How to show $\ln x < \int_0^{2\pi (x+\frac{1}{2})} \frac{1- \cos t}{t} dt$?

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.

• Your first line has an integrand with $t$ in the numerator and $t$ in the denominator. Should the numerator be $\sin^2(tx+t/2)$? Commented Feb 27, 2022 at 17:13
• @DionelJaime What you said is right. I have added brackets to avoid confusion.
– Bio
Commented Feb 27, 2022 at 17:19

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 $$u > 0$$.