# Using Rouche's theorem

Let $p>1$.

Consider $\phi(p)=\int_0^{\infty}\left|\frac{\sin t}{t}\right|^pdt$. Function $\phi(p)$ is analytic on its domain.

It's derivative, $\phi'(p)=\int_0^{\infty}\left|\frac{\sin t}{t}\right|^p\ln\left|\frac{\sin t}{t}\right|dt$.

Let $g(p)=\frac{1}{2\sqrt p}\phi(p)$ and $f(p)=\sqrt p \phi'(p)$. Using Rouche's theorem http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem , show that $f(p)+g(p)$ has one zero.

Thank you.

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I cannot understand why Rouche's theorem should be used.$f(p)+g(p)=\frac{d}{dp}p^{1/2}\phi(p)$, then use Rolle's theorem to prove the existence of real zeros. –  y zhao May 21 '12 at 4:41
I've tried Roll's theorem and its very hard to prove with the help of it. –  David May 22 '12 at 12:56
try to calculate $2^{1/2}\phi(2),4^{1/2}\phi(4),6^{1/2}\phi(6)$ with a computer,then you will find something interesting. –  y zhao May 24 '12 at 15:41