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 $\overline{s}$.
 A: 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.
