Simplifing double integral. Is it possible to evaluate analytically?

Question

I have the following integral

$$I(\theta, r) = \int_0^R \int_0^{2\pi} \frac{ abr}{(c^2 + b^2/4 + r^2 -2cr \cos (\theta))^{3/2}}d \theta dr,$$

where $a$, $b$ and $c$ are positive numbers and $rdr d\theta$ represents the elementary area on the disk. I want to simplify this double integral to simple integral. This integral is important for my research and it's describing the magnetic field produced by a disc at a point outside. I used Maple to simplify this integral and I obtain something with a elliptic integral and on different versions of Maple I obtain different results. I am having trouble seeing how it can be put into the elliptic form. I am new to elliptic integrals and did not learn about them in a formal setting. Is there a way to show that mathematically in a rigorous way? Is that true? Any help / hints / indication would be highly appreciated.

Answer 1

---

Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the double integral

$$\mathcal{I}{\left(b,c,R\right)}:=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}\mathrm{d}r\,\frac{br}{\left[b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}\right]^{3/2}}.$$

Note: The integral posed in your question can be expressed in this notation as

$$2a\,\mathcal{I}{\left(\frac{b}{2},c,R\right)}=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}\mathrm{d}r\,\frac{abr}{\left[\frac14b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}\right]^{3/2}}.$$

---

For any continuous function $f:\left[-1,1\right]\rightarrow\mathbb{R}$, we have the following integration symmetry:

$$\begin{align} \int_{0}^{2\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)} &=\int_{0}^{\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)}+\int_{\pi}^{2\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)}\\ &~~~~~+\int_{0}^{\pi}\mathrm{d}\vartheta\,f{\left(\cos{\left(2\pi-\vartheta\right)}\right)};~~~\small{\left[\theta=2\pi-\vartheta\right]}\\ &=\int_{0}^{\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)}+\int_{0}^{\pi}\mathrm{d}\vartheta\,f{\left(\cos{\left(\vartheta\right)}\right)}\\ &=2\int_{0}^{\pi}\mathrm{d}\theta\,f{\left(\cos{\left(\theta\right)}\right)}.\\ \end{align}$$

We can show similarly that for any continuous function $g:\left[0,1\right]\rightarrow\mathbb{R}$, we have the following integration symmetries:

$$\int_{0}^{\pi}\mathrm{d}\theta\,g{\left(\sin{\left(\theta\right)}\right)}=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,g{\left(\sin{\left(\theta\right)}\right)},$$

$$\int_{0}^{\pi}\mathrm{d}\theta\,g{\left(\cos^{2}{\left(\theta\right)}\right)}=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,g{\left(\cos^{2}{\left(\theta\right)}\right)}.$$

---

Given fixed but arbitrary $(p,q)\in\mathbb{R}^{2}$ such that $0<q-p^{2}$, the quadratic expression $x^{2}-2px+q$ is positive-definite, i.e.,

$$\forall x\in\mathbb{R}:0<x^{2}-2px+q.$$

Consider the derivative of the following expression: for all $x\in\mathbb{R}$,

$$\begin{align} \frac{d}{dx}\left[\frac{px-q}{\sqrt{x^{2}-2px+q}}\right] &=\frac{1}{\sqrt{x^{2}-2px+q}}\frac{d}{dx}\left[px-q\right]+\left(px-q\right)\frac{d}{dx}\left[\frac{1}{\sqrt{x^{2}-2px+q}}\right]\\ &=\frac{p}{\sqrt{x^{2}-2px+q}}-\frac{\left(px-q\right)}{2\left(x^{2}-2px+q\right)^{3/2}}\frac{d}{dx}\left[x^{2}-2px+q\right]\\ &=\frac{p}{\sqrt{x^{2}-2px+q}}-\frac{\left(px-q\right)\left(2x-2p\right)}{2\left(x^{2}-2px+q\right)^{3/2}}\\ &=\frac{p\left(x^{2}-2px+q\right)}{\left(x^{2}-2px+q\right)^{3/2}}-\frac{\left(px-q\right)\left(x-p\right)}{\left(x^{2}-2px+q\right)^{3/2}}\\ &=\frac{p\left(x^{2}-2px+q\right)-\left(px-q\right)\left(x-p\right)}{\left(x^{2}-2px+q\right)^{3/2}}\\ &=\frac{\left(px^{2}-2p^{2}x+pq\right)-\left(px^{2}-p^{2}x-qx+pq\right)}{\left(x^{2}-2px+q\right)^{3/2}}\\ &=\frac{\left(q-p^{2}\right)x}{\left(x^{2}-2px+q\right)^{3/2}},\\ \end{align}$$

$$\implies\frac{r}{\left(q-2pr+r^{2}\right)^{3/2}}=\frac{d}{dr}\left[-\frac{q-pr}{\left(q-p^{2}\right)\sqrt{q-2pr+r^{2}}}\right];~~~\small{0<q-p^{2}}.$$

Setting $p=c\cos{\left(\theta\right)}\land q=b^{2}+c^{2}$ where $0<b\land0<c$, we have

$$q-p^{2}=b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}=b^{2}+c^{2}\sin^{2}{\left(\theta\right)}>0,$$

and then

$$\frac{br}{\left[b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}\right]^{3/2}}=\frac{d}{dr}\bigg{[}-\frac{b\left[b^{2}+c^{2}-cr\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}}}\bigg{]}.$$

---

Suppose $\left(b,c,R\right)\in\mathbb{R}_{>0}^{3}$. Using the antiderivative established above to perform the integration with respect to $r$, we find

$$\begin{align} \mathcal{I}{\left(b,c,R\right)} &=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}\mathrm{d}r\,\frac{br}{\left[b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}\right]^{3/2}}\\ &=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}\mathrm{d}r\,\frac{d}{dr}\bigg{[}-\frac{b\left[b^{2}+c^{2}-cr\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}-2cr\cos{\left(\theta\right)}+r^{2}}}\bigg{]}\\ &=\int_{0}^{2\pi}\mathrm{d}\theta\,\bigg{[}\frac{b\sqrt{b^{2}+c^{2}}}{b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}}\\ &~~~~~-\frac{b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}-2cR\cos{\left(\theta\right)}+R^{2}}}\bigg{]}\\ &=2\int_{0}^{\pi}\mathrm{d}\theta\,\bigg{[}\frac{b\sqrt{b^{2}+c^{2}}}{b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}}\\ &~~~~~-\frac{b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}-2cR\cos{\left(\theta\right)}+R^{2}}}\bigg{]};~~~\small{symmetry}\\ &=\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}}{b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}}\\ &~~~~~-\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}-2cR\cos{\left(\theta\right)}+R^{2}}}\\ &=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}}{b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}};~~~\small{symmetry}\\ &~~~~~-\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}+R^{2}-2cR\cos{\left(\theta\right)}}}\\ &=2\pi-\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}+R^{2}-2cR\cos{\left(\theta\right)}}},\\ \end{align}$$

where we used the following integration formula to compute the first integral in the second-to-last line above:

$$\begin{align} \int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}}{b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}\sec^{2}{\left(\theta\right)}}{\left(b^{2}+c^{2}\right)\sec^{2}{\left(\theta\right)}-c^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}\sec^{2}{\left(\theta\right)}}{\left(b^{2}+c^{2}\right)\left[1+\tan^{2}{\left(\theta\right)}\right]-c^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2b\sqrt{b^{2}+c^{2}}\sec^{2}{\left(\theta\right)}}{b^{2}+\left(b^{2}+c^{2}\right)\tan^{2}{\left(\theta\right)}}\\ &=\int_{0}^{\infty}\mathrm{d}x\,\frac{2b\sqrt{b^{2}+c^{2}}}{b^{2}+\left(b^{2}+c^{2}\right)x^{2}};~~~\small{\left[\tan{\left(\theta\right)}=x\right]}\\ &=\int_{0}^{\infty}\mathrm{d}y\,\frac{2b}{b^{2}+y^{2}};~~~\small{\left[x=\frac{y}{\sqrt{b^{2}+c^{2}}}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{2}{1+t^{2}};~~~\small{\left[y=bt\right]}\\ &=\pi.\\ \end{align}$$

---

Now, suppose $\left(b,c,R\right)\in\mathbb{R}_{>0}^{3}$ and set

$$p:=\frac{R}{\sqrt{b^{2}+c^{2}}}\land q:=\frac{c}{\sqrt{b^{2}+c^{2}}}.$$

Then, $0<p\land0<q<1$ and

$$\frac{2q}{q-1}<0<\frac{2q}{1+q}<1,$$

$$0<p^{2}+2pq+1\land0<\frac{4pq}{p^{2}+2pq+1}<1.$$

Setting $\eta:=\frac{2q}{q+1}\land\lambda:=\frac{2q}{q-1}\land\kappa:=\sqrt{\frac{4pq}{p^{2}+2pq+1}}$, we obtain the following expression for the remaining integral in terms of standard elliptic integrals:

$$\begin{align} \mathcal{I}{\left(b,c,R\right)} &=2\pi-\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2b\left[b^{2}+c^{2}-cR\cos{\left(\theta\right)}\right]}{\left[b^{2}+c^{2}-c^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{b^{2}+c^{2}+R^{2}-2cR\cos{\left(\theta\right)}}}\\ &=2\pi-\int_{0}^{\pi}\mathrm{d}\theta\,\frac{2\sqrt{1-q^{2}}\left[1-pq\cos{\left(\theta\right)}\right]}{\left[1-q^{2}\cos^{2}{\left(\theta\right)}\right]\sqrt{1+p^{2}-2pq\cos{\left(\theta\right)}}}\\ &=2\pi-\int_{0}^{\pi}\mathrm{d}\theta\,\left[\frac{1-p}{1-q\cos{\left(\theta\right)}}+\frac{1+p}{1+q\cos{\left(\theta\right)}}\right]\frac{\sqrt{1-q^{2}}}{\sqrt{1+p^{2}-2pq\cos{\left(\theta\right)}}};~~~\small{P.F.D.}\\ &=2\pi-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,2\left[\frac{1-p}{1-q\cos{\left(2\theta\right)}}+\frac{1+p}{1+q\cos{\left(2\theta\right)}}\right]\\ &~~~~~\times\frac{\sqrt{1-q^{2}}}{\sqrt{1+p^{2}-2pq\cos{\left(2\theta\right)}}};~~~\small{\left[\theta\mapsto2\theta\right]}\\ &=2\pi-2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\left[\frac{1-p}{1+q-2q\cos^{2}{\left(\theta\right)}}+\frac{1+p}{1-q+2q\cos^{2}{\left(\theta\right)}}\right]\\ &~~~~~\times\frac{\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1-4pq\cos^{2}{\left(\theta\right)}}}\\ &=2\pi-2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\left[\frac{1-p}{1+q-2q\sin^{2}{\left(\theta\right)}}+\frac{1+p}{1-q+2q\sin^{2}{\left(\theta\right)}}\right]\\ &~~~~~\times\frac{\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1-4pq\sin^{2}{\left(\theta\right)}}};~~~\small{\left[\theta\mapsto\frac{\pi}{2}-\theta\right]}\\ &=2\pi-2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\left[\frac{1-p}{1+q}\cdot\frac{1}{1-\frac{2q}{q+1}\sin^{2}{\left(\theta\right)}}+\frac{1+p}{1-q}\cdot\frac{1}{1-\frac{2q}{q-1}\sin^{2}{\left(\theta\right)}}\right]\\ &~~~~~\times\frac{\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}\sqrt{1-\frac{4pq}{p^{2}+2pq+1}\sin^{2}{\left(\theta\right)}}}\\ &=2\pi-\frac{2\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\left[\frac{1-p}{1+q}\cdot\frac{1}{1-\eta\sin^{2}{\left(\theta\right)}}+\frac{1+p}{1-q}\cdot\frac{1}{1-\lambda\sin^{2}{\left(\theta\right)}}\right]\\ &~~~~~\times\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\theta\right)}}}\\ &=2\pi-\frac{1-p}{1+q}\cdot\frac{2\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{1}{1-\eta\sin^{2}{\left(\theta\right)}}\cdot\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\theta\right)}}}\\ &~~~~~-\frac{1+p}{1-q}\cdot\frac{2\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{1}{1-\lambda\sin^{2}{\left(\theta\right)}}\cdot\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\theta\right)}}}\\ &=2\pi-\frac{1-p}{1+q}\cdot\frac{2\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}}\,\Pi{\left(\eta,\kappa\right)}-\frac{1+p}{1-q}\cdot\frac{2\sqrt{1-q^{2}}}{\sqrt{p^{2}+2pq+1}}\,\Pi{\left(\lambda,\kappa\right)},\\ \end{align}$$

where for our purposes here the elliptic integrals of the third kind are defined by

$$\Pi{\left(\phi,\eta,\kappa\right)}:=\int_{0}^{\phi}\mathrm{d}\theta\,\frac{1}{1-\eta\sin^{2}{\left(\theta\right)}}\cdot\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\theta\right)}}};~~~\small{\phi\in\mathbb{R}\land\eta\in\left(-\infty,1\right)\land\kappa\in\left(0,1\right)},$$

$$\Pi{\left(\eta,\kappa\right)}:=\Pi{\left(\frac{\pi}{2},\eta,\kappa\right)}.$$

---