MHB Trigonometric absolute value integral

[jacobi1]

Evaluate [math]\lim_{n \to \infty} \int_0^1 | \sin(nx)| \ dx. [/math]

 

[chisigma]

Evaluate [math]\lim_{n \to \infty} \int_0^1 | \sin(nx)| \ dx. [/math]

[sp]Starting from the definite integral...

$\displaystyle \int_{0}^{\frac{\pi}{n}} \sin (n x)\ d x = \frac{2}{n}\ (1)$

... You can conclude that...

$\displaystyle \int_{0}^{1} |\sin (n x)| \ d x \sim \lfloor \frac{n}{\pi} \rfloor\ \frac{2}{n} = \frac{2}{\pi}\ (2)$

... so that the requested limit is $\frac{2}{\pi}$... [/sp]

Kind regards

$\chi$ $\sigma$

 

[Opalg]

[sp]Notice first that $$\int_0^\pi|\sin x|\,dx = \int_0^\pi\sin x\,dx = -\cos\pi + \cos0 = 2.$$ Thus the area under the graph of $|\sin x|$ occupies a fraction $\dfrac2\pi$ of the rectangle $[0,\pi] \times [0,1].$

Since $|\sin x|$ is periodic with period $\pi$, the same result will be true in any rectangle of the form $[x,x+\pi] \times [0,1].$

If we shrink the $x$-axis by a factor $1/n$ then the same argument shows that the area under the graph of $|\sin (nx)|$ occupies a fraction $\dfrac2\pi$ of any rectangle of the form $\bigl[x,x+\frac\pi n\bigr] \times [0,1].$

The unit square can be partitioned as the union of a number of such rectangles, together with a bit left over at the end. As $n$ increases, the area of this leftover bit will decrease to $0$. Thus when $n$ is large, the area under the graph of $|\sin (nx)|$ occupies a fraction close to $\dfrac2\pi$ of the unit square.

Therefore [math]\lim_{n \to \infty} \int_0^1 | \sin(nx)| \, dx = \frac2\pi. [/math]

[That argument could be adapted so as to sound more rigorous. But in this case I think that the informal, geometric approach works best.][/sp]

 

[jacobi1]

Yes, Opalg and chisigma, you are both correct. :)
My solution is slightly different:

Spoiler

The function sin(nx) has many oscillations over [0,1]. For our present purpose, an 'oscillation' is defined as a 'hump' of the sine function above the x axis-where it starts from 0, increases to a maximum, and then decreases back to 0. Finding the area of one of those oscillations, multiplying that area by the number of oscillations, multiplying by two (because the number of 'humps' is the same above and below the x axis) and then taking the limit will give us the answer.
The first [math] x \neq 0[/math] where [math] \sin(nx)=0 [/math] is at [math] x= \frac{\pi}{n}[/math]. Therefore, the area of anyone of the oscillations is

[math] \int_0^{\frac{\pi}{n}} \sin(nx) \ dx = \frac{2}{n} [/math].

The number of oscillations is just the frequency- [math] f = \frac{1}{T}[/math], where T is period. The period of [math] \sin(bx) [/math] is [math] \frac{2 \pi}{b}[/math], and so the frequency of sin(nx) is [math] \frac{n}{2 \pi} [/math]. Putting it all together, we have
[math] 2 \times \frac{2}{n} \times \frac{n}{2 \pi} = \frac{2}{\pi} [/math].