# If $f$ is not continuous on $[1,\infty)$ but $\int_1^{\infty} f(x) \, dx$ converges, does $\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$ converge? [closed]

It is true I think that if $$f$$ is continuous on $$[1,\infty)$$ and $$\int_1^{\infty} f(x) \, dx$$ converges, then $$\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$$ converges.

But what if $$f$$ is not continuous but $$\int_1^{\infty} f(x) \, dx$$ converges. Must it necessarily be true that $$\int_1^{\infty} \frac{|f(x)|}{x^3} \, dx$$ converges?

• I have a feeling the answer is no; my guess is to use some high enough degree rational function. Dec 1, 2020 at 4:16
• @nilradical1 Can you explain what you mean by degree of a rational function or provide an example? Dec 1, 2020 at 4:28
• Sorry, for polynomials $p(x)$ and $q(x)$, the degree of the rational function $p(x)/q(x)$ is $\max\{\deg p(x), \deg q(x)\}$. Dec 1, 2020 at 4:32

The Statement is False Even in the Continuous Case

If $$f(x)=x^3\sin\left(x^5\right)$$, then \begin{align} \int_1^\infty x^3\sin\left(x^5\right)\,\mathrm{d}x &=\frac15\int_1^\infty x^{-1/5}\sin(x)\,\mathrm{d}x\tag{1a}\\ &=\frac15\cos(1)-\frac1{25}\int_1^\infty x^{-6/5}\cos(x)\,\mathrm{d}x\tag{1b} \end{align} Explanation:
$$\text{(1a)}$$: substitute $$x\mapsto x^{1/5}$$
$$\text{(1b)}$$: integrate by parts

The integral in $$(1)$$ converges, but \begin{align} \int_1^\infty\frac{\left|\,x^3\sin\left(x^5\right)\,\right|}{x^3}\,\mathrm{d}x &=\frac15\int_1^\infty x^{-4/5}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{2a}\\ &\ge\frac15\sum_{k=2}^\infty\int_{(k-1)\pi}^{k\pi}x^{-4/5}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{2b}\\ &\ge\frac15\sum_{k=2}^\infty2(k\pi)^{-4/5}\tag{2c}\\ &=\frac25\pi^{-4/5}\sum_{k=2}^\infty k^{-4/5}\tag{2d} \end{align} Explanation:
$$\text{(2a)}$$: substitute $$x\mapsto x^{1/5}$$
$$\text{(2b)}$$: remove the integral on $$[1,\pi)$$ and
$$\phantom{\text{(2b):}}$$ break up the integral on $$[\pi,\infty)$$ into pieces
$$\text{(2c)}$$: $$\int_{(k-1)\pi}^{k\pi}|\!\sin(x)|\,\mathrm{d}x=2$$
$$\phantom{\text{(2c):}}$$ $$x^{-4/5}\ge(k\pi)^{-4/5}$$ on $$[(k-1)\pi,k\pi]$$
$$\text{(2d)}$$: pull common factors out front

the integral in $$(2)$$ does not converge.

Motivation for $$\boldsymbol{x^3\sin\left(x^5\right)}$$

Since $$|f(x)|$$ is being divided by $$x^3$$, we want the function to grow at least as fast as $$x^3$$; thus, the factor of $$x^3$$.

The average of $$|\!\sin(x)|$$ over each period is $$\frac2\pi$$, so the integral of $$\left|\sin\left(x^n\right)\right|$$ over a large interval should be approximately $$\frac2\pi$$ times the size of the interval.

The integral of $$\sin(x)$$ over each period is $$0$$, but when multiplied by $$x^3$$, the integral oscillates between large and small values $$\int x^3\sin(x)\,\mathrm{d}x =-x^3\cos(x)+3x^2\sin(x)+6x\cos(x)-6\sin(x)+C\tag3$$ If we accelerate the oscillation by using $$\sin\left(x^n\right)$$, the integral doesn't have time to get big before being cancelled. Letting $$u=x^n$$, we have $$\int x^3\sin\left(x^n\right)\,\mathrm{d}x=\frac1n\int u^{\frac4n-1}\sin(u)\,\mathrm{d}u\tag4$$ To get convergence on the right side, we need $$n\gt4$$, so $$n=5$$ seems a good choice.

An Extension to the Preceding

Consider $$f(x)=x^\alpha\sin\left(x^{\alpha+\beta}\right)$$. For $$\beta\gt1$$ and $$\alpha\gt-\beta$$, \begin{align} \int_1^\infty f(x)\,\mathrm{d}x &=\int_1^\infty x^\alpha\sin\left(x^{\alpha+\beta}\right)\,\mathrm{d}x\tag{5a}\\ &=\frac1{\alpha+\beta}\int_1^\infty x^{-\frac{\beta-1}{\alpha+\beta}}\sin(x)\,\mathrm{d}x\tag{5b}\\ &=\frac1{\alpha+\beta}\cos(1)-\frac{\beta-1}{(\alpha+\beta)^2}\int_1^\infty x^{-\frac{\beta-1}{\alpha+\beta}-1}\cos(x)\,\mathrm{d}x\tag{5c} \end{align} Explanation:
$$\text{(5a)}$$: definition
$$\text{(5b)}$$: substitute $$x\mapsto x^{\frac1{\alpha+\beta}}$$
$$\text{(5c)}$$: integrate by parts

The integral in $$\text{(5a)}$$ converges absolutely for $$\alpha\lt-1$$ and the integral in $$\text{(5c)}$$ converges for $$\beta\gt1$$ and $$\alpha\ge-1$$. That is, the integral in $$(1)$$ converges for $$\beta\gt1$$ and all $$\alpha$$.

However, \begin{align} \int_1^\infty\frac{|f(x)|}{x^3}\,\mathrm{d}x &=\int_1^\infty\frac{\left|\,x^\alpha\sin\left(x^{\alpha+\beta}\right)\,\right|}{x^3}\,\mathrm{d}x\tag{6a}\\ &=\frac1{\alpha+\beta}\int_1^\infty x^{-\frac{\beta+2}{\alpha+\beta}}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{6b}\\ &\ge\frac1{\alpha+\beta}\sum_{k=2}^\infty\int_{(k-1)\pi}^{k\pi}x^{-\frac{\beta+2}{\alpha+\beta}}\left|\,\sin(x)\,\right|\mathrm{d}x\tag{6c}\\ &\ge\frac1{\alpha+\beta}\sum_{k=2}^\infty2(k\pi)^{-\frac{\beta+2}{\alpha+\beta}}\tag{6d}\\ &=\frac2{\alpha+\beta}\pi^{-\frac{\beta+2}{\alpha+\beta}}\sum_{k=2}^\infty k^{-\frac{\beta+2}{\alpha+\beta}}\tag{6e} \end{align} Explanation:
$$\text{(6a)}$$: definition
$$\text{(6b)}$$: substitute $$x\mapsto x^{\frac1{\alpha+\beta}}$$
$$\text{(6c)}$$: remove the integral on $$[1,\pi)$$ and
$$\phantom{\text{(6c):}}$$ break up the integral on $$[\pi,\infty)$$ into pieces
$$\text{(6d)}$$: $$\int_{(k-1)\pi}^{k\pi}|\!\sin(x)|\,\mathrm{d}x=2$$
$$\phantom{\text{(6d):}}$$ $$x^{-\frac{\beta+2}{\alpha+\beta}}\ge(k\pi)^{-\frac{\beta+2}{\alpha+\beta}}$$ on $$[(k-1)\pi,k\pi]$$
$$\text{(6e)}$$: pull common factors out front

for $$\beta\gt1$$, the sum in $$\text{(6e)}$$ does not converge for $$\alpha\ge2$$.

A Family of Counterexamples

Thus, $$f(x)=x^\alpha\sin\left(x^{\alpha+\beta}\right)\tag7$$ is a counterexample for $$\alpha\ge2$$ and $$\beta\gt1$$.

• What's the intuition behind picking such a function? I mean how did you come up with that? Dec 1, 2020 at 11:34
– robjohn
Dec 1, 2020 at 12:26
• Can you elaborate on the coming up with the second inequality where you go from $x^{-4/5}|\sin(x)|$ to $2(k\pi)^{-4/5}$? I think the 2 comes from $2/\pi$ times the length of each interval which is $\pi$ but I don't see the $(k\pi)^{-4/5}$ and the "$\geq$" part. Dec 1, 2020 at 15:20
– robjohn
Dec 1, 2020 at 15:35

Both statements are false. Consider $$f(x) = e^{x/2} \sin(e^x)$$. Then

$$\int_{1}^{\infty} f(x) \, \mathrm{d}x \stackrel{(u=e^x)}{=} \int_{e}^{\infty} \frac{\sin u}{\sqrt{u}} \, \mathrm{d}u = \frac{\cos e}{\sqrt{e}} - \frac{1}{2} \int_{e}^{\infty} \frac{\cos u}{u^{3/2}} \, \mathrm{d}u$$

converges. On the other hand,

$$\int_{1}^{\infty} \frac{\left| f(x) \right|}{x^3} \, \mathrm{d}x \stackrel{(u=e^x)}{=} \int_{e}^{\infty} \frac{\left| \sin u \right|}{\sqrt{u} \log^3 u} \, \mathrm{d}u$$

and it is not hard to show that this integral diverges.

Almost the same proof goes if we set $$f(x) = e^{x/2} \phi(e^x)$$, where $$\phi$$ is a non-zero $$T$$-periodic function such that $$\int_{0}^{T} \phi(x) \, \mathrm{d}x = 0$$. (The above proof used $$\phi(x)=\sin x$$, but there is a plethora of other examples, such as $$\phi(x) = x - \lfloor x\rfloor - \frac{1}{2}$$.)