# Show that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|\ dx$ diverges

I read that the improper Riemann integral $$\int_0^1 \Bigg|\frac{1}{x}\sin\frac{1}{x}\Bigg|\ dx$$ diverges.

I have tried comparison criteria for $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx$, but I cannot find a function $f$ with a divergent integral such that $0\leq f(x)\leq|\frac{1}{x}\sin\frac{1}{x}|$. I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral. I have also tried to think about some use of complex analysis, but found none useful to prove the desired result. Coud anybody give a proof of this divergence? $\infty$ thanks!

• Using the change-of-variables version, you can argue in a manner similar to what I did in my answer here. – David Mitra Oct 26 '14 at 12:25

I also notice, by using a change of variable, that $\int_0^1 |\frac{1}{x}\sin\frac{1}{x}|dx=\int_1^\infty |\frac{1}{x}\sin x|dx$, but I have not found a use of this equality to prove the divergence of the integral.

Subdivide $(1,\infty)$ into an infinite number of intervals of the form $\Big(k\pi,~(k+1)\pi\Big)$, and notice that for each of them, their graphic is tangent to a hyperbola, and that the area of each slice is greater than that of a triangle of height $\dfrac1{\pi~\bigg(k+\dfrac12\bigg)}$, due to the convexity/concavity of the sine function. This shows that the value of our integral is greater than a constant multiple of the harmonic series, which is known to be divergent.

• @everybody: Thank you so much!!! Lucian: excuse me, but I'm not able to see that the sum of the areas of the triangles divergers, nor that the maxima, satisfying $x\cos x=\sin x$ lie on a hyperbola... :-( $\infty$ thanks again! – Self-teaching worker Oct 26 '14 at 13:38
• @DavideZena: Plot $\dfrac{\sin}x$ and $\dfrac1x$ onto the same graphic, and tell me what you notice. – Lucian Oct 26 '14 at 13:43
• Thank you again! I know that $|\frac{\sin x}{x}|\leq\frac{1}{|x|}$ and, by plotting, I notice that the graph is tangent to the hyperbola, although I wouldn't be able to prove it rigourously... – Self-teaching worker Oct 26 '14 at 13:53
• @DavideZena: The maxima don't lie on the hyperbola $($see the edited version$)$, nor is this relevant. All that matters is that each triangle lies completely inside the graphic, which is a direct implication of the function's concavity on each such interval. – Lucian Oct 26 '14 at 13:59
• And $\sum_{k=1}^\infty\frac{k}{2\pi(k+1/2)}$ diverges. Thank you so much!!! – Self-teaching worker Oct 26 '14 at 14:03

Hint: Calculate the integral only for those intervals where $\sin 1/x ≥ 1/2$.

Performing the changes of variables $x= 1/u$ you get

$$\int_0^1 \bigg|\frac{1}{x}\sin\frac{1}{x}\bigg|\ dx=\int_1^\infty \bigg|\frac{\sin x}{x}\ \bigg| dx\ge \int_π^\infty \Bigg|\frac{\sin x}{x}\ \bigg|\\= \lim_{n\to \infty}\int_π^{nπ} \bigg|\frac{\sin x}{x}\ \bigg|dx= \sum_{n=1}^{\infty}\int_{nπ}^{(n+1)π} \bigg|\frac{\sin x}{x}\ \bigg|dx\\ \ge \sum_{n=1}^{\infty}\frac{1}{ (n+1)π} \int_{nπ}^{(n+1)π} |\sin x|dx =\color{red}{\sum_{n=1}^{\infty}\frac{2}{ (n+1)π}=\infty}$$

indeed for all $n$ $$\color{blue}{\int_{nπ}^{(n+1)π} |\sin x|dx=2}$$