How to show the function achieves maximum at $a = 0$ Consider this integration:
$$ f(a) = \int_0^1 \frac{-\log(x^2)}{|x-a|^\beta} dx$$ with $0 < \beta <1$ fixed. 
I know that $f(0)$ is finite. Is it true that $f(0) \ge f(a)$ for all $a \in \mathbb R$? 
I feel that this is true since when $a = 0$ the two singularities coincides therefore the integration should be the largest, but I don't know how to show it. I know this is in a form of a convolution but I can't recall any theories to help me to deal with it.
Any help is appreciated.

The claim does not hold as pointed out by @Jack D'Aurizio. But how about if the integration is from $-1$ to $1$ instead?
i.e. if 
$ f(a) = \int_{-1}^1 \frac{-\log(x^2)}{|x-a|^\beta} dx$, does it hold that $f(0) \ge f(a)$ for all $a \in \mathbb R$?
 A: We have $f(0)=\frac{2}{(1-\beta)^2},\,f(1)=\frac{2 H_{1-\beta}}{1-\beta}$ and for any $A>a>1$
$$ f(A)=\int_{0}^{1}\frac{-\log(x^2)}{(A-x)^\beta}\,dx < \int_{0}^{1}\frac{-\log(x^2)}{(a-x)^\beta}\,dx =f(a)$$
is trivial, so $f$ is decreasing on $(1,+\infty)$ and similarly it can be proved that $f$ is increasing on $\mathbb{R}^-$.
The only non-trivial part is to study the behaviour of $f$ over $(0,1)$. On the other hand, if $a\in(0,1)$
$$\begin{eqnarray*} f(a)&=&\int_{0}^{a}\frac{-2\log(x)}{(a-x)^{\beta}}\,dx+\int_{a}^{1}\frac{-2\log x}{(x-a)^{\beta}}\,dx\\&=&a^{1-\beta}\int_{0}^{1}\frac{-2\log(au)}{(1-u)^{\beta}}\,du+\int_{0}^{1-a}\frac{-2\log(1-x)}{(1-a-x)^{\beta}}\,dx\\&=&-2\int_{0}^{1}\frac{a^{1-\beta}\log(au)+(1-a)^{1-\beta}\log(1-(1-a)u)}{(1-u)^{\beta}}\,du\end{eqnarray*}$$
and $\lim_{a\to 0^+}f'(a)$ is not zero but $+\infty$, so the claim does not hold. $f(a)$ attains its maximum on the right of $a=0$.
For instance, this is the graph of $f(a)$ over $\left(0,\frac{1}{10}\right)$ for $\beta=\frac{1}{2}$:
$\hspace{1cm}$
On the other hand, the maximum of $f(a)$ is not expected to be much larger than $f(0)=\frac{2}{(1-\beta)^2}$.
