On integration involving elliptic integral

Question

I'm working on the following integral involving an elliptic integral. $E(m)$ is the complete elliptic integral of second kind with parameter $m=k^2$. Is there a way to solve this? Or, is it impossible to expressed in closed form? $$\int^y_0{\frac{\sqrt{2ax}}{(a+x)^2(a-x)}E\Bigl(\frac{-4ax}{(a-x)^{2}}\Bigl)dx}$$

Answer 1

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Define the function $\mathcal{I}:\{(a,y)\in\mathbb{R}^{2}\mid0<y<a\}\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{I}{\left(a,y\right)}:=\int_{0}^{y}\mathrm{d}x\,\frac{\sqrt{2ax}}{\left(a+x\right)^{2}\left(a-x\right)}E{\left(-\frac{4ax}{\left(a-x\right)^{2}}\right)},$$

where $E{\left(m\right)}$ is the complete elliptic integral of the second kind, which for our purposes here is defined as a function of the elliptic parameter $m$ by the integral representation

$$E{\left(m\right)}:=\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{1-mt^{2}}}{\sqrt{1-t^{2}}};~~~\small{m\le1}.$$

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It's not difficult to show that the elliptic integral $E$ obeys the following Landen relation:

$$E{\left(m\right)}=\sqrt{1-m}\,E{\left(\frac{m}{m-1}\right)};~~~\small{m<1}.$$

Proof:

$$\begin{align} E{\left(m\right)} &=\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{1-mt^{2}}}{\sqrt{1-t^{2}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{1-m\left(1-t^{2}\right)}}{\sqrt{1-t^{2}}};~~~\small{\left[t\mapsto\sqrt{1-t^{2}}\right]}\\ &=\sqrt{1-m}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{1-\frac{m}{m-1}t^{2}}}{\sqrt{1-t^{2}}}\\ &=\sqrt{1-m}\,E{\left(\frac{m}{m-1}\right)}.\\ \end{align}$$

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Suppose $(a,y)\in\mathbb{R}^{2}\land0<y<a$. Rescaling the parameters of $\mathcal{I}$ as well as the integration variable by the same arbitrary scalar factor $p\in\mathbb{R}_{>0}$ yields the following result:

$$\begin{align} \mathcal{I}{\left(pa,py\right)} &=\int_{0}^{py}\mathrm{d}x\,\frac{\sqrt{2pax}}{\left(pa+x\right)^{2}\left(pa-x\right)}E{\left(-\frac{4pax}{\left(pa-x\right)^{2}}\right)}\\ &=\int_{0}^{y}\mathrm{d}u\,\frac{p\sqrt{2papu}}{\left(pa+pu\right)^{2}\left(pa-pu\right)}E{\left(-\frac{4papu}{\left(pa-pu\right)^{2}}\right)};~~~\small{\left[x=pu\right]}\\ &=p^{-1}\int_{0}^{y}\mathrm{d}u\,\frac{\sqrt{2au}}{\left(a+u\right)^{2}\left(a-u\right)}E{\left(-\frac{4au}{\left(a-u\right)^{2}}\right)}\\ &=p^{-1}\mathcal{I}{\left(a,y\right)}.\\ \end{align}$$

Hence,

$$\mathcal{I}{\left(a,y\right)}=p\,\mathcal{I}{\left(pa,py\right)};~~~\small{0<y<a\land0<p},$$

and by choosing $p=a^{-1}$ we can automatically set the first parameter to unity:

$$\mathcal{I}{\left(a,y\right)}=a^{-1}\mathcal{I}{\left(1,a^{-1}y\right)};~~~\small{0<a^{-1}y<1}.$$

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Setting $m=-\frac{4x}{\left(1-x\right)^{2}}$ inside the Landen relation $E{\left(m\right)}=\sqrt{1-m}\,E{\left(\frac{m}{m-1}\right)}$, we have

$$E{\left(-\frac{4x}{\left(1-x\right)^{2}}\right)}=\frac{\left(1+x\right)}{\left(1-x\right)}\,E{\left(\frac{4x}{\left(1+x\right)^{2}}\right)};~~~\small{-1\le x<1}.$$

Then, given $0<z<1$, we have $0<\frac{4z}{(1+z)^{2}}<1$. Setting $b:=\frac{4z}{(1+z)^{2}}$,

we obtain

$$\begin{align} \mathcal{I}{\left(1,z\right)} &=\int_{0}^{z}\mathrm{d}x\,\frac{\sqrt{2x}}{\left(1+x\right)^{2}\left(1-x\right)}E{\left(-\frac{4x}{\left(1-x\right)^{2}}\right)}\\ &=\int_{0}^{z}\mathrm{d}x\,\frac{\sqrt{2x}}{\left(1+x\right)^{2}\left(1-x\right)}\cdot\frac{\left(1+x\right)}{\left(1-x\right)}\,E{\left(\frac{4x}{\left(1+x\right)^{2}}\right)}\\ &=\int_{0}^{z}\mathrm{d}x\,\frac{\sqrt{2}\sqrt{x}}{\left(1+x\right)\left(1-x\right)^{2}}\,E{\left(\frac{4x}{\left(1+x\right)^{2}}\right)}\\ &=\frac{1}{\sqrt{2}}\int_{0}^{z}\mathrm{d}x\,\frac{\left(\frac{2\sqrt{x}}{1+x}\right)}{\left(1-x\right)^{2}}\,E{\left(\left(\frac{2\sqrt{x}}{1+x}\right)^{2}\right)}\\ &=\frac{1}{\sqrt{2}}\int_{0}^{\frac{2\sqrt{z}}{1+z}}\mathrm{d}y\,\frac{2y}{\left(1+\sqrt{1-y^{2}}\right)^{2}\sqrt{1-y^{2}}}\\ &~~~~~\times\frac{\left(1+\sqrt{1-y^{2}}\right)^{2}y}{4\left(1-y^{2}\right)}\,E{\left(y^{2}\right)};~~~\small{\left[\frac{2\sqrt{x}}{1+x}=y\right]}\\ &=\frac{1}{2\sqrt{2}}\int_{0}^{\frac{2\sqrt{z}}{1+z}}\mathrm{d}u\,\frac{u^{2}}{\left(1-u^{2}\right)^{3/2}}\,E{\left(u^{2}\right)};~~~\small{\left[\frac{2\sqrt{x}}{1+x}=u\right]}\\ &=\frac{1}{4\sqrt{2}}\int_{0}^{\frac{4z}{(1+z)^{2}}}\mathrm{d}x\,\frac{\sqrt{x}}{\left(1-x\right)^{3/2}}\,E{\left(x\right)};~~~\small{\left[u=\sqrt{x}\right]}\\ &=\frac{1}{4\sqrt{2}}\int_{0}^{b}\mathrm{d}x\,\frac{\sqrt{x}}{\left(1-x\right)^{3/2}}\,E{\left(x\right)};~~~\small{\left[b:=\frac{4z}{(1+z)^{2}}\in(0,1)\right]}\\ &=\frac{b\sqrt{b}}{4\sqrt{2}}\int_{0}^{1}\mathrm{d}t\,\frac{\sqrt{t}}{\left(1-bt\right)^{3/2}}\,E{\left(bt\right)};~~~\small{\left[x=bt\right]}\\ &=\frac{b^{3/2}}{4\sqrt{2}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{1/2}E{\left(bt\right)}}{\left(1-bt\right)^{3/2}}.\\ \end{align}$$

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At this point I'm not sure how to proceed or if this is as simple as it gets. Perhaps there's something to be done with hypergeometrics or maybe very clever integration by parts? Anyway, hopefully my answer will be useful to you, if not satisfying. Best of luck!