TITLE: How to show $\ln x < \int_0^{2\pi (x+\frac{1}{2})} \frac{1- \cos t}{t} dt$?
QUESTION [5 upvotes]: This was after a change of variables from the initial problem to show
$$
\ln x < \int_0^\pi \frac{\sin^2 ([x+\frac{1}{2}] t)}{t/2} dt
$$
I have also tried to write
$$
\ln x = \int_{2\pi}^{2\pi x} \frac{1}{t} dt
$$
And then I was able to show the stronger statement
$$
\ln x < \int_0^{2\pi x} \frac{1- \cos t}{t} dt
$$
By the following
$$
\int_{2\pi}^{2\pi x} \frac{\cos t}{t} dt < \int_0^{2\pi} \frac{1-\cos t}{t} dt
$$
My reasoning is that the LHS is bounded (less than 1, observed graphically) while the RHS is a definite integral ($\approx 2.44$ according to Wolfram Alpha).
Is there a cleaner way to do this? This is a textbook problem, and if it helps I am only asked to prove for the case when $x$ is a positive integer.

REPLY [4 votes]: We only need to prove the case when $x > 1$.
Let $k = \lfloor x \rfloor$.
We have
\begin{align*}
 \int_0^{2\pi x} \frac{1 - \cos t}{t}\mathrm{d} t
 &\ge \int_0^{2\pi k} \frac{1 - \cos t}{t} \mathrm{d} t\\
 &= \sum_{i=0}^{k-1} \int_{2\pi i}^{2\pi (i+1)} \frac{1 - \cos t}{t} \mathrm{d} t\\
 &\ge \sum_{i=0}^{k-1} \int_{2\pi i}^{2\pi (i+1)} \frac{1 - \cos t}{2\pi(i + 1)} \mathrm{d} t\\
 &= \sum_{i=0}^{k-1} \frac{1}{i + 1}\\
 &> \sum_{i=0}^{k-1}\ln \left(1 + \frac{1}{i + 1}\right)\\
 &= \ln (k + 1)\\
 &\ge \ln x
\end{align*}
where we have used $\ln(1 + u) < u$ for all $u > 0$.
We are done.