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. 
 A: You can do some contour integral $I_2$ along a semi-circle with the diameter along the x-axis and extending into the upper half-plane. Then $I_2$ would include the pole at $-\xi+i\epsilon$ and would be homotopic to $I_3$ along a small circle (diameter $r<\epsilon$) around that pole with
$$I_3 = \int_{S_r(-\xi+i\epsilon)} \frac{dz}{(z+\xi-i\epsilon)(z-\xi+i\epsilon)} = \mathrm{res}\, f(-\xi+i\epsilon)\overset{\mathrm{homotopic}}{=}I_2$$
Finally you could split integral $I_2$ into the part along the x-axis ($=I$) and the upper arc ($I_4$).
$I_4$ is then without any poles along its integration path an can be calculated as
$$I_4=\int_0^\pi \frac{i\,dt}{(e^{it}+\xi-i\epsilon)(e^{it}-\xi+i\epsilon)}$$
And finally
$$I=I_2-I_4$$
A: 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
$$\lim_{\epsilon \to 0} \, \frac{i \left(\log \left(\epsilon ^2\right)-\log \left(4 x^2+\epsilon ^2\right)\right)+2 \tan ^{-1}\left(\frac{x \epsilon }{2 x^2+\epsilon ^2}\right)+2 \tan ^{-1}\left(\frac{x}{\epsilon }\right)}{4 (\epsilon +i x)} = \left(-\frac{1}{\text{sgn}(x)}\right) \infty$$
