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

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?

\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}