1
$\begingroup$

Suppose $\rho: [0,1] \rightarrow [0,\infty)$ with the following two properties:

$$\int_0^1 \rho(x) dx = 1$$

and

$$\int_0^1 \rho(x) x dx =\frac{1}{2} $$

Now let $$w(s) \equiv \int_0^1 \rho(x) x^s dx$$

for $s\in \mathbb{C}$.

And define $F(s) = \frac{1}{\frac{1}{2} - w(s)}$. What can we say about the poles of $F$? Clearly $s=1$ is a pole because of $w(1) = 1/2$. I have read, without proof, that

  • The $s=1$ pole is the one with largest real-part
  • All other poles come in pairs of complex conjugates, that is if $s$ is a pole then also $\overline{s}$ (the complex conjugate)

I would like to know why this is the case, i.e. see a proof. But unfortunately I am not able to do so. The only thing I got was an inequality for the integral summand in the denumerator of $F$, i.e. when $s=\alpha+i\beta$

$$\begin{eqnarray} \vert w(s) \vert &\equiv& \vert \int_0^1 \rho(x) x^s dx\vert\\ &\leq& \int_0^1 \vert \rho(x) x^s \vert dx \\ &\leq& \int_0^1 r^{\alpha} \rho(x)dx \\ &\leq& \xi^{\alpha} \int_0^1 \rho(x)dx\\ &\leq& \xi^{\alpha} \end{eqnarray} $$

In the second-last step I used the first mean value theorem for integration and therefore $\xi\in[0,1]$. Now any $s$ that is a pole of $F(s)$ has $w(s)=\frac{1}{2}$ and hence

$$\frac{1}{2} \leq \xi^{\alpha}$$

In case of the $s=1$ pole the equality is given, because for the $s=1$ case we have $\xi=\frac{1}{2}$. But I am not sure if that is of use in general to show that any other $\alpha<1$. It would only help if I could show that $\xi>\frac{1}{2}$ for any $s\neq 1$ and hence one gets a contraction if $\alpha>1$:

$$\frac{1}{2} \leq \xi^{\alpha} < \xi < \frac{1}{2}$$

But I am not sure if that works.

Any help is appreciated. Many thaks!

EDIT

I understand the second point, that is all poles have to come in complex conjugate pairs. Suppose $s$ is a pole then $w(s) = \frac{1}{2} \in \mathbb{R}$ and hence $\overline{w(s)} = \frac{1}{2}$. And

$$\overline{w(s)} = \overline{\int_0^1 \rho(x) x^s dx} = \int_0^1 \rho(x) \overline{x^s} dx = w(\overline{s})$$

That is if $s$ is a pole then also $\overline{s}$.

$\endgroup$
  • $\begingroup$ It looks like $\rho(x) \equiv 1$ satisfies the assumptions, but then what happens say when $s = -2$? Is $w(-2)$ even defined? $\endgroup$ – bryanj Jun 22 '14 at 15:38
  • $\begingroup$ Indeed the whole setup is of probabilistic nature and $\rho(x)=1$ is the uniform case. $F(s)$ is actually a laplace transform of a function $F(\alpha)$. The goal is to determine the asymptotic behavior for $\alpha \rightarrow -\infty$ from the poles of $F(s)$ by applying Mellin's inverse formula. I am not sure if one can not simply say that for values like $s=-2$ $F$ is simply 0. $\endgroup$ – antarcticfox Jun 22 '14 at 15:54
1
$\begingroup$

To show that the real of a pole is not more than $1$:

If the real part $\alpha$ of $s = \alpha + i \beta$ is more than one, then for $0<x<1$ you get $|x^s| = x^\alpha < x$, so $$ \Bigg |\int _0 ^1 p(x) x^s dx \Bigg| \le \int _0 ^1 p(x)|x^s| dx = \int _0 ^1 p(x)x^{\alpha} dx < \int _0 ^1 p(x)x dx = 1/2 $$ so $|w(s)| < 1/2$, therefore the denominator of $F(s)$ does not vanish and $s$ is not a pole.

To show that $s = 1$ is the only pole whose real part is $1$:

Say $s = 1 + i \beta$ is a pole, so $w(s) = 1/2$. Then $$ \int _0 ^1 p(x) x e^{i \beta \log x} dx = \int _0 ^1 p(x) x \cos(\beta \log x) dx + i \int _0 ^1 p(x) x \sin(\beta \log x) dx = 1/2 $$ The imaginary part has to vanish, so you're left with $$ 1/2 = \int _0 ^1 p(x) x \cos(\beta \log x) dx $$ If $\beta \ne 0$, the for some value of $x$ in $0 < x < 1$, we have $|\cos(\beta \log x)| < 1$ and you get $$ 1/2 = \Bigg| \int _0 ^1 p(x) x \cos(\beta \log x) dx \Bigg | < \Bigg| \int _0 ^1 p(x) x dx \Bigg | = 1/2 $$ So there can't be a pole when the real part is one and the imaginary part is non-zero.

$\endgroup$
  • $\begingroup$ Thanks @bryanj this all make sense and I can almost follow all steps of it. The only possible obligation I have is that $\vert x^s\vert <x$ only if $0<x<1$ but the integral also evaluates the points $x=0$ and $x=1$... So we would need a $\leq$, wouldn't we? If so that would result in $1/2\leq 1/2$ and hence that is not necessarily a reason to say that there cannot be any pole. $\endgroup$ – antarcticfox Jun 22 '14 at 14:41
  • 1
    $\begingroup$ Yes that's a detail. But, you can chop the integral into, say, three pieces, like $0 \le x \le 1/3$, $1/3 \le x \le 2/3$, and $2/3 \le x \le1$. On the outer two pieces, you can allow yourself $\le$, but on the middle piece you have strict inequality. Adding the three integrals gives you strict inequality. :) $\endgroup$ – bryanj Jun 22 '14 at 14:53
  • $\begingroup$ Bam! That works :-) Thank you @bryanj. $\endgroup$ – antarcticfox Jun 22 '14 at 15:07
  • $\begingroup$ Can we also say something about the multiplicity of the poles? Are all poles simple and isolated? $\endgroup$ – antarcticfox Jun 22 '14 at 15:11
  • 1
    $\begingroup$ I think a pole at $s$ will be simple if $w'(s) \ne 0$. I haven't worked out the details, but differentiating under the integral looks promising. $\endgroup$ – bryanj Jun 22 '14 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.