Integral with $+i\epsilon$ prescription involving residue theorem? Consider the integral $$I = \int_{-1}^{1} \frac{\text{d}x}{(x + \xi - i\epsilon) (x- \xi + i \epsilon)}$$ where $\xi$ is valued in $[-1,1]$. If I want to note the contribution of this integral at the point $x=\xi$ does the $+i\epsilon$ prescription allow me to simply write that $$I_{x=\xi} = \frac{1}{2\xi}?$$ I just said that $x-\xi$ is then zero while $x+\xi$ is $2\xi$ and the $+i\epsilon$ prescription avoids the pole at this point. 
1) Is this answer correct?
2) If so, is there a more mathematically rigourous way of showing this result and if the result is not correct how can one proceed to find this $x=\xi$ contribution to $I$? I'm thinking the result is ok as the residue theorem tells me that the residue of the pole term is just one but would be nice to check this. 

It is not clear to me what your question means. 
But why not calculate the integral 
$$f(\xi ,\epsilon )=\int_{-1}^1 \frac{1}{(\xi +x-i \epsilon ) (-\xi +x+i \epsilon )} \, dx$$
explicitly?
Writing the integrand as
$$\frac{1}{(\xi +x-i \epsilon ) (-\xi +x+i \epsilon )}=\frac{1}{a^2+x^2}$$
with
$$a\to \epsilon +i \xi$$
the integral can be done using the well known relation
$$\int_{-1}^1 \frac{1}{a^2+x^2} \, dx=\frac{2 \tan ^{-1}\left(\frac{1}{a}\right)}{a}$$
Hence the explicit solution of the integral is
$$f(\xi ,\epsilon )=\frac{2 \tan ^{-1}\left(\frac{1}{\epsilon +i \xi }\right)}{\epsilon +i \xi }$$
Let us plot Re, Im, and Abs of the complex function $f$ as a function of $\xi$ for two small values of $\epsilon$


EDIT #1 
Your question sounds as if you are talking about the indefinite integral.  
Ok this is given by
$$\int \frac{1}{(\xi +x-i \epsilon ) (-\xi +x+i \epsilon )} \, dx\\= \frac{1}{4 (\epsilon +i \xi )}\left( i \left(\log \left(\xi ^2+x^2-2 \xi  x+\epsilon ^2\right)-\log \left(\xi ^2+x^2+2 \xi  x+\epsilon ^2\right)\right)\\+2 \tan ^{-1}\left(\frac{x \epsilon }{\xi  (\xi -x)+\epsilon ^2}\right)+2 \tan ^{-1}\left(\frac{x \epsilon }{\xi  (\xi +x)+\epsilon ^2}\right)\right)$$
and in the limit $\xi \to x$ this is
$$\frac{i \left(\log \left(\epsilon ^2\right)-\log \left(4 x^2+\epsilon ^2\right)\right)+2 \tan ^{-1}\left(\frac{x \epsilon }{2 x^2+\epsilon ^2}\right)+2 \tan ^{-1}\left(\frac{x}{\epsilon }\right)}{4 (\epsilon +i x)}$$
The limit $\epsilon \to 0$ of which is $\infty$ in this sense
$$\lim_{\epsilon \to 0} \, \frac{i \left(\log \left(\epsilon ^2\right)-\log \left(4 x^2+\epsilon ^2\right)\right)+2 \tan ^{-1}\left(\frac{x \epsilon }{2 x^2+\epsilon ^2}\right)+2 \tan ^{-1}\left(\frac{x}{\epsilon }\right)}{4 (\epsilon +i x)} = \left(-\frac{1}{\text{sgn}(x)}\right) \infty$$
