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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.

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  • $\begingroup$ Unbalanced parentheses in denominator. Did you at least compute the inner integral ? $\endgroup$ Commented Sep 27, 2021 at 10:35
  • $\begingroup$ yes, in maple, but it gives me something too complicated $\endgroup$
    – Mihaela
    Commented Sep 27, 2021 at 10:39
  • $\begingroup$ I need a result that depends on a, b and c $\endgroup$
    – Mihaela
    Commented Sep 27, 2021 at 10:47
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    $\begingroup$ The quickest way might be to consult the Handbook of Elliptic Integrals for Engineers and Physicists by Byrd & Friedman. $\endgroup$ Commented Sep 28, 2021 at 9:19
  • 1
    $\begingroup$ Thanks. I will look on that book also $\endgroup$
    – Mihaela
    Commented Sep 28, 2021 at 11:18

1 Answer 1

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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

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

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


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  • $\begingroup$ Can you explain why for very high c, like c = 40000, if I do numerically the integral $I(b,c,R)$ with the double integral and with the elliptic integral, I do not obtain the same result? It seems that it's not valid this approach anymore. $\endgroup$
    – Mihaela
    Commented Dec 15, 2021 at 19:37
  • $\begingroup$ @Mihaela I did a numerical check and didn't find any problem. Are you sure you used the right argument convention for the elliptic integrals? The conventions for the software you used might differ from mine. $\endgroup$
    – David H
    Commented Dec 17, 2021 at 13:37
  • $\begingroup$ I am using Maple where the elliptic integral of third kind is defined as $$EllipticPi(z, v, k) =\int_0^z( \frac {1}{ (1-vt^2) (\sqrt( 1 -t^2)) (\sqrt( 1 -k^2 t^2 )) })$$ and I use $EllipticPi(\eta, \kappa), EllipticPi(\lambda, \kappa)$. $\endgroup$
    – Mihaela
    Commented Dec 17, 2021 at 17:31
  • $\begingroup$ For $R:= 20, b := 6000, c := 50000$, $I(b, c, R) = 1.5 \cdot 10^{-8} + (3.120980860 \cdot 10^{-13})I$ using elliptics integrals and $I(b, c, R) =5.903877203*10^{-8} $ using the double integral. $\endgroup$
    – Mihaela
    Commented Dec 17, 2021 at 17:34
  • $\begingroup$ It can be something numerically? $\endgroup$
    – Mihaela
    Commented Dec 17, 2021 at 17:37

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