Divergent integrals $\int_{a}^{\infty}\frac{dx}{f(x)}$ and $\int_{a}^{\infty}\frac{dx}{f(x)+b}$ for positive $f$ and $b$

Question

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.

Answer 1

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

$$\int_0^\infty \frac{1}{g(x)+b}dx=\sum_{n=1}^\infty \frac{1}{g(n)+b}$$

$$=\sum_{n=1}^\infty\frac{n}{1+bn}\cdot \frac{1}{n^2} +\frac{1}{1+b}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)$$

$$=\sum_{n=1}^\infty\frac{n}{bn^3+n^2} +\frac{1}{1+b}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)<\infty$$

Answer 2

QC_QAQA's answer with modifications, for future reference. We may assume that $a=0$, since for $$\tilde{f}(x)=f(x-a),\ x\in[a,\infty),$$ we have $$\int_{a}^{\infty}\frac{dx}{\tilde{f}(x)}=\lim_{z\to \infty}\int_{a}^{z}\frac{dx}{f(x-a)}=\lim_{z\to\infty}\int_{0}^{z-a}\frac{dy}{f(y)}=\int_{0}^{\infty}\frac{dx}{f(x)},$$ $$\int_{a}^{\infty}\frac{dx}{\tilde{f}(x)+b}=\lim_{z\to \infty}\int_{a}^{z}\frac{dx}{f(x-a)+b}=\lim_{z\to\infty}\int_{0}^{z-a}\frac{dy}{f(y)+b}=\int_{0}^{\infty}\frac{dx}{f(x)+b}.$$ We will give an example of a positive function that is discontinuous at an infinite countable set $S$ with no accumulation point in $[0,\infty)$ (but otherwise continuous), but whose limits exist as $x$ approaches $m\in S$, that is, $h\colon[0,\infty)\to(0,\infty)$ is continuous in $[0,\infty)\setminus S$, $S\subset(0,\infty)$ is infinite countable with no accumulation point in $[0,\infty)$ and for any $m\in S$ there exist $L_{m^{-}}>0$ and $L_{m^{+}}>0$ such that $$\lim_{x\to m^{-}}h(x)=L_{m^{-}}\text{ and }\lim_{x\to m^{+}}h(x)=L_{m^{+}},$$ which satisfies $$\int_{0}^{\infty}\frac{dx}{h(x)}=\infty\text{ and for any }b>0\text{ we have }\int_{0}^{\infty}\frac{dx}{h(x)+b}<\infty.$$ Let us consider the function $$h(x)=\begin{cases} \frac{1}{n},&\ x\in [n,n+1/n^2),\ n\in\mathbb{N},\\ \lfloor x \rfloor ^2+1,&\ \text{otherwise}. \end{cases}$$ Note that $h\colon[0,\infty)\to(0,\infty)$, $h$ is continuous in $[0,\infty)\setminus S$ with $$S=\{n\colon n\geq 2\}\cup\{n+\frac{1}{n^2}\colon n\geq 2\}=S_1\cup S_2$$ and for any $m=n\in S_1$ we have $$\lim_{x\to m^{-}}h(x)=(n-1)^2+1\text{ and }\lim_{x\to m^{+}}h(x)=\frac{1}{n}$$ and for any $m=n+\frac{1}{n^2}\in S_2$ we have $$\lim_{x\to m^{-}}h(x)=\frac{1}{n}\text{ and }\lim_{x\to m^{+}}h(x)=n^2+1.$$ Moreover, we have $$\int_0^\infty \frac{dx}{h(x)}=\sum_{n=0}^\infty \int_{n}^{n+1}\frac{dx}{h(x)}=1+\sum_{n=1}^\infty\left(\int_{n}^{n+\frac{1}{n^2}}\frac{dx}{h(x)}+\int_{n+\frac{1}{n^2}}^{n+1}\frac{dx}{h(x)}\right)$$ $$=1+\sum_{n=1}^{\infty}n\cdot\frac{1}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2+1}\left(1-\frac{1}{n^2}\right)=\infty.$$ However, for any $b>0$ we get $$\int_0^\infty \frac{dx}{h(x)+b}=\sum_{n=0}^\infty\int_{n}^{n+1}\frac{dx}{h(x)+b}=\frac{1}{1+b}+\sum_{n=1}^\infty\frac{n}{1+bn}\cdot\frac{1}{n^2}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)$$ $$=\frac{1}{1+b}+\sum_{n=1}^\infty\frac{1}{bn^2+n}+\sum_{n=1}^\infty\frac{1}{n^2+1+b}\left(1-\frac{1}{n^2}\right)<\infty.$$ Finally, using the above function $h\colon[0,\infty)\to(0,\infty)$, we construct a continuous function $f\colon[0,\infty)\to (0,\infty)$ such that $$\int_{0}^{\infty}\frac{dx}{f(x)}=\infty\text{ and for any }b>0\text{ we have }\int_{0}^{\infty}\frac{dx}{f(x)+b}<\infty.$$

To this end, define $g(x)=\frac{1}{h(x)}$, $x\in [0,\infty)$, which is continuous in $[0,\infty)\setminus S$, and for $m\in S$ note that $$\lim_{x\to m^{-}}g(x)=\frac{1}{L_{m^{-}}}\text{ and }\lim_{x\to m^{+}}g(x)=\frac{1}{L_{m^{+}}}.$$ Since $m\in S$ is not an accumulation point of $S$, there exists $a_{m}>0$ such that $$[m-a_m,m+a_{m}]\subset [0,\infty)\text{ and }[m-a_m,m+a_{m}]\cap S=\{m\}.$$ In particular, there exists $0<M_{m}<\infty$ such that $$\max_{x\in[m-a_{m},m+a_{m}]}g(x)=M_{m}.$$ For $m\in S$ and $0<\varepsilon<a_{m}$ we connect $(m-\varepsilon,g(m-\varepsilon))$ and $(m+\varepsilon,g(m+\varepsilon))$ by a~straight line and integrate it over $[m-a_{m},m+a_{m}]$. We observe that $$\int_{m-\varepsilon}^{m+\varepsilon}\left[\frac{g(m+\varepsilon)-g(m-\varepsilon)}{2\varepsilon}x+g(m+\varepsilon)-(m+\varepsilon)\frac{g(m+\varepsilon)-g(m-\varepsilon)}{2\varepsilon}\right]dx$$ $$\leq\varepsilon |g(m+\varepsilon)-g(m-\varepsilon)|+\varepsilon(g(m+\varepsilon)+g(m-\varepsilon))\to 0\text{ as }\varepsilon\to 0^{+},$$ where the first is the area of a right triangle with catheti of length $|g(m+\varepsilon)-g(m-\varepsilon)|$ and $2\varepsilon$ and the other is the area of the rectangle with sides of length $\frac{g(m+\varepsilon)+g(m-\varepsilon)}{2}$ and $2\varepsilon$. We also note that $$\int_{m-\varepsilon}^{m+\varepsilon}g(x)dx\leq 2\varepsilon \max_{x\in [m+a_m,m+a_{m}]}g(x)=2\varepsilon M_{m}\to 0\text{ as }\varepsilon\to 0^{+}.$$ Let $S=\{m_{n}\colon n\in\mathbb{N}\}$ and choose $0<\alpha_{n}<a_{m_n}$ such that $$\int_{m_n-\alpha_n}^{m_n+\alpha_n}\!\bigg[\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}x+g(m_n+\alpha_n)-(m_n+\alpha_n)\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}\bigg]dx\leq\frac{1}{n^2}$$ and $$\int_{m_n-\alpha_n}^{m_n+\alpha_n}g(x)dx\leq \frac{1}{n^2}.$$ Let $N_{n}=[m_n-\alpha_n,m_n+\alpha_n]$, $n\in\mathbb{N}$, and define $$f(x)=\begin{cases} h(x),&x\not\in N_{n},\\ 1/\left[\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}x+g(m_n+\alpha_n)-(m_n+\alpha_n)\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}\right],&x\in N_{n}. \end{cases}$$ First note that $f$ is positive and $f$ is continuous, since $f$ is equal to $h$ except for the reciprocal of a~linear positive function around discontinuities, which connects $(m_n-\alpha_n,h(m_n-\alpha_n))$ and $(m_n+\alpha_n,h(m_n+\alpha_n))$.

For the ease of notation, let $T=\bigcup_{n\in\mathbb{N}} N_{n}$ and observe that since $$\infty=\int_{0}^{\infty}\frac{dx}{h(x)}=\int_T \frac{dx}{h(x)}+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)}=\int_T g(x)dx+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)}$$ $$\leq \sum_{n=1}^\infty \frac{1}{n^2}+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)}=\frac{\pi^2}{6}+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)},$$ it follows that $$\int_{[0,\infty)\setminus T}\frac{dx}{h(x)}=\infty.$$ Since $$\int_0^\infty \frac{dx}{f(x)}=\int_T \frac{dx}{f(x)}+\int_{[0,\infty)\setminus T}\frac{dx}{f(x)}=\int_T \frac{dx}{f(x)}+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)}\geq \int_{[0,\infty)\setminus T}\frac{dx}{h(x)},$$ we conclude that $$\int_{0}^{\infty}\frac{dx}{f(x)}=\infty.$$ Now we have to show that $\int_{0}^{\infty}\frac{dx}{f(x)+b}<\infty$ for any $b>0$. Observe that $$\int_T \frac{dx}{f(x)+b}\leq\int_{T}\frac{dx}{f(x)}\leq\sum_{n=1}^{\infty}\int_{m_n-\alpha_n}^{m_n+\alpha_n}\bigg[\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}x$$ $$+g(m_n+\alpha_n)-(m_n+\alpha_n)\frac{g(m_n+\alpha_n)-g(m_n-\alpha_n)}{2\alpha_n}\bigg]dx\leq\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$$ Hence we obtain $$\int_0^\infty \frac{dx}{f(x)+b}=\int_T \frac{dx}{f(x)+b}+\int_{[0,\infty)\setminus T}\frac{dx}{f(x)+b}=\int_T \frac{dx}{f(x)+b}+\int_{[0,\infty)\setminus T}\frac{dx}{h(x)+b}$$ $$\leq\frac{\pi^{2}}{6}+\int_{0}^{\infty}\frac{dx}{h(x)+b}<\infty.$$ If $h\in C^{k}([0,\infty)\setminus S)$ (as above), then instead of joining points $(m-\varepsilon,\frac{1}{h(m-\varepsilon)})$ and $(m+\varepsilon,\frac{1}{h(m+\varepsilon)})$ by a straight line, we correct $g(x)=\frac{1}{h(x)}$ to the function of class $C^{k}$ in such a way that the graph of the correction lies in the rectangle $[m-\varepsilon,m+\varepsilon]\times[0,\frac{1}{h(m-\varepsilon)}+\frac{1}{h(m+\varepsilon)}]$. Then $f$ is also $C^{k}$.

Answer 3

The opposite would be true: since $$ f+b >f \Rightarrow \frac{1}{f+b} < \frac{1}{f} \rightarrow\int\frac{dx}{f+b} \to \infty \Rightarrow\int \frac{dx}{f} \to \infty $$