# Poles of a function defined in terms of an integral

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 $\overline{s}$.

• It looks like $\rho(x) \equiv 1$ satisfies the assumptions, but then what happens say when $s = -2$? Is $w(-2)$ even defined? – bryanj Jun 22 '14 at 15:38
• 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. – antarcticfox Jun 22 '14 at 15:54

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.

• 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. – antarcticfox Jun 22 '14 at 14:41
• 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. :) – bryanj Jun 22 '14 at 14:53
• Bam! That works :-) Thank you @bryanj. – antarcticfox Jun 22 '14 at 15:07
• Can we also say something about the multiplicity of the poles? Are all poles simple and isolated? – antarcticfox Jun 22 '14 at 15:11
• 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. – bryanj Jun 22 '14 at 15:32