Contour Integral with Real Singularity

Question

So, if I'm working in spherical coordinates, how would I evaluate the following integral? I know that I'm supposed to use contour integration and Jordan's lemma, but the fact that the singularity is located on the real axis is really throwing me off. Any advice for solving this improper integral via contour integration only?

$$\int_0^\infty{\frac{r \sin{(r\rho)}}{r^2-\alpha^2}} dr$$ where alpha is a real number

Answer 1

Hint: begin by noting that $$ \int_0^\infty{\frac{r \sin{(r\rho)}}{r^2-\alpha^2}}\,dr = \Im\left\{ \int_0^\infty{\frac{r e^{(ir\rho)}}{r^2-\alpha^2}}\,dr \right\} $$

For the indentation contour around $\alpha$: let $\gamma_\epsilon$ be a semicircle of radius $\epsilon$ that goes counterclockwise around $\alpha$. We wish to evaluate $$ \int_{\gamma_\epsilon} \frac{z e^{(iz\rho)}}{z^2-\alpha^2}\,dz = \int_{\gamma_\epsilon} \frac{1}{z - \alpha} \cdot \frac{z e^{(iz\rho)}}{z+\alpha}\,dz $$ Noting that $g(z) = \frac{z e^{(iz\rho)}}{z+\alpha}$ is continuous around $z = \alpha$, we can take $\epsilon$ sufficiently small so that $g(z) \approx g(\alpha)$ when $|z- \alpha| < \epsilon$. We conclude $$ \lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} \frac{g(z)}{z-\alpha}\,dz = \int_{\gamma_\epsilon} \frac{g(\alpha)}{z-\alpha}\,dz = \pi i g(\alpha) $$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{r\sin\pars{r\rho} \over r^{2} - \alpha^{2}}\,\dd r:\ {\large ?}.\quad}$ These integrals are very common in Quantum Mechanics as related to scattering theory. In that cases, $\ds{\alpha}$ has a 'small' imaginary part which indicates whether a wave is going to a scattering center or is leaving the scattering center.

The 'history' is more or less like this: \begin{align} &\color{#c00000}{\int_{0}^{\infty}{r\sin\pars{r\rho}\over r^{2} - \pars{\verts{\alpha} \pm \ic 0^{+}}^{2}}\,\dd r} \\[3mm]&=\int_{0}^{\infty}r\sin\pars{r\rho}\, {1 \over 2\pars{\verts{\alpha} \pm \ic 0^{+}}}\bracks{% {1 \over r - \pars{\verts{\alpha} \pm \ic 0^{+}}} -{1 \over r + \pars{\verts{\alpha} \pm \ic 0^{+}}}}\,\dd r \\[3mm]&={1 \over 2\pars{\verts{\alpha} \pm \ic 0^{+}}}\,2\ic\, \Im\int_{0}^{\infty}{r\sin\pars{r\rho} \over r - \verts{\alpha} \mp \ic 0^{+}}\,\dd r \\[3mm]&={\ic \over \verts{\alpha} \pm \ic 0^{+}} \int_{0}^{\infty}r\sin\pars{r\rho} \bracks{\pm\,\pi\,\delta\pars{r - \verts{\alpha}}}\,\dd r ={\ic \over \verts{\alpha} \pm \ic 0^{+}} \,\bracks{\pm\pi\verts{\alpha}\sin\pars{\verts{\alpha}\rho}} \\[3mm]&=\pm\ic\pi\verts{\alpha}\sin\pars{\verts{\alpha}\rho}\, \bracks{\pp{1 \over \verts{\alpha}} \mp \ic\pi\delta\pars{\verts{\alpha}}} \end{align} where $\ds{\delta\pars{x}}$ is the Dirac Delta 'Function' and we used the identities ( 'under' the integral sign ) $\ds{{1 \over x \pm \ic 0^{+}} = \pp{1 \over x} \mp \ic\pi\delta\pars{x}}$. $\ds{\pp}$ denotes the Principal Value.

$$ \color{#00f}{\large\int_{0}^{\infty} {r\sin\pars{r\rho} \over r^{2} - \pars{\verts{\alpha} \pm \ic 0^{+}}^{2}}\,\dd r =\pm\,\ic\pi\sin\pars{\verts{\alpha}\rho}} $$