Find the sign of $\int_{0}^{2 \pi}\frac{\sin x}{x} dx$ I'd love your help with finding the sign of the following integral: $$\int_{0}^{2 \pi}\frac{\sin x}{x} dx$$
I know that computing it is impossible.  I tried to use integration by parts and maybe to learn about the sign of each part and conclude something but It didn't work for me.
Any suggestions?
 A: \begin{align*}
\int_0^{2\pi}\frac{\sin x}{x}\,dx&=\int_0^{\pi}\frac{\sin x}{x}\,dx+\int_\pi^{2\pi}\frac{\sin x}{x}\,dx\\
&=\int_0^{\pi}\frac{\sin x}{x}\,dx+\int_0^{\pi}\frac{\sin(x+\pi)}{x+\pi}\,dx\\
&=\int_0^{\pi}\Bigl(\frac{1}{x}-\frac{1}{x+\pi}\Bigr)\sin x\,dx\\
&=\pi\int_0^{\pi}\frac{\sin x}{x(x+\pi)}\,dx\\
&>0
\end{align*}
A: Regarding the sign, it is easy the check that every area in each $\pi$ interval is always smaller than the preceeding one. The sign is positive.
For the value, integrate in the same interval 
$$y = \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$$
The difference between the sinc function and that is at most $\approx 0.015$ in that interval.
Adding $$\cos \frac{x}{16}$$ makes the error at most $\approx 0.003$
For the first one you have.
$$I = \frac{104}{105} \sqrt{2} \sim\sqrt{2} $$
For the latter:
$$I = \frac{{1568}}{{2145}}\frac{{\sqrt {2 + \sqrt 2 } }}{2} \sim \sqrt{2}$$
Maybe the area is $\sqrt{2}$ after all.
A: Let $f(x)=\sin(x)/x$. So $f(x)=0$ when $\sin(x)=0$. So the only solution in the interval $(0,2\pi)$ is $x=\pi$. Plugging in test points shows that $f(x)$ is positive to the left of $x=\pi$ and negative to the right. 
Next, if you accept that $1-x/3 \leq \sin(x)/x$ (make some argument using MacLaurin series), then $\int_0^{\pi} f(x)\,dx \geq \frac{1}{2}(3)(1)=3/2$. On the other hand $|\sin(x)|/x \leq |\sin(x)|/3$ for $x \geq 3$ so $\int_{\pi}^{2\pi} |f(x)|\,dx \leq \int_{\pi}^{2\pi} \frac{|\sin(x)|}{3}\,dx = 2/3$.
So $\int_0^{2\pi} f(x)\,dx \geq 3/2-2/3>0$
