1

Am I required to know if it is positive or negative? This expression is not really integrable, not within the boundaries of the course at least. I tried comparing it to other integrals but I get zero all the time. I am not sure I have fully understood what I should do. I would really appreciate any help in this.

An attempt using an advice here: $\sin x$ between $0$ to $\pi$ equals $-\sin x$ between $\pi $ to $2\pi$. Since we are looking at $\sin x\over x$, $\int_{\pi}^{2\pi}{\sin x\over x}$ might be negative, but still is smaller than $\int_{0}^{\pi}{\sin x\over x}$, therefore, the sign is +1. Is that sufficient?

Empiricist
  • 7,933
Meitar
  • 2,911
  • 2
    It's a fine bounded and continuous function, so the integral exists. It helps to split it up at $\pi$. – Daniel Fischer Jun 26 '15 at 11:54
  • split the integral at $\pi$ argue that the part with boundarys $[\pi ,2\pi]$ is smaller in absolute value then the part with boundarys $[0, \pi]$ – tired Jun 26 '15 at 11:56
  • I don't know how to use the "absolute value" argument, but I edited my answer. How is it? Actually it is supposed to be a question in a test of a strict professor, and it couldn't be that it is that simple. Is there any hard work needs to be done that I missed? – Meitar Jun 26 '15 at 12:01

2 Answers2

8

It is positive: $$\int_{0}^{2\pi}\frac{\sin x}{x}\,dx = \int_{0}^{\pi}\frac{\sin x}{x}\,dx+\int_{0}^{\pi}\frac{-\sin(x)}{x+\pi}\,dx = \pi\int_{0}^{\pi}\frac{\sin x}{x(x+\pi)}\,dx.$$

Jack D'Aurizio
  • 353,855
3

https://en.wikipedia.org/wiki/Sinc_function

Look at the graph and compare the intervals $x \in [0,\pi]$ and $[\pi,2\pi]$: The sign is clearly positive.

You can just argue

$$\forall x \in [0,\pi] \ : \ \ \ \frac{\sin(t)}{t} > -\frac{\sin(t+\pi)}{t + \pi} = \frac{\sin(t)}{t + \pi}$$

which yields that the first half of your integral has greater magnitude than the second half.

GDumphart
  • 2,250