Solution (continued)

We see that the eigenvalue of the eigenfunction $ f_{(r,s)} $ is given by

$$\lambda_{(r,s)} = \int_{-\pi}^{\pi} \sum_{k=0}^{\infty} a^{-k} e^{i(2k+1)(r\cos(t)+s\sin(t))} dt = \int_{-\pi}^{\pi} \sum_{k=0}^{\infty} a^{-k} e^{i(2k+1)\sqrt{r^2+s^2}\cos(t+\phi)} dt$$

for an appropriately chosen $ \phi $. Thus we need only actually consider $ \lambda_{(r,0)} $, which we from now on denote $ \lambda(r) $. Then some calculation gives us that the integral is

$$\int_{-\pi}^{\pi} \frac{a(a-1)\cos(r\cos(t))}{(a-1)^2+4a\sin^2(r\cos(t))} dt$$

We can show that (and this will suffice) that when $ a $ is near $ 1 $,

$$\int_{0}^{\frac{\pi}{2}} \frac{(a-1)\cos(r\cos(t))}{(a-1)^2+4a\sin^2(r\cos(t))} dt \geq -4(a-1)^{-\frac{3}{4}}-\frac{\pi}{2}$$


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