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Let $a, b$ be arbitrary vectors in $\mathbb{R}^3$ and define $\partial B_1(0):=\{r\in\mathbb{R}^3:|r|=1\}$ as the boundary of the three-dimensional unit sphere.

I am trying to evaluate the following integral:

$$I = \iint_{\partial B_1(0)}\frac{d\Omega}{\left(1+a\cdot r\right)\left(1+b\cdot r\right)}$$

So far, I've said that we can, given a vector $a$ and a fixed vector $b$, write the following:

$$a = c b + x$$

where $c = \frac{a\cdot b}{|b|}$ and $x\in\mathbb{R}^3$ such that $x$ orthogonal to $b$.

Then, if we choose our coordinate system so that the positive $z$-axis runs parallel to the vector $b$, then, considering spherical coordinates, the angle between $r$ and $b$ is given by $\theta$, and we have:

\begin{align} I = \int_0^\pi d\theta\sin\theta\int_0^{2\pi}d\phi\frac{1}{\left(1+|a|\cos\theta\right)\left(1+c\cos\theta + x\cdot r\right)} \end{align}

Now, I know that, since $x\cdot b = 0$, $x$ must lie in the $x-y-$plane, so we can express it as $(\rho\cos\zeta, \rho\sin\zeta, 0)$ for some modulus $\rho$ and some polar angle $\zeta$. Thus:

$$x\cdot r = \rho\sin\theta\left(\cos\zeta\cos\phi+\sin\zeta\sin\phi\right) = \rho\sin\theta\cos\left(\phi-\zeta\right)$$

Letting $\phi-\zeta=:\xi$, we have $d\xi = d\phi$ and thus, since all periods of the cosine function are equal, we have:

\begin{align} I&=\int_0^\pi d\theta\frac{\sin\theta}{1+|a|\cos\theta}\int_{-\zeta}^{2\pi-\zeta}\frac{d\xi}{1+c\cos\theta+\rho\sin\theta\cos\xi}\\ &=\int_{-1}^1 \frac{du}{1+|a|u}\int_{0}^{2\pi}\frac{d\xi}{1+cu+\rho\sqrt{1-u^2}\cos\xi} \end{align}

And this is where I'm stumped. I've been thinking of maybe converting the second integral into a contour integral over the unit circle, which reduces to evaluating

$$\oint_{|z|=1}\frac{dz}{z^2+\alpha z + 1}$$

where $\alpha = 2\frac{1+cu}{\rho\sqrt{1-u^2}}$, but this then leads to a very complicated dependancy of the residues on $u$, so I was maybe wondering if there was a simpler way. Is this even the right way to attack the integral $I$? Or is $I$ perhaps so complicated that it admits no simpler solution?

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

lets $\ds{\mathbf{A} \equiv \pars{\mathbf{a} + \mathbf{b}}/2}$ and $\ds{\mathbf{B} \equiv \pars{\mathbf{a} - \mathbf{b}}/2}$. With $\ds{\mrm{\mathbf{R}}\pars{u} \equiv \mathbf{A} + \mathbf{B}u}$:

\begin{align} I & \equiv \iint_{\partial B_1\pars{0}}{\dd\Omega \over \pars{1 + \mathbf{a}\cdot\mathbf{r}}\pars{1 + \mathbf{b}\cdot\mathbf{r}}} \\[5mm] & = \iint_{\partial B_1\pars{0}} 2\int_{-1}^{1}{\dd u \over \bracks{\pars{1 + \mathbf{a}\cdot\mathbf{r}}\pars{1 + u} + \pars{1 + \mathbf{b}\cdot\mathbf{r}}\pars{1 - u}}^{\, 2}} \,\dd\Omega \\[5mm] & = {1 \over 2}\int_{-1}^{1}\iint_{\partial B_1\pars{0}}{\dd\Omega \over \bracks{1 + \mrm{\mathbf{R}}\pars{u}\cdot\mathbf{r}}^{\, 2}}\,\dd u = {1 \over 2}\int_{-1}^{1}\int_{0}^{2\pi}\int_{0}^{\pi}{\sin\pars{\theta}\,\dd\theta\,\dd\phi \over \bracks{\rule{0pt}{5mm}% 1 + \verts{\mrm{\mathbf{R}}\pars{u}}\cos\pars{\theta}}^{\, 2}}\,\dd u \\[5mm] & = \pi\int_{-1}^{1}\bracks{{1 \over 1 - \verts{\mrm{\mathbf{R}}\pars{u}}} - {1 \over 1 + \verts{\mrm{\mathbf{R}}\pars{u}}}}\,\dd u \\[5mm] & = {\pi \over B}\int_{-B}^{B}{\dd u \over 1 - \root{u^{2} + 2\mathbf{A}\cdot\hat{\mathbf{B}}\, u + A^{2}}} - {\pi \over B}\int_{-B}^{B}{\dd u \over 1 + \root{u^{2} + 2\mathbf{A}\cdot\hat{\mathbf{B}}\, u + A^{2}}} \end{align}

where I used the notation $\ds{P \equiv \verts{\mathbf{P}}}$. Both integrals can be evaluated by means of an Euler Substitution.

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  • $\begingroup$ This is excellent. Many thanks! I would have never come up with the idea of writing the initial product as the integral of a different quantity. $\endgroup$ – Tom May 8 '18 at 12:37
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    $\begingroup$ @Tom The first step is one of the Feynman Parametrizations which is widely used to calculate his Diagrams. Check the link for more information. It's quite useful. $\endgroup$ – Felix Marin May 8 '18 at 21:58
  • $\begingroup$ Thanks for the resource @Felix Marin. $\endgroup$ – Tom May 8 '18 at 22:51
  • $\begingroup$ @Tom $\large{\bullet\quad\bullet \brace \smile}$ $\endgroup$ – Felix Marin May 8 '18 at 23:00

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