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I am wondering how to evaluate this integral:

$$I = -i\int_{-\infty}^{\infty}\textrm{d}p\dfrac{pe^{ipr}}{\sqrt{p^2+m^2}}$$

In my physics text book the substitution $p=i\rho$ is made together with "pushing the contour up to wrap around the upper branch cut".

$$I = 2\int_{m}^{\infty}\textrm{d}\rho\dfrac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$

Apparently then $I \propto e^{-mr}$ holds.

Coud somebody explain me a.) The exact process of "pushing the contour up to wrap around the upper branch cut" and b.) the method to solve the latter integral?

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  • $\begingroup$ Given that the Fourier transform of $\frac{1}{\sqrt{x^2+1}}$ is a multiple of $K_0(|s|)$ (modified Bessel function of the second kind), you may just exploit the properties of the Fourier transform. $\endgroup$ Commented May 14, 2016 at 17:55

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When dealing with branch cuts, it is sometimes useful to find a suitable technique that removes them. In our case, $$ J(m)=\int_{\mathbb{R}}\frac{p e^{ipr}}{\sqrt{p^2+m^2}}\,dp =m\int_{\mathbb{R}}\frac{x e^{imr x}}{\sqrt{x^2+1}}\,dx=m\int_{\mathbb{R}}\frac{x\sin(mr x)}{\sqrt{x^2+1}}\,dx$$ is not a convergent integral in the usual sense, since $\frac{x}{\sqrt{x^2+1}}\to 1$ as $|x|\to +\infty$, but its principal value is given by: $$ PV\,J(m) = -2m\int_{0}^{+\infty}\left(1-\frac{x}{\sqrt{x^2+1}}\right)\sin(mr x)\,dx $$ that is a convergent integral by Dirichlet's test, since $\sin$ has a bounded primitive while $1-\frac{x}{\sqrt{1+x^2}}$ is rapidly decreasing ($\sim\frac{1}{2x^2}$) to zero as $x\to +\infty$. Integration by parts then gives:

$$ PV\, J(m) = -\frac{2}{r}\int_{0}^{+\infty}\frac{\cos(mr x)}{(1+x^2)^{3/2}}\,dx=-2|m|\,K_1(|mr|)\tag{1}$$

where $K_1$ is a modified Bessel function of the second kind. $K_1(z)$ behaves like $\sqrt{\frac{\pi}{2z}} e^{-z}$ for large values of $z$ and like $\frac{1}{z}$ in a right neighbourhood of the origin.

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