# Contour Integral with Real Singularity

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

• This doesn't converge. – Potato Apr 27 '14 at 22:15
• @Potato I think the Cauchy principal value exists, though – Omnomnomnom Apr 27 '14 at 22:17
• @Potato thanks for the response. It should converge, as my resource is asserting that it is equivalent to a specific value (what I'm trying to figure out). Could I do a contour with 2 indentations around alpha and -alpha? – Incognito Apr 27 '14 at 22:19
• @Incognito It does not converge! Around its singularity, it looks like $1/(r-\alpha)$, which is not integrable if $\alpha$ is real. – Potato Apr 27 '14 at 22:20
• As @Omnomnomnom insinuated, you need to take the principal value. If you do that, you get a good value. – Daniel Fischer Apr 27 '14 at 22:21

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 $$\lim_{\epsilon \to 0^+} \int_{\gamma_\epsilon} \frac{g(z)}{z-\alpha}\,dz = \int_{\gamma_\epsilon} \frac{g(\alpha)}{z-\alpha}\,dz = \pi i g(\alpha)$$
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.
$$\color{#00f}{\large\int_{0}^{\infty} {r\sin\pars{r\rho} \over r^{2} - \pars{\verts{\alpha} \pm \ic 0^{+}}^{2}}\,\dd r =\pm\,\ic\pi\sin\pars{\verts{\alpha}\rho}}$$