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I was wondering about evaluating the following definite integral analytically: \begin{equation} \int_{-\infty}^{\infty}\frac{1}{\sqrt{k-p}\sqrt{k+p}}\,\mathrm dp \end{equation}

Does someone know how to approach this?

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    Do you know the trick $(a - b)(a + b) = a^2 - b^2$?2012-06-06
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    I would like to go around the singularity and possibly apply the Residue theorem to calculate it. Is that possible?2012-06-06
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    I'm not sure how you're going to go around the singularity when your integral seems to go right through it...2012-06-06
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    And your integrand is non-real for some $p$ ... and anyway not absolutely convergent. So you need to provide some explanation for us to make any sense of it.2012-06-06
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    1.Substitute p=z to work in the complex plane 2. Move one pole up to $k+i\gamma$ and after the integration add a limit of $\gamma$ going to zero and to the same thing for the other singularity. 3. Which results in two contour integrals in the complex plane one around the singularity in the upper plane plus an over the singularity in lower plane.2012-06-06
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    $\int_{-\infty}^{\infty}\frac{1}{\sqrt{k+i \gamma+z}\sqrt{k-i \gamma-z}}$2012-06-06
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    and then take the limit such $\lim_{\gamma \rightarrow 0} \int_{-\infty}^{\infty}\frac{1}{\sqrt{k+i \gamma+z}\sqrt{k-i \gamma-z}}dz$2012-06-06
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    @Micheal To begin with, $p$ and $z$ are just letters.2012-06-06
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    you have issues with the complex square root, but both pieces of this integral, $\int_{-k}^k and \int _k^{\infty} $ are standard trig integrals, substitute $x=kcos(\theta)$ in first and $x = ktan(\theta)$ in 2nd, and if you can't get anything sensible out of it, it is because of the problems limned above.2012-06-06

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