# Prove that for every $T > \frac{\pi}{2}$, $\int_{\frac{\pi}{2}}^T \frac{cos(x)}{x}dx < 0$

I tried doing integration by parts a few times, after doing it 3 times I get the following expression: $$\int_{\frac{\pi}{2}}^T \frac{cos(x)}{x}dx = \frac{sin(T)}{T} - \frac{1}{\frac{\pi}{2}} - \frac{cos(T)}{T^2} - 2*\frac{sin(T)}{T^3} + \frac{2}{\frac{\pi^3}{8}} - 6*\int_{\frac{\pi}{2}}^T \frac{sin(x)}{x^4}dx$$

From here it seems obvious, the non integral terms keep getting smaller, while the first two terms are smaller than 0. And if I theoretically keep doing integration by parts to infinity, the function inside the integral is a function sequence so that: f_n(x) = \left\{\begin{aligned} &\frac{cos(x)}{x^n} &&: n :even\\ &\frac{sin(x)}{x^n} &&: n :odd \end{aligned} \right.

Which obviously converges to 0 uniformly, so $\lim_{n \rightarrow \infty }\int_{\frac{\pi}{2}}^T f_n(x)dx = 0$

That's the idea at least. I'm not entirely sure how to explain this formally. This whole expression looks like a series, but I can't see the rule for it, but even so, I have a feeling it will look terrible as a series. I think I might have just over complicated this whole thing, but it seems the most straightforward for me. Any help will be greatly appreciated.

$\cos x$ is negative on $[\pi/2,3\,\pi/2]$ and positive on $[3\,\pi/2,5\,\pi/2]$. $$\Bigl|\int_{\pi/2}^{3\pi/2}\frac{\cos x}{x}\,dx\Bigr|=\int_{\pi/2}^{3\pi/2}\frac{|\cos x|}{x}\,dx\ge\frac{2}{3\,\pi}\int_{\pi/2}^{3\pi/2}|\cos x|\,dx=\frac{4}{3\,\pi}\tag{1}$$ $$\int_{3\pi/2}^{5\pi/2}\frac{\cos x}{x}\,dx\le\frac{2}{3\,\pi}\int_{3\pi/2}^{5\pi/2}\cos x\,dx=\frac{4}{3\,\pi}\tag{2}$$ It follows from (1) and (2) that $$\int_{\pi/2}^{T}\frac{\cos x}{x}\,dx\le0,\quad \frac{\pi}{2}\le x\le\frac{5\,\pi}{2}.$$ The argument can be repeated on intervals $[(4\,k+1)\,\pi/2,(4\,k+5)\,\pi/2]$, $k\ge1$.
• Thank you for your answer, this seems quite simple. Though just one last question. The reason the argument can be applied on the other intervals is because of $cosx$ periodicity of $2\pi$ right? – Xsy Jan 27 '15 at 19:20
• Yes. You can repeat the argument with $4k+1$ and $4k+5$ instad of 1 and 5. – Julián Aguirre Jan 27 '15 at 21:12