Proving that $\int^1_0 \frac{1}{x}cos(\frac{1}{x})dx$ is conditionally convergent.

I think that the first part, proving that $\int^1_0 \frac{1}{x}cos(\frac{1}{x})dx$ converges is relatively easy to do.

Letting $u=\frac{1}{x}$, the integral becomes,

$\int^\infty_1\frac{cosu}{u}du$=$\frac{sinu}{u}|^M_1+\int^\infty_1\frac{sinu}{u^2}du$=$\frac{sinM}{M}-sin1$+k, where k is a constant as we know that $\int^\infty_1\frac{sinu}{u^2}du <\int^\infty_1\frac{1}{u^2}du $ which is an absolutely convegent sum. As M goes to infinity, the sum goes to $k-sin1$, so the integral is convergent.

But I'm stuck at the part where I have to prove conditional convergence, ie. $\int^1_0 |\frac{1}{x}cos(\frac{1}{x})|dx$ is divergent. This is what I've tried so far,

$\int^1_0 |\frac{1}{x}cos(\frac{1}{x})|dx$=$\int^\infty_1|\frac{cosu}{u}|du$=$\sum^\infty_{n=1}\int^{n\pi+\pi}_{n\pi}|(-1)^n\frac{cosu} {u}|du+\int^\pi_1\frac{cosu}{u}du$

The lone term we know exists, so we can ignore it and focus on proving that the complicated summation is divergent. Letting $u=v+n\pi$,

$\sum^\infty_{n=1}\int^{n\pi+\pi}_{n\pi}\frac{cosu}{u}du$= $\sum^\infty_{n=1}\int^{\pi}_{0}\frac{cos(v+n\pi)}{v+n\pi}dv$= $\sum^\infty_{n=1}\int^{\pi}_{0}\frac{cos(v)}{v+n\pi}dv$

For 0≤v≤$\pi$, $v+n\pi≤\pi+n\pi$

$\int^{\pi}_{0}\frac{cos(v)}{v+n\pi}dv$ > $\int^{\pi}_{0}\frac{cos(v)}{\pi+n\pi}dv$=0

Here's where I don't know if my reasoning is correct. Does this mean that I can find some $\epsilon$>0 such that $\int^{\pi}_{0}\frac{cos(v)}{v+n\pi}dv$=$\epsilon$>0? If that is so, does that mean that $\sum^\infty_{n=1}\int^{\pi}_{0}\frac{cos(v)}{v+n\pi}dv$=$\sum^\infty_{n=1}\epsilon=\infty$, thus proving that the series is conditionally convergent?

Comments appreciated!


$$\begin{align} \int^\infty_1\frac{|\cos u|}{u}du &\ge \int^\infty_1\frac{\cos^2 u}u du \\ &=\int^\infty_1\frac{1+\cos 2u}{2u}du \\ &=\underbrace{\int^\infty_1\frac{1}{2u}du}_{=\infty}+\underbrace{\int^\infty_1\frac{\cos 2u}{2u}du}_{\text{convergent}}\\ \end{align} $$

  • $\begingroup$ Hmm what a pretty way to show it! $\endgroup$ – Yip Jung Hon Oct 28 '18 at 14:09
  • $\begingroup$ @YipJungHon Please accept this answer if you think it is useful. $\endgroup$ – Szeto Oct 28 '18 at 14:10

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