Approximation of an integral on a certain limit In Peskin & Schroeder - Chapter 6 - the authors make the following approximation when $-q^2\rightarrow\infty$
$$\int_0^1 \!\!d\xi\, \frac{-q^2/2}{-q^2\xi(1-\xi)+m^2} \simeq \frac{1}{2}\int_0d\xi\, \frac{-q^2}{-q^2\xi+m^2} + \begin{pmatrix}\!\text{equal contribution}\!\\ \text{from } \xi\approx 1 \end{pmatrix}.\tag{6.64} $$
It might be silly but I don't 'get' all the steps. It is clear to me that the function goes as $\frac{1}{2\xi(1-\xi)}$ in the $-q^2\rightarrow\infty$ limit and the dominant contributions are then about $\xi=0$ and $\xi=1$, but... I don't see clearly what they did in the right hand side. It doesn't seem like a Taylor expansion, then , what is it?
For context: this is a relevant integral for the analysis of IR divergences.
 A: Here's what I could conjure up. I don't like it cause of sweeping infinities under the rug so feel free to downvote/not accept.
Let
\begin{align}
f\left(\frac{m^2}{-q^2} \right) = \frac{1}{2} \int_0^1 d\xi\; \frac{1}{\xi(1-\xi)+\frac{m^2}{-q^2}}.
\end{align}
Then
\begin{align}
f'\left(\frac{m^2}{-q^2} \right)=-\frac{1}{2}\int_0^1 d\xi\;\left(\frac{1}{\xi(1-\xi)+\frac{m^2}{-q^2}} \right)^2.
\end{align}
Taylor expanding around $0$ (since large $-q^2$, corresponds to small $\frac{m^2}{-q^2}$)
\begin{align}
f(x)&\approx f(0) + f'(0) x+...\\
&=\frac{1}{2}\int_0^1 d\xi\;\frac{1}{\xi (1-\xi)} - \frac{1}{2}\int_0^1 d\xi\;\frac{1}{\xi^2 (1-\xi)^2} \frac{m^2}{-q^2} \\
&=\frac{1}{2}\int_0^1 d\xi\;\left(\frac{1}{\xi} +\frac{1}{1-\xi} \right) -\frac{1}{2}\int_0^1 d\xi\;\left(\frac{1}{\xi^2} + \frac{1}{(1-\xi)^2}+\frac{2}{\xi(1-\xi)} \right)\frac{m^2}{-q^2},
\end{align}
where the term $\left( \frac{2}{\xi(1-\xi)}\right)$ is subleading, despite being formally infinite, because  $\frac{1}{\xi^2}>\frac{1}{\xi}$, for $\xi\in(0,1)$. If we ignore this term,  we get the Taylor series for
$$\frac{1}{2}\int_0^1 d\xi \frac{-q^2}{-q^2\xi +m^2} + \frac{1}{2}\int_0^1 d\xi \frac{-q^2}{-q^2(1-\xi) +m^2}.$$
In general, comparing the higher order terms of the series, we'd neet to ignore terms in less negative power of $\xi$ to arrive at the result.
