Post History
#3: Post edited
by
Snoopy
·
2024-01-07T16:03:35Z (11 months ago)
fixed typos
- > **Question**: Why is
- $$
- \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ?
- $$
- There are several other ways to state the fact in the question depending on the contexts. For examples:
- The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not *absolutely integrable* on $[0,\infty]$.- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not in $L^1([0,\infty))$.
- The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable on $[0,\infty]$.- I will share my own answer below, and I would like to see other approaches if there are any.
- **Notes:** This is an [example](https://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals) of improper integrals that are convergent but not absolutely convergent.
- > **Question**: Why is
- $$
- \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ?
- $$
- There are several other ways to state the fact in the question depending on the contexts. For examples:
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not *absolutely integrable* on $[0,\infty)$.
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not in $L^1([0,\infty))$.
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable on $[0,\infty)$.
- I will share my own answer below, and I would like to see other approaches if there are any.
- **Notes:** This is an [example](https://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals) of improper integrals that are convergent but not absolutely convergent.
#2: Post edited
by
Snoopy
·
2024-01-07T16:03:01Z (11 months ago)
added more
> **Question**: Why is- $$
- \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ?
- $$
An alternative way to phrase this equation is why the function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not absolutely integrable on $[0,\infty]$.- I will share my own answer below, and I would like to see other approaches if there are any.
- **Notes:** This is an [example](https://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals) of improper integrals that are convergent but not absolutely convergent.
- > **Question**: Why is
- $$
- \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ?
- $$
- There are several other ways to state the fact in the question depending on the contexts. For examples:
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not *absolutely integrable* on $[0,\infty]$.
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not in $L^1([0,\infty))$.
- - The function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable on $[0,\infty]$.
- I will share my own answer below, and I would like to see other approaches if there are any.
- **Notes:** This is an [example](https://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals) of improper integrals that are convergent but not absolutely convergent.
#1: Initial revision
by
Snoopy
·
2024-01-07T15:51:57Z (11 months ago)
Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$?
> **Question**: Why is
$$
\int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ?
$$
An alternative way to phrase this equation is why the function $\displaystyle f(x)=\frac{\sin(x)}{x}$ is not absolutely integrable on $[0,\infty]$.
I will share my own answer below, and I would like to see other approaches if there are any.
**Notes:** This is an [example](https://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals) of improper integrals that are convergent but not absolutely convergent.