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I've seen the following claim in a book without proof and don't know why it holds. Let $a<b \in\mathbb{R}$. Then the integral $$\frac{1}{2\pi i}\int_a^b\left(\frac{1}{t-i\epsilon-\lambda}-\frac{1}{t+i\epsilon-\lambda}\right)dt$$ converges to $0$ if $\lambda\not\in(a,b)$, $\frac{1}{2}$ if $\lambda=a$ or $b$ and $1$ if $\lambda\in(a,b)$ as $\epsilon\searrow0$. The $2\pi i$ made me think that this could be an application of the residue theorem but I don't know which integration path to choose.

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up vote 2 down vote accepted

Hint: just split the integral up in real and imaginary part similar this proof.

Spoiler below:

Expand: $$ \frac{1}{t-\lambda \pm i \epsilon} = \frac{1}{t-\lambda \pm i \epsilon} \frac{t-\lambda \mp i \epsilon}{t-\lambda \mp i \epsilon} = \frac{t-\lambda \mp i \epsilon}{(t-\lambda)^2 + \epsilon^2}.$$ Thus $$ \frac{1}{t-\lambda-i\epsilon}-\frac{1}{t-\lambda+i\epsilon} = \frac{2i \epsilon}{(t-\lambda)^2 + \epsilon^2} .$$ and $$\frac{1}{2\pi i}\int_a^b\left(\frac{1}{t-i\epsilon-\lambda}-\frac{1}{t+i\epsilon-\lambda}\right)dt = \frac{\epsilon}{\pi} \int_a^b \frac{dt}{(t-\lambda) + \epsilon^2} = \frac{1}{\pi}\left[\arctan\left(\frac{b-\lambda}{\epsilon}\right) - \arctan\left(\frac{a-\lambda}{\epsilon}\right)\right].$$ Now, we know that $$\begin{align} \lim_{x\to \pm \infty} \arctan(x) &=\pm\frac\pi2, & \arctan(0)&=0.\end{align}$$ from which the result stated follows immediately.

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Thank you! Thought to much about complex integration! –  Julian Jan 11 '13 at 10:56
    
@Julien: yes, I know this. One is tempted to use complex analysis methods but they make the problem just more difficult. –  Fabian Jan 11 '13 at 15:36
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