Vrftsjtryk

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?

$$
begin{align*}
int_0^{2pi}frac{sin x}{x},dx&=int_0^{pi}frac{sin x}{x},dx+int_pi^{2pi}frac{sin x}{x},dx\
&=int_0^{pi}frac{sin x}{x},dx+int_0^{pi}frac{sin(x+pi)}{x+pi},dx\
&=int_0^{pi}Bigl(frac{1}{x}-frac{1}{x+pi}Bigr)sin x,dx\
&=piint_0^{pi}frac{sin x}{x(x+pi)},dx\
&>0
end{align*}
$$

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.

$$I = frac{104}{105} sqrt{2} simsqrt{2} $$

For the latter:

$$I = frac{{1568}}{{2145}}frac{{sqrt {2 + sqrt 2 } }}{2} sim sqrt{2}$$

Maybe the area is $sqrt{2}$ after all.

Let $f(x)=sin(x)/x$. So $f(x)=0$ when $sin(x)=0$. So the only solution in the interval $(0,2pi)$ 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 $int_{pi}^{2pi} |f(x)|,dx leq int_{pi}^{2pi} frac{|sin(x)|}{3},dx = 2/3$.

So $int_0^{2pi} f(x),dx geq 3/2-2/3>0$