Why does this limit exist? I want to show that $$\lim_{\epsilon \to 0} \int_\mathbb {R} \frac{1}{x^2-(1+i\epsilon)^2} dx$$ converges. I evaluated this integral numerically and it seams like it converges to some value for small $\epsilon$. Any ideas how to show that it converges?
 A: Nice example for the importance of the statement "the limit of an integral is not (necessarily) the integral of the limit." 
Anyways, we can find an explicit, although admittedly messy, anti-derivative using partial fractions (difference of two squares in the denominator). Evaluating the improper integral (edit: for $\epsilon > 0$) gives
$$\int_\mathbb{R} \frac{1}{x^2 - (1+ i \epsilon)^2 } ~\mathrm{d}x = \frac{\pi}{\epsilon - i} = \frac{ \epsilon \pi + i \pi}{1 + \epsilon^2}.$$
Taking a limit as $\epsilon \to 0$ (edit: $\epsilon\to 0^+$) yields an answer of $i \pi$. 
Edit!!
I originally forgot to account for a few branch cut issues. As pointed out in the comments, there are issues depending on the sign of $\epsilon$. The above argument holds for $\epsilon > 0$ only. However, for $\epsilon < 0$ we have 
$$\int_\mathbb{R} \frac{1}{x^2 - (1+ i \epsilon)^2 } ~\mathrm{d}x = - \frac{\pi}{\epsilon - i} =  - \frac{ \epsilon \pi + i \pi}{1 + \epsilon^2}.$$
Thus taking the limit as $\epsilon \to 0^-$ gives $- i \pi$, and therefore the limit does not exist. We can also see this sign reversal by means of the residue theorem (which gives a compact way to evaluate the improper integral as well). 
Apologies, but as the old joke goes, "How do you horrify a mathematician? Let $\epsilon < 0$..."
Edit 2: Alternate Solution via Residue Theorem
Let $$f: \mathbb{C} \to \hat{\mathbb{C}}, \quad f(z) = \frac{1}{z^2 - (1+ i \epsilon)^2} = \frac{1}{(z + 1 + i \epsilon)(z - 1 - i \epsilon)}.$$
$f$ is meromorphic with simple poles at $z_1 = 1 + i \epsilon$ and $z_2 = -1 - i \epsilon$. Define the simple closed curve $\gamma = \gamma_1 + \gamma_2$ with counter-clockwise orientation, where $\gamma_1$ goes from $-R$ to $R$ along the real line and $\gamma_2$ is the semi-circular arc from $+R$ to $iR$ to $-R$. By the Residue Theorem
   $$\int_{\gamma_1} f(z) ~\mathrm{d}z + \int_{\gamma_2} f(z) ~\mathrm{dz} = \oint_\gamma f(z) ~\mathrm{d}z = 2 \pi i \sum Res(f, a_k)$$
where $\{a_k \}$ is the set of poles in the region inside our Jordan curve. Applying an ML-estimate to the integral along $\gamma_2$ gives
   $$\left| \int_{\gamma_2} f(z) ~\mathrm{d}z \right| \approx \frac{\pi R}{R^2} \overset{R \to \infty}{\longrightarrow} 0$$
and therefore
  $$\int_{-\infty}^\infty f(z) ~\mathrm{d}z  = 2 \pi i \sum Res(f, a_i).$$
We have two cases: $\epsilon >0$ and $\epsilon <0$. 


*

*Case 1: If $\epsilon>0$, then the only pole inside our region is $z_1 = 1+ i \epsilon$ as $\Im z_2 < 0$. Since $f$ has a simple pole at $z_1$, 
\begin{align*}
Res(f, z_1) &= \lim_{z \to z_1} (z-z_1) f(z) \\
&= \lim_{z \to z_1} \frac{1}{z - z_2}\\
& = \frac{1}{z_1 - z_2} \\
&= \frac{1}{1 + i \epsilon - (-1 - i \epsilon)} \\
&= \frac{1}{2 (1+i \epsilon)}.
 \end{align*}
Thus we have 
$$\int_{-\infty}^\infty f(z) ~\mathrm{d} z = 2 \pi i \frac{1}{2 (1+i \epsilon)} \to \pi i$$
as $\epsilon \to 0+$. 

*Case 2: If $\epsilon <0$, then the only pole inside our region is $z_2 = -1- i \epsilon$ as $\Im z_1 < 0$. Since $f$ has a simple pole at $z_2$, we have
\begin{align*}
Res(f, z_2) &= \lim_{z \to z_2} (z-z_2) f(z) \\
 &= \lim_{z \to z_2} \frac{1}{z-z_1} \\ 
 &= \frac{1}{z_2 - z_1} \\
 &= \frac{1}{-1 - i \epsilon - (1 + i \epsilon)} \\
 &= - \frac{1}{2 (1 + i \epsilon)}.
\end{align*}
Thus we have
$$\int_{-\infty}^\infty f(z) ~\mathrm{d} z = 2 \pi i \frac{-1}{2 (1+i \epsilon)} \to -\pi i$$
as $\epsilon \to 0^-$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\lim_{\epsilon \to 0^{\Large\color{#f00}{\pm}}}\int_\mathbb{R}
{\dd x \over x^{2} - \pars{1 + \ic\epsilon}^2} =
\lim_{\epsilon \to 0}\bracks{\pars{%
\int_{-\infty}^{\infty}{\dd x \over x - 1 - \ic\epsilon} -
\int_{-\infty}^{\infty}{\dd x \over x + 1 + \ic\epsilon}}
{1 \over 2 + \ic\epsilon}}
\\[1cm] = &\
{1 \over 2}\bracks{\mrm{P.V.}\int_{-\infty}^{\infty}{\dd x \over x - 1} + \ic\pi\,\mrm{sgn}\pars{\epsilon}\int_{-\infty}^{\infty}\delta\pars{x - 1}}
\\[5mm] - &\
{1 \over 2}\bracks{\mrm{P.V.}\int_{-\infty}^{\infty}{\dd x \over x + 1} - \ic\pi\,\mrm{sgn}\pars{\epsilon}\int_{-\infty}^{\infty}\delta\pars{x + 1}
\,\dd x} = \ic\pi\,\mrm{sgn}\pars{\epsilon}
\end{align}

$$
\bbx{\ds{\begin{array}{rcl}
\ds{\lim_{\epsilon \to 0^{\LARGE\color{#f00}{-}}}\int_\mathbb{R}
{\dd x \over x^{2} - \pars{1 + \ic\epsilon}^2}} & \ds{=} & \ds{-\ic\pi}
\\[2mm]
\ds{\lim_{\epsilon \to 0^{\LARGE\color{#f00}{+}}}\int_\mathbb{R}
{\dd x \over x^{2} - \pars{1 + \ic\epsilon}^2}} & \ds{=} &
\ds{\phantom{-}\ic\pi}
\end{array}}}
$$
