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I want to calculate the integral for $n \in \mathbb{N}$ and $p \in \mathbb{R}$

$ \lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} dx \frac{\log^n(x-i\epsilon)}{x-i\epsilon} e^{ixp} $

For $n = 0$ I simply get the heavyside function $\Theta(p)$. Is there a similar result for $n=1,2$?

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1 Answer 1

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I assume that you want the right limit (the two-sided limit doesn't exist) and that $\ln$ is the principal value of the logarithm. Then, since $$\frac {\ln^n(x - i \epsilon)} {x - i \epsilon} = \frac 1 {n + 1} \frac \partial {\partial x} \frac {\partial^{n + 1}} {\partial \alpha^{n + 1}} (x - i \epsilon)^\alpha \bigg\rvert_{\alpha = 0},$$ we just need the Fourier transform of $x_+^\alpha$, which is known: $$f(x) = (x - i 0)^\alpha = x_+^\alpha + e^{-\pi i \alpha} (-x)_+^\alpha,\\ \mathcal F[f](p) = i (e^{\pi i \alpha/2} - e^{-3 \pi i \alpha/2}) \Gamma(\alpha + 1) p_+^{-\alpha - 1}, \\ \lim_{\epsilon \to 0^+} \int_{\mathbb R} \frac {\ln^n(x - i \epsilon)} {x - i \epsilon} e^{i p x} dx = -\frac {i p} {n + 1} \frac {\partial^{n + 1}} {\partial \alpha^{n + 1}} \mathcal F[f](p) \bigg\rvert_{\alpha = 0} = \\ \frac {H(p)} {n + 1} \sum_{j + k + l = n + 1} \binom {n + 1} {j, k, l} (-1)^{l} \left( \! \left( \frac {\pi i} 2 \right)^j - \left( -\frac {3 \pi i} 2 \right)^j \right) \Gamma^{(k)}(1) \ln^{l} p.$$

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