Asymptotic expansion of double integral

Question

Define $$G(\theta) = \int\limits_0^\infty \int\limits_0^{2\pi} \frac{r\,e^{-r^2/2t}}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}} \mathrm{d} r \,\mathrm{d} \varphi$$ Clearly, for $\theta=0$, this does not converge. However, I would like to obtain an asymptotic expansion for $\theta\searrow 0$.

How could I approach this problem? The function $$F(r, \theta) = \int_0^{2\pi} \frac{\mathrm{d}\varphi}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}}$$ looks somewhat like an elliptic integral, but I have no experience in dealing with these and I have no idea how this could help.

Any thoughts on how to approach this problem?

Answer 1

Part 1: Expression of inner integral as an elliptic integral

You are right about the inner integral which you name $F{(r,\theta)}$ resembling an elliptic integral, and indeed it can expressed (with some difficulty) in terms of an elliptic integral of the first kind. For this reason I would avoid naming your function with a capital $F$, since this is usually reserved for a specific elliptic integral. In my work below, I will use a lowercase $f$ instead (and I've also elected to change $r$ to $\rho$ out of stylistic preferences). Argument conventions for defining elliptic integrals vary, but I will here be adopting the one used by Gradshteyn & Ryzhik since I will be invoking identities from their tome as part of my answer; they define:

$$F{\left(\varphi,k\right)}:=\int_{0}^{\varphi}\frac{\mathrm{d}\alpha}{\sqrt{1-k^2\sin^2{\alpha}}}=\int_{0}^{\sin{\varphi}}\frac{\mathrm{d}x}{\sqrt{(1-x^2)(1-k^2x^2)}},~~~\text{where }k^2<1;\\ K{\left(k\right)}:=F{\left(\frac{\pi}{2},k\right)}.$$

With that long-winded preamble out of the way, let's get to work solving the problem at hand. For real parameters $\theta,\rho\in\mathbb{R}\setminus\{n\pi|n\in\mathbb{Z}\}$ such that $\theta\neq\rho$,

$$\begin{align} f{(\theta,\rho)} &:=\int_{0}^{2\pi}\frac{\mathrm{d}\varphi}{\sqrt{1-\left(\cos{\theta}\cos{\rho}+\sin{\theta}\sin{\rho}\cos{\varphi}\right)^2}}\\ &=\int_{0}^{\pi}\frac{2\,\mathrm{d}\varphi}{\sqrt{1-\left(\cos{\theta}\cos{\rho}+\sin{\theta}\sin{\rho}\cos{\varphi}\right)^2}}\\ &=\int_{1}^{-1}\frac{-2\,\mathrm{d}x}{\sqrt{1-x^2}\sqrt{1-\left(\cos{\theta}\cos{\rho}+x\sin{\theta}\sin{\rho}\right)^2}};~~~\cos{\varphi}\rightarrow x\\ &=\int_{-1}^{1}\frac{2\,\mathrm{d}x}{\sqrt{1-x^2}\sqrt{1-\left(\cos{\theta}\cos{\rho}+x\sin{\theta}\sin{\rho}\right)^2}}\\ &=\int_{-1}^{1}\frac{2\left|\csc{\theta}\right|\left|\csc{\rho}\right|\,\mathrm{d}x}{\sqrt{1-x^2}\sqrt{\csc{\theta}\csc{\rho}-\left(\cot{\theta}\cot{\rho}+x\right)^2}}\\ &=\int_{-1}^{1}\frac{2\left|\csc{\theta}\right|\left|\csc{\rho}\right|\,\mathrm{d}x}{\sqrt{1-x^2}\sqrt{\left(\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}-x\right)\left(\csc{\theta}\csc{\rho}+\cot{\theta}\cot{\rho}+x\right)}}\\ &=\small{2\left|\csc{\theta}\right|\left|\csc{\rho}\right|\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^2}\sqrt{\left(\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}-x\right)\left(\csc{\theta}\csc{\rho}+\cot{\theta}\cot{\rho}+x\right)}}}.\\ \end{align}$$

Since the plan is to ultimately consider the limit as $\theta\searrow0$, we may go ahead and restrict $\theta$ to the interval $0<\theta<\pi$. Because we shall need to integrate over $\rho$ from $\rho=0$ to $\rho\to+\infty$, it's not necessarily obvious at first that we can likewise restrict $\rho$ to the same interval, but since it is true that $f{(\theta,\rho+\pi)}=f{(\theta,\rho)}$ it follows that we can suppose $0<\rho<\pi$ without loss of generality.

Note that with the further restrictions $0<\theta,\rho<\pi$, we have the inequalities

$$\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}>1\\ \iff 1-\cos{\theta}\cos{\rho}>\sin{\theta}\sin{\rho}\\ \iff 1>\cos{\theta}\cos{\rho}+\sin{\theta}\sin{\rho}\\ \iff 1>\cos{\left(\theta-\rho\right)},$$

and

$$-1>-\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}\\ \iff -\sin{\theta}\sin{\rho}>-1-\cos{\theta}\cos{\rho}\\ \iff \cos{\theta}\cos{\rho}-\sin{\theta}\sin{\rho}>-1\\ \iff \cos{\left(\theta+\rho\right)}>-1.$$

In order to continue, I shall merely quote without proof a definite integral from Gradshteyn's Table of Integrals, Series, and Products, specifically proposition 3.147(4) on p.324 which states:

Given $a>b\ge u>c>d$, $$\int_{c}^{u}\frac{\mathrm{d}x}{\sqrt{(a-x)(b-x)(x-c)(x-d)}}=\frac{2}{\sqrt{(a-c)(b-d)}}F{(\delta,q)},$$ where $\delta=\arcsin{\sqrt{\frac{(b-d)(u-c)}{(b-c)(u-d)}}},~~q=\sqrt{\frac{(b-c)(a-d)}{(a-c)(b-d)}}$.

Letting $b=u=1$, $c=-1$, $a=\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}$, and $d=-\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}$, we have $\delta=\frac{\pi}{2}$ and

$$\begin{align} q &=\sqrt{\frac{2(a-d)}{(1+a)(1-d)}}\\ &=\sqrt{\frac{2((\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho})-(-\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}))}{(1+(\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}))(1-(-\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}))}}\\ &=\sqrt{\frac{4\csc{\theta}\csc{\rho}}{(1+\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho})(1+\csc{\theta}\csc{\rho}+\cot{\theta}\cot{\rho})}}\\ &=\sqrt{\frac{4\sin{\theta}\sin{\rho}}{\left(\sin{\theta}+\sin{\rho}\right)^2}}\\ &=\frac{2\sqrt{\sin{\theta}\sin{\rho}}}{\left|\sin{\theta}+\sin{\rho}\right|}.\\ \end{align}$$

Thus, for $0<\theta\neq\rho<\pi$, we have:

$$\begin{align} f{(\theta,\rho)} &=\small{2\left|\csc{\theta}\right|\left|\csc{\rho}\right|\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^2}\sqrt{\left(\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho}-x\right)\left(\csc{\theta}\csc{\rho}+\cot{\theta}\cot{\rho}+x\right)}}}\\ &=2\left|\csc{\theta}\right|\left|\csc{\rho}\right|\cdot\frac{2\,F{\left(\frac{\pi}{2},\frac{2\sqrt{\sin{\theta}\sin{\rho}}}{\left|\sin{\theta}+\sin{\rho}\right|}\right)}}{\sqrt{(1+\csc{\theta}\csc{\rho}-\cot{\theta}\cot{\rho})(1+\csc{\theta}\csc{\rho}+\cot{\theta}\cot{\rho})}}\\ &=\frac{4\left|\csc{\theta}\right|\left|\csc{\rho}\right|}{\sqrt{(\csc{\theta}+\csc{\rho})^2}}\,K{\left(\frac{2\sqrt{\sin{\theta}\sin{\rho}}}{\left|\sin{\theta}+\sin{\rho}\right|}\right)}\\ &=\frac{4}{\sin{\theta}+\sin{\rho}}\,K{\left(\frac{2\sqrt{\sin{\theta}\sin{\rho}}}{\sin{\theta}+\sin{\rho}}\right)}\\ &=4\csc{\theta}\,\frac{1}{1+\frac{\sin{\rho}}{\sin{\theta}}}\,K{\left(\frac{2\sqrt{\frac{\sin{\rho}}{\sin{\theta}}}}{1+\frac{\sin{\rho}}{\sin{\theta}}}\right)}\\ &=\begin{cases} 4\csc{\theta}\,K{\left(\frac{\sin{\rho}}{\sin{\theta}}\right)};~~~\rho<\theta\\ 4\csc{\rho}\,K{\left(\frac{\sin{\theta}}{\sin{\rho}}\right)};~~~\theta<\rho.\\ \end{cases}\\ \end{align}$$

---

Part 2: Representation of double integral as a single integral over bounded domain

For $t,\theta\in\mathbb{R}^{+}$ with $0<\theta<\pi$,

$$\begin{align} G{\left(t,\theta\right)} &:=\int_{0}^{\infty}\mathrm{d}r\int_{0}^{2\pi}\mathrm{d}\varphi\,\frac{r\,e^{-\frac{r^2}{2t}}}{\sqrt{1-\left(\cos{\theta}\cos{r}+\sin{\theta}\sin{r}\cos{\varphi}\right)^2}}\\ &=\int_{0}^{\infty}\mathrm{d}r\,r\,e^{-\frac{r^2}{2t}}\int_{0}^{2\pi}\frac{\mathrm{d}\varphi}{\sqrt{1-\left(\cos{\theta}\cos{r}+\sin{\theta}\sin{r}\cos{\varphi}\right)^2}}\\ &=\int_{0}^{\infty}\rho\,e^{-\frac{\rho^2}{2t}}\,f{\left(\theta,\rho\right)}\,\mathrm{d}\rho\\ &=\sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}r\,e^{-\frac{r^2}{2t}}\,f{\left(\theta,r\right)}\,\mathrm{d}r\\ &=\sum_{n=0}^{\infty}\int_{0}^{\pi}\left(\rho+n\pi\right)\,e^{-\frac{\left(\rho+n\pi\right)^2}{2t}}\,f{\left(\theta,\rho+n\pi\right)}\,\mathrm{d}\rho;~~~r\rightarrow\rho+n\pi\\ &=\sum_{n=0}^{\infty}\int_{0}^{\pi}\left(\rho+n\pi\right)\,e^{-\frac{\left(\rho+n\pi\right)^2}{2t}}\,f{\left(\theta,\rho\right)}\,\mathrm{d}\rho\\ &=\int_{0}^{\pi}\sum_{n=0}^{\infty}\left[\left(\rho+n\pi\right)\,e^{-\frac{\left(\rho+n\pi\right)^2}{2t}}\right]\,f{\left(\theta,\rho\right)}\,\mathrm{d}\rho\\ &=:\int_{0}^{\pi}h{\left(t,\rho\right)}\,f{\left(\theta,\rho\right)}\,\mathrm{d}\rho,\\ \end{align}$$

where in the last line we've defined the new auxiliary function $h{\left(t,\rho\right)}$ by the infinite series

$$h{\left(t,\rho\right)}:=\sum_{n=0}^{\infty}\left(\rho+n\pi\right)\,e^{-\frac{\left(\rho+n\pi\right)^2}{2t}}.$$

I'm not completely sure, but I believe the function $h{\left(t,\rho\right)}$ can be written in terms of elliptic theta functions somehow.

Answer 2

The following is not a proof, however numerics shows that the expressions seem to be correct:

In your integral for $G(\theta)$, the denominator $\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}$ is responsible for the divergence. In fact it becomes minimal for $r \to r_n = n\pi + \theta \cos(\varphi)$ (to first order in $\theta$), where it approaches the value $\theta |\sin(\varphi)|$. We expand the denominator close to this minimal value and have the approximation $$ 1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2 \approx \theta^2 \sin^2(\varphi) + (r- r_n)^2.$$ As the denominator is only large close to the points $r=r_n$, we can put for a first approximation the corresponding value in the numerator.

It is thus useful to write the integral $G(\theta)$ as $G(\theta) =\int\limits_0^{2\pi}\! d\varphi \sum_n g_n$ with $$ \begin{align} g_0 &= \int_0^{\pi/2}\!dr \frac{r\,e^{-r^2/2t}}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}}\\ g_{n\geq1} &= \int_{(n-1/2) \pi}^{(n+1/2)\pi}\!dr \frac{r\,e^{-r^2/2t}}{\sqrt{1-(\sin\theta\sin r \cos\varphi + \cos \theta \cos r)^2}}. \end{align} $$

The individual integrals, we evaluate with logarithmic accuracy $$ \begin{align} g_{n\geq 1} &\approx \int_{-\pi/2}^{\pi/2}\!d\delta \frac{r_n\,e^{-r_n^2/2t}}{ \sqrt{ \theta^2 \sin^2(\varphi) + \delta^2} } \approx 2 r_n\,e^{-r_n^2/2t} \int_{ \theta \sin(\varphi)}^1 \frac{1}{\delta}= 2 r_n\,e^{-r_n^2/2t} \left|\ln\theta |\sin(\varphi)| \right|\\ &\approx 2 r_n\,e^{-r_n^2/2t} (|\ln\theta| + |\ln \sin(\varphi)|) \end{align} $$ For $n=0$, the value is only half of the one above as the range of integration is halved.

In total, we obtain the estimate $$ G(\theta) \sim 2\pi \sum_{n\geq1} n e^{-\pi^2 n^2/2 t} \int_0^{2\pi}\!d\varphi \, (|\ln\theta| + |\ln \sin(\varphi)|) \sim 4\pi^2 |\ln\theta| \sum_{n\geq 1} n e^{-\pi^2 n^2/2 t}. $$

The sum can only be evaluated in terms of elementary functions in the limit $t\ll1$ and $t\gg1$. In the former case, we have (taking only the term with $n=1$) $$ G(\theta) \sim 4 \pi^2 e^{-\pi^2/2 t} |\ln \theta|, \qquad t \ll 1.$$ In the later case, we have $$ G(\theta) \sim 4 \pi^2 |\ln\theta| \int_0^\infty\!dn\,ne^{-\pi^2 n^2/2 t} = 4 t |\ln\theta|, \qquad t \gg 1.$$