5

Could you help me with the following question?

Show that the following numbers are positive for $j$ odd and negative otherwise: $$\frac{1}{\pi}\int_0^{j\pi} \frac{\sin t}{t}\,dt - \frac{1}{2}$$

Michael Hardy
  • 1
User4356
  • 53

2 Answers2

4

We have

$$ \int_0^\infty\frac{\sin t}t\mathrm dt=\frac\pi2 $$

and thus

$$ \frac1\pi\int_0^{j\pi}\frac{\sin t}t\mathrm dt-\frac12=-\frac1\pi\int_{j\pi}^\infty\frac{\sin t}t\mathrm dt\;. $$

Integrating by parts yields

$$ \begin{align} \int_{j\pi}^\infty\frac{\sin t}t\mathrm dt &= \left[-\frac{\cos t}t\right]_{j\pi}^\infty-\int_{j\pi}^\infty\frac{\cos t}{t^2}\mathrm dt \\ &= \frac{(-1)^j}{j\pi}-\int_{j\pi}^\infty\frac{\cos t}{t^2}\mathrm dt\;, \end{align} $$

and then

$$ \left|\int_{j\pi}^\infty\frac{\cos t}{t^2}\mathrm dt\right|\lt\int_{j\pi}^\infty\frac1{t^2}\mathrm dt=\frac1{j\pi} $$

gives the desired result.

joriki
  • 238,052
1

There may be a better way to do this, but you could use the properties of the Sine integral: $$\frac{1}{\pi}\int_0^{n\pi} \frac{\sin t}{t} \, dt = \frac{\operatorname{Si}(n\pi)}{\pi}$$ Expanding $\operatorname{Si}(x)$ around infinity: $$\operatorname{Si}(x)\sim \sin(x) \left( -\frac{1}{x^2}+\frac{6}{x^5}-\cdots\right) + \cos(x)\left( -\frac{1}{x}+\frac{2}{x^3}-\cdots\right)+\frac{\pi}{2}$$ So that: $$\frac{\operatorname{Si}(n\pi)}{\pi}-\frac{1}{2} \sim \frac{\cos(n\pi)} \pi \left( -\frac{1}{n\pi}+\frac{2}{(n\pi)^3}-\cdots\right)$$ Since the term in brackets is always smaller than zero, the result follows.

Michael Hardy
  • 1
Nathaniel Bubis
  • 33,018
  • The result follows for $n$ large enough, but we still don't know if it's true for small $n$ or how big "large enough" is. – Antonio Vargas Feb 20 '13 at 02:58
  • 2
    This is an asymptotic, divergent series (see Wikipedia); the series in parentheses also diverges by itself (since the numerator grows factorially and the denominator only grows exponentially) and can't be said to have a definite sign. – joriki Feb 20 '13 at 07:28