous at a countable set $S$ (but otherwise continuous) but whose limits exist as $x$ approaches $m\in S$. That is, if $h(x)$ is positive discontinuous at $x\in S$ (and only there) with
$$\lim_{x\to m^{+}}h(x)=L_1\text{ and }\lim_{x\to m^{-}}h(x)=L_2$$
existing, we can construct $f(x)$ from $h(x)$ such that $f(x)$ is positive, continuous, and
$$\int_0^\infty \frac{1}{h(x)}dx<\infty \Leftrightarrow \int_0^\infty \frac{1}{f(x)}dx<\infty $$
To do this, let $g(x)=h(x)^{-1}$ and define
$$\beta_m^{+}=\lim_{x\to n^{+}}g(x)\text{ and }\beta_m^{-}=\lim_{x\to n^{-}}g(x)$$
$$\beta_m=\beta_m^{+}-\beta_m^{-}$$
(with $m\in S$). Now, for $m$ in $S$, choose $0<\alpha_m\leq \frac{1}{4}$ small enough such that
$$\int_{m-\alpha_m}^{m+\alpha_m}\left[\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}x+g(m+\alpha_m)-(m+\alpha_m)\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}\right]dx$$
$$\leq \frac{1}{n^2}$$
and
$$\int_{m-\alpha_m}^{m+\alpha_m}g(x)dx\leq \frac{1}{n^2}$$
(where $n$ is the natural number associated with $m$ in the bijection between $S$ and $\mathbb{N}$). To show that this is indeed possible, first note that
$$\lim_{\alpha_m\to 0^{+}}g(m+\alpha_m)=\beta_m^{+}\text{ and }\lim_{\alpha_m\to 0^{+}}g(m-\alpha_m)=\beta_m^{-}$$
This then implies the integral
$$\lim_{\alpha_m\to 0^{+}}\int_{m-\alpha_m}^{m+\alpha_m}\left[\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}x+g(m+\alpha_m)-(m+\alpha_m)\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}\right]dx$$
$$\leq\lim_{\alpha_m\to 0^{+}}\left[\alpha_m (g(m+\alpha_m)-g(m-\alpha_m))+\alpha_m \frac{g(m+\alpha_m)+g(m-\alpha_m)}{2}\right]=0$$
For the second integral, note that
$$\int_{m-\alpha_m}^{m+\alpha_m}g(x)dx\leq 2\alpha_m \max_{x\in [m+1/4,m+1/4]}g(x)$$
which goes to zero as $\alpha_m$ goes to zero. With this, define
$$f(x)=\left\{\begin{matrix}
h(x) && x\not\in N_{\alpha_m}(m)\\
\left[\frac{g(m+\alpha_m)-g(m-\alpha_m}{2\alpha_m}x+g(m+\alpha_m)-(m+\alpha_m)\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}\right]^{-1} && x\in N_{\alpha_m}(m)
\end{matrix}\right.$$
(where $N_{\alpha_m}(m)=(m-\alpha_m,m+\alpha_m)$) for all $m\in S$. Also, for ease of notation let
$$T=\bigcup_{m\in S} N_{\alpha_m}(m)$$
First, it is not to hard to show that $f(x)$ is continuous. This is because the $f(x)$ is equal to $h(x)$ except for an inverse linear function around the discontinuities. Second, note that
$$\infty=\int_T \frac{1}{h(x)}dx+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx=\int_T g(x)dx+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx$$
$$\leq \sum_{n=1}^\infty \frac{1}{n^2}+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx=\frac{\pi^2}{6}+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx$$
This implies
$$\int_{\mathbb{R}/T}\frac{1}{h(x)}dx=\infty$$
Then we get
$$\int_0^\infty \frac{1}{f(x)}dx=\int_T \frac{1}{f(x)}dx+\int_{\mathbb{R}/T}\frac{1}{f(x)}dx$$
$$=\sum_{m\in S}\int_{m-\alpha_m}^{m+\alpha_m}\left[\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}x+g(m+\alpha_m)-(m+\alpha_m)\frac{g(m+\alpha_m)-g(m-\alpha_m)}{2\alpha_m}\right]dx$$
$$+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx\geq  -\sum_{n=1}^\infty\frac{1}{n^2}+\int_{\mathbb{R}/T}\frac{1}{h(x)}dx=\infty$$
We conclude
$$\int_0^\infty \frac{1}{f(x)}dx=\infty$$
If, on the other hand,
$$\int_0^\infty \frac{1}{h(x)}dx<\infty$$
then in a similar manner we can show that
$$\int_0^\infty \frac{1}{f(x)}dx<\infty$$
Having shown that it is sufficient to consider a positive, discontinuous function with direction limits that exist at all discontinuities, define
$$g(x)=\left\{\begin{matrix}
\frac{1}{n}&& x\in [n,n+1/n^2]\\ 
\lfloor x \rfloor ^2+1&&\text{otherwise}
\end{matrix}\right.$$
It is easy to show that $g(x)$ satisfies all the conditions necessary for the work above to apply. Then
$$\int_0^\infty \frac{1}{g(x)}dx=\sum_{n=1}^\infty \frac{1}{g(n)}=\sum_{n=1}^\infty \frac{1}{n}+1+\sum_{n=1}^\infty\frac{1}{n^2+1}\left(1-\frac{1}{n^2}\right)=\infty$$
However, for all $b>0$ we have
$$\int_0^\infty \frac{1}{g(x)+b}dx=\sum_{n=1}^\infty \frac{1}{g(n)+b}$$
$$=\sum_{n=1}^\infty\frac{n}{1+bn}\cdot \frac{1}{n^2} +\frac{1}{1+b}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)$$
$$=\sum_{n=1}^\infty\frac{n}{bn^3+n^2} +\frac{1}{1+b}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)<\infty$$
