A list the integrals and identities used in solving the problem in this post.
Throughout this post we consider $0<\alpha, \gamma,r \leq 1$ and $a, b, c, k$ is a non-negative integer.
(1) I'm not sure where I saw this. I'll update the reference if I find it. UPDATE: I found the reference. It's given in Advances in Combinatorial Methods and Applications to Probability and Statistics
$\displaystyle\sum_{a,b,c \geq 0 \\ a+b+c=n \\ a-c=k}\binom{n}{a,b,c}\alpha^a(1-\alpha-\gamma)^b\gamma^c=\frac{(\alpha/\gamma)^{k/2}}{\pi}\int_0^\pi \cos k\theta\text{ } (1-\alpha-\gamma+2\sqrt{\alpha\gamma}\cos \theta)^n \, d\theta$
(2) This is given in Generality of algebra. Anyway, this is easy to prove if we recognize $r^k \cos kx=\text{Re}(r^k e^{ikx})$ and it's just a geometric series.
$1+r\cos x+r^2 \cos 2x+r^3 \cos 3x+\cdots=\displaystyle\frac{1-r\cos x}{1-2r\cos x+r^2}$
(3) This is just pattern recognition by using different values in Mathematica.
$\displaystyle\int \frac{1-r \cos x}{1-2r \cos x +r^2}\,dx=\frac{x}{2}+\tan^{-1}\left(\frac{1+r}{1-r}\tan\left(\frac{x}{2}\right)\right)+\text{constant}$
Interestingly,
$\displaystyle\frac{1}{\pi}\int_0^\pi \frac{1-r \cos x}{1-2r \cos x +r^2}\,dx=H[1-r]$
where $H[x]$ is the Heaviside Unit step function.
(4) I definitely saw this Wikipedia. But again I couldn't find it now. UPDATE: Found this too in List of definite integrals.
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{\cos kx}{\sec \phi-\cos x}\,dx=\frac{(\sec \phi - \tan \phi)^k}{\tan \phi}$
(5) Using $k=0$ in (4),
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{1}{\sec \phi-\cos x}\,dx=\cot\phi$
(6) Multiplying (4) by $r^k$ and summing across all $k$,
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{1+r\cos x+r^2 \cos 2x+r^3 \cos 3x+\cdots}{\sec \phi-\cos x}\,dx=\frac{\cot \phi}{1-r(\sec \phi - \tan \phi)}$
Until then
Yours Aye
Me
Throughout this post we consider $0<\alpha, \gamma,r \leq 1$ and $a, b, c, k$ is a non-negative integer.
(1) I'm not sure where I saw this. I'll update the reference if I find it. UPDATE: I found the reference. It's given in Advances in Combinatorial Methods and Applications to Probability and Statistics
$\displaystyle\sum_{a,b,c \geq 0 \\ a+b+c=n \\ a-c=k}\binom{n}{a,b,c}\alpha^a(1-\alpha-\gamma)^b\gamma^c=\frac{(\alpha/\gamma)^{k/2}}{\pi}\int_0^\pi \cos k\theta\text{ } (1-\alpha-\gamma+2\sqrt{\alpha\gamma}\cos \theta)^n \, d\theta$
(2) This is given in Generality of algebra. Anyway, this is easy to prove if we recognize $r^k \cos kx=\text{Re}(r^k e^{ikx})$ and it's just a geometric series.
$1+r\cos x+r^2 \cos 2x+r^3 \cos 3x+\cdots=\displaystyle\frac{1-r\cos x}{1-2r\cos x+r^2}$
(3) This is just pattern recognition by using different values in Mathematica.
$\displaystyle\int \frac{1-r \cos x}{1-2r \cos x +r^2}\,dx=\frac{x}{2}+\tan^{-1}\left(\frac{1+r}{1-r}\tan\left(\frac{x}{2}\right)\right)+\text{constant}$
Interestingly,
$\displaystyle\frac{1}{\pi}\int_0^\pi \frac{1-r \cos x}{1-2r \cos x +r^2}\,dx=H[1-r]$
where $H[x]$ is the Heaviside Unit step function.
(4) I definitely saw this Wikipedia. But again I couldn't find it now. UPDATE: Found this too in List of definite integrals.
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{\cos kx}{\sec \phi-\cos x}\,dx=\frac{(\sec \phi - \tan \phi)^k}{\tan \phi}$
(5) Using $k=0$ in (4),
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{1}{\sec \phi-\cos x}\,dx=\cot\phi$
(6) Multiplying (4) by $r^k$ and summing across all $k$,
$\displaystyle\frac{1}{\pi}\int_0^\pi\frac{1+r\cos x+r^2 \cos 2x+r^3 \cos 3x+\cdots}{\sec \phi-\cos x}\,dx=\frac{\cot \phi}{1-r(\sec \phi - \tan \phi)}$
Until then
Yours Aye
Me
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