I have the following integral equation stemming from a physical problem (connected with BCS theory): $$ 1 = \frac{2}{\pi}u \int_0^1 dx \sqrt{\frac{1-x^2}{u^2\phi^2 + x^2}}, $$ where $u$ is a positive parameter and I have to solve the equation for the variable $\phi$, which is acceptable only within the range $-1/2 < \phi < 1/2$. I guess that asking for an analytic expression $\phi(u)$ is probably too much, but at least I would like an approximated solution when $u\ll 1$. From physical considerations and numerical analysis I know that the solution should be something like $\phi \sim e^{-1/u}$, however I have no clue about how to prove it. Can anyone point me to the right direction?
I am not able to transform this integral into elliptic functions, and Wolfram Mathematica can't solve it either. Moreover, I am not sure how to expand for small values of $u$... I am thinking that there's no way out, except numerics.
Assuming $a>0$ and $x>0$, using the elliptic integral of the second kind $$I=\int \sqrt{\frac{1-x^2}{a^2+x^2}}\,dx=-i E\left(i \sinh ^{-1}\left(\frac{x}{a}\right)|-a^2\right)$$ $$J=\int_0^1 \sqrt{\frac{1-x^2}{a^2+x^2}}\,dx=-i E\left(i \sinh ^{-1}\left(\frac{1}{a}\right)|-a^2\right)$$ So, with $a=u\phi$, the equation to be solved for $\phi$ is $$1+\frac{2 i }{\pi }\, u\, E\left(i \sinh ^{-1}\left(\frac{1}{u \phi }\right)|-u^2 \phi ^2\right) =0$$
This equation cannot be inversed but any numerical method (such as Newton) should work as a charm $\cdots$ provided a reasonable estimate.
I had no problem using $\phi_0=e^{-1/u}$
But, as usual, @Svyatoslav's estimate $$\phi_0=\frac 4 u \exp\left( -\left(\frac{\pi }{2 u}+1\right)\right)$$ is extremely good.
Edit
Doing what @user170231 suggested in comments (sorry fot not having paid a sufficient attention), we have somthing simpler (even if there is one hypergeometric function). But, it can be simplified as $$J=\frac{\left(a^2+1\right) }{a}K\left(-\frac{1}{a^2}\right)-a E\left(-\frac{1}{a^2}\right)$$
The equation is then $$1-\frac{2 \left(u^2 \phi ^2+1\right) K\left(-\frac{1}{u^2 \phi ^2}\right)}{\pi \phi }+\frac{2 u^2 \phi E\left(-\frac{1}{u^2 \phi ^2}\right)}{\pi }=0$$ which can be expanded as a series
$$1+\sum_{n=0}^\infty A_n\, u^{2n+1}=0$$ where $$A_0=-\frac{2 \left(\log \left(\frac{u}{4}\right)+\log (\phi)+1\right)}{\pi }$$ $$A_1=\frac{\phi ^2 \left(\log \left(\frac{u}{4}\right)+\log (\phi)\right)}{2 \pi }$$
$A_0$ is the term given by @Svyatoslav.
I suppose that we are not very far from a solution in terms of the generalized Lambert function.
Using $$\phi_0=\frac 4 u \exp\left( -\left(\frac{\pi }{2 u}+1\right)\right)$$ use Newton method $$\phi_{n+1}=\phi_n-\frac {f(\phi_n)}{f'(\phi_n)}$$ with $$f'(\phi)=\frac{2 u^2 }{\pi }\left(E\left(-\frac{1}{u^2 \phi ^2}\right)-K\left(-\frac{1}{u^2 \phi ^2}\right)\right)$$
Update
Let $\phi=\frac{1}{u \sqrt{x}}$ and we look for the inverse of $$\frac 2 \pi u=\frac{ \sqrt{x}}{ (x+1) K(-x)- E(-x)}\tag 1$$ where the rhs is a nice function (its reciprocal looks like a Lambert function).
Expand for large $x$ to $O\left(\frac{1}{x^{3/2}}\right)$ and let $x=e^t$ to obtain $$e^{-t}=4\, \frac {t+2 (2 \log (2)-1) }{t+ \left(\frac{u}{\pi }+4 \log (2)\right) }$$
and the solution is given in terms of the generalized Lambert function (have a look at equation $(4)$).
Expanding the rhs of $(1)$ as a series, we have avery good asymptotics $$\frac 2 \pi u=\frac 1 t \sum_{n=0}^\infty (-1)^{n}\, {P_n(t)}\,\left(\frac{x}{t}\right)^n \quad \text{where} \quad x=\frac{1}{16} e^{2( t+1)}$$ the first polynomials being $$\left( \begin{array}{cc} n & P_n(t) \\ 0 & 1 \\ 1 & \frac{t+1}{4} \\ 2 & \frac{14 t^2+17 t+8}{128} \\ 3 & \frac{90 t^3+113 t^2+78 t+24}{1536} \\ 4 & \frac{6948 t^4+8551 t^3+6860 t^2+3360 t+768}{196608} \\ 5 & \frac{90900 t^5+107183 t^4+91810 t^3+55380 t^2+21120 t+3840}{3932160}\\ \end{array} \right)$$