How can I prove that the length of $y=x\sin(π/x)$ in $(0,1]$ is infinite?

Question

How can I prove that the length of $y=x\sin(π/x)$ in $(0,1]$ is infinite? Is evaluating sum of length of straight line and use comparison test right?

Answer 1

The arc-length formula says the length is $\displaystyle \int_0^{1}\sqrt{1+\left (\sin\left(\frac{π}{x}\right)-\frac{π}{x}\cos\left(\frac{π}{x}\right)\right)^2}\,dx$.

Can you prove that this doesn't converge, by comparing it to a simpler integral?

(A simpler integral would be the integrand without the $1+$ so just $\int|y'|dx.)$

Answer 2

HINT.-Take the sequence $x_n=\dfrac{2}{2n+1}$ so you do have $y(x_n)=\dfrac{2}{2n+1}\sin\dfrac{(2n+1)\pi}{2}=\pm\dfrac{2}{2n+1}$.

It follows

$$\int_{\frac{2}{2(n+1)+1}}^{\frac{2}{2n+1}}ds\gt\dfrac{2}{2n+1}+\dfrac{2}{2(n+1)+1}\gt\dfrac{2}{2n+1}$$because the arc of the curve between the two points $x_1=\frac{2}{2n+1}$ and $x_2=\frac{2}{2(n+1)+1}$ is longer than $|f (x_1)|+ |f (x_2)|$.

Now the series $$\sum^{\infty}_2\dfrac{2}{2n+1}=\sum^{\infty}_2\dfrac{1}{n+0.5}$$ diverges (calculate the integral $\int_2^{\infty}\frac{1}{n+0.5}$ or compare with the harmonic series).