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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?

Thanks!

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Do you want an approximate? –  Pedro Tamaroff Feb 8 '12 at 16:34
    
Wolfram alpha will give you an approximation. (by the way it's positive) –  Bill Cook Feb 8 '12 at 16:34
2  
No, as I said I want to find only the sign, I want to find it by a "calculus- investigation", and not by checking it on W.A :) Thanks! –  Jozef Feb 8 '12 at 16:38

3 Answers 3

up vote 19 down vote accepted

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

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Thanks Julian! :) very nice solution! –  Jozef Feb 8 '12 at 16:49

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.

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

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