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]

Question

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?

Answer 1

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.

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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.

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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 $\alpha\ge2$.

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A Family of Counterexamples

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

Answer 2

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 $\phi(x) = x - \lfloor x\rfloor - \frac{1}{2}$.)