Comments on Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$?
Parent
Why is $ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty$?
Question: Why is $$ \int_0^{\infty}\left|\frac{\sin x}{x}\right|\ dx=\infty\quad ? $$
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 of improper integrals that are convergent but not absolutely convergent.
Post
Since, $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$, the singularity of the integral is not at $x=0$. On the other hand, one can rewrite the integral as $ \int_{0}^\infty\frac{1}{x}\cdot |\sin(x)|\ dx. $ It suffices to analyze the sum $$ \int_0^\pi\frac{\sin(x)}{x}\ dx+\int_{\pi}^{N\pi}\frac{1}{x}\cdot |\sin(x)|\ dx. $$ and show that it goes to $\infty$ as $N\to\infty$. Why integrate up to $N\pi$ instead of $N$? Well, partly due to technical convenience: the function $|\sin(x)|$ has the period of $\pi$. The graph of the integrand suggests that we may estimate the integral over each period and then add them together.
As mentioned earlier, one can write the first integral as one for the continuous function. So we focus on the second one.
On the interval $[\pi,2\pi]$, one has $$ \int_\pi^{2\pi}\frac1x|\sin(x)|\ dx\ge \frac1{2\pi}\int_\pi^{2\pi}|\sin(x)|\ dx= \frac{1}{2\pi}\cdot 2 $$ where we use the simple fact that $\int_a^b f(x)\ dx\ge \int_a^b g(x)\ dx$ if $f(x)\ge g(x)$ on $[a,b]$. We work with the lower bounds because in order to show something diverges, big lower bounds would help.
Similarly, on the interval, $[2\pi,3\pi]$, we have $$ \int_{2\pi}^{3\pi}\frac1x|\sin(x)|\ dx\ge \frac1{3\pi}\int_{2\pi}^{3\pi}|\sin(x)|\ dx= \frac{1}{3\pi}\cdot 2 $$
and in general, $$ \int_{(n-1)\pi}^{n\pi}\frac1x|\sin(x)|\ dx\ge \frac1{n\pi}\int_{(n-1)\pi}^{n\pi}|\sin(x)|\ dx = \frac{1}{n\pi}\cdot 2 $$
By adding the estimates together and using the fact that the harmonic series diverges, we have the desired proof.
1 comment thread
0 comment threads