Integral equation $\int_0^{2 \pi} \frac{1}{\sqrt{1+R^2 \sin^2(x)}}f(R \cos(x)) d x = 1$

Question

Can we prove that there does not exist a function $f$, which satisfies this equation for all $R>0$: $$\int_0^{2 \pi} \frac{1}{\sqrt{1+R^2 \sin^2(x)}} f(R \cos(x))\, dx= 1.$$

Answer 1

Here's some considerations. Change of variables $z=R\cos x$ transforms the integral into $$ 2\int_{-R}^R\frac{f(z)dz}{\sqrt{(1+R^2-z^2)(R^2-z^2)}}. $$ Using Mma it is possible to calculate integrals $$ a_n=2\int_{-R}^R\frac{z^ndz}{\sqrt{(1+R^2-z^2)(R^2-z^2)}}, $$ for example, $a_0=\frac{4 K\left(\frac{R^2}{R^2+1}\right)}{\sqrt{R^2+1}}$, $a_2=4 \sqrt{R^2+1} \left(K\left(\frac{R^2}{R^2+1}\right)-E\left(\frac{R^2}{R^2+1}\right)\right)$, $a_{2n+1}=0\,$, where $K$ and $E$ are complete elliptic integrals of the first and second kind respectively. Maclaurin series expansion for $a_{2n}$ begins with $R^{2n}$ (it can be seem from the initial integral there $n$th term in expansion of $f$ has $R^n \cos^n x$.) This allows to obtain coefficients for $$ f(z)=\sum_{n=0}^\infty c_nz^n $$ from the equality $$ \sum_{n=0}^\infty c_n a_n(R)\equiv1 $$ one by one: $c_0$ from $c_0a_0(R)=1+O(R^2)$, then $c_2$ from $c_0a_0(R)+c_2a_2(R)=1+O(R^4)\,$ etc. The result is $$ f(z)=\frac1{2\pi}\bigg( 1+\frac{z^2}{2}-\frac{7 z^4}{24}+\frac{59 z^6}{240}-\frac{3013 z^8}{13440}+\frac{10147 z^{10}}{48384}-\frac{191833 z^{12}}{967680}+ $$ $$ \frac{52145347 z^{14}}{276756480}-\frac{11939071981 z^{16}}{66421555200}+\frac{55491350861 z^{18}}{322618982400}-\frac{28327222665017 z^{20}}{171633298636800}+\ldots\bigg)= $$ $$ 0.159155+0.079577 z^2-0.0464202 z^4+0.0391256 z^6-0.0356796 z^8+ 0.0333777 z^{10} $$ $$ -0.0315509 z^{12}+ 0.0299873 z^{14}-0.0286076 z^{16}+0.0273751 z^{18}-0.0262677 z^{20}+\ldots $$ So coefficients signs seems to be alternating beginning from the second. Absolute values are monotonously decreasing, but slowly, so if it goes further that way the convergence radius should be 1.