A question regarding Czarnowski's cat function

Solution 1:

We seek a reduction formula for integrands of the form $$\sin^{2n} x.$$ Integration by parts with the choice $$u = \sin^{2n-1} x, \quad du = (2n-1) \sin^{2n-2} x \cos x \, dx, \\ dv = \sin x \, dx, \quad v = -\cos x, $$ gives $$\begin{align*} I_n (x) &= \int \sin^{2n} x \, dx = -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} x \cos^2 x \, dx \\ &= -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} (1 - \sin^2 x) \, dx \\ &= -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) - (2n-1)I_n(x). \end{align*}$$ Consequently, $$I_n(x) = \frac{1}{2n} \left( -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) \right).$$ On the interval $[0,\pi]$, we then have $$I_n = \int_{x=0}^\pi I_n(x) \, dx = \frac{2n-1}{2n} I_{n-1}.$$ And for $n = 0$, we trivially have $$I_0(x) = 1, \quad I_0 = \pi.$$ It follows that $$I_n = \pi \prod_{k=1}^n \frac{2k-1}{2k} = \pi \prod_{k=1}^n \frac{(2k-1)(2k)}{(2k)^2} = \frac{(2n)!}{4^n (n!)^2} \pi = \binom{2n}{n}\frac{\pi}{4^n}.$$ Therefore, since $c(x,n+2) \le c(x,n)$, $$\begin{align*} \int_{x=-\pi}^\pi c(x,18) - c(x,26) \, dx &= 2 \int_{x=0}^\pi \sin^{18} x - \sin^{26} x \, dx \\ &= 2(I_{9} - I_{13}) \\ &= 2\pi \left(\binom{18}{9}\frac{1}{4^{9}} - \binom{26}{13} \frac{1}{4^{13}}\right) \\ &= \frac{\pi}{2^{25}}\left(\binom{18}{9} 4^4 - \binom{26}{13}\right) \\ &= \frac{255765 \pi }{4194304}. \end{align*}$$