# Integral $\int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx$

I have been struggling with the integral

$$I_n(\alpha) = \frac{1}{\pi} \int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx,$$

where $\alpha$ is real and $n$ is a non-negative integer. It is relatively easy to get the values for specific $n$;

$$I_0(\alpha) = 1,~~I_1(\alpha) = \alpha/2,~~I_2(\alpha) =\frac{1}{8} \left(3 \alpha ^2+4\right), \ldots$$

But how do I get the expression for general $I_n(\alpha)$? I have tried building a recursion but did not quite succeed. I also tried taking the derivatives of 3.661.3 from Gradshteyn-Ryzhik but did not get anything nice.

Edit I

Attempt using the binomial theorem:

\begin{align} \frac{1}{\pi} \int_0^{\pi} dx~ \big( 1 + \alpha\cos (x) \big)^n \cos (x)^n &= \frac{1}{\pi} \int_0^{\pi} dx~ \cos (x)^n \sum_{m=0}^n{ {n}\choose{m}} \alpha^m \cos(x)^m \\ &= \sum_{m=0}^n{ {n}\choose{m}} \alpha^m \frac{1}{\pi} \int_0^{\pi} dx~ \cos (x)^{n+m} \\ &= \sum_{m=0}^n{ {n}\choose{m}} \alpha^m \frac{ 2^{n+m} \pi }{(n+m)! \Gamma\left( \frac{1}{2}(1-n-m) \right)^2} \begin{cases} 1,~~n+m ~ {\rm even}\\ 0,~~n+m ~ {\rm odd} \end{cases} \end{align}

But I did not quite manage to express the last sum in some nice form... the best I got is:

\begin{align} I_n(\alpha) &= \left(1 + (-1)^n\right) \frac{\Gamma (n+1)} {2^{n+1}\Gamma \left(\frac{n+2}{2}\right)^2} \,_3F_2\left(\tfrac{1}{2}(1-n),\tfrac{1}{2}(1+n),-\tfrac{n}{2};\tfrac{1}{2},1+\tfrac{n}{2};\alpha ^2\right) \\ &~~~~+ \left(1 + (-1)^{n+1}\right) \frac{n \Gamma (n+1)} {2^n(n+1) \Gamma \left(\frac{n+1}{2}\right)^2} \alpha \,_3F_2\left(\tfrac{1}{2}(1-n),1-\tfrac{n}{2},1+\tfrac{n}{2};\tfrac{3}{2},\tfrac{1}{2}(3+n);\alpha ^2\right) \end{align}

But it is not very instructive...

P.S. The same can be also be obtained from the solution suggested by orlp.

Edit II

Following a nice suggestion of Igor's I got the result

$$I_n(\alpha) = \frac{2i}{(2n)!} \lim_{z\to i} \frac{d^{2n}}{dz^{2n}} \frac{\left(1 +\alpha + (1-\alpha) z^2 \right)^n \left(1 - z^2\right)^n}{(z+i)^{2n+1}}.$$ However, it is quite tricky now (at least for me) to explicitly compute this derivative. Any suggestions?

• Binomial theorem and reduction formula for $\cos^n x$. – A---B May 3 '17 at 2:30
• Yes, I tried that one and got a sum with an 'odd/even' cases, which could be converted to Hypergeometric fnc (Mathematica told me that)... but this result seemed a bit convolved... I was hoping there would be something more elegant. But possibly I didn't quite do the optimal tricks along the way. – z.v. May 3 '17 at 2:43
• I've added my binomial theorem attempt. – z.v. May 3 '17 at 2:53

This is a proof through small steps without really explaining my thought process, because at this point I spent way too much time on it and my thoughts are lost in the wind. The end result is a series of coefficients for $\alpha$.

$$I_n(\alpha) = \frac{1}{\pi} \int_0^\pi (1+\alpha \cos(x))^n \cos(x)^n dx$$

$$I_n(\alpha) = \frac{1}{\pi} \int_0^\pi \sum_{k=0}^n\binom{n}{k}(\alpha\cos(x))^k \cos(x)^n dx$$

$$I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}\alpha^k\int_0^\pi \cos(x)^{k+n} dx$$

$$I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}\alpha^k\int_0^\pi 2^{-k-n}(e^{ix}+e^{-ix})^{k+n} dx$$

$$I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n\binom{n}{k}2^{-k-n}\alpha^k\int_0^\pi (e^{ix}+e^{-ix})^{k+n} dx$$

From this point on $k+n$ must be even, otherwise the integral is zero:

$$I_n(\alpha) = \frac{1}{\pi} \sum_{k=0}^n \frac{1+(-1)^{k+n}}{2}\binom{n}{k}2^{-k-n}\alpha^k \pi \binom{k+n}{\frac{k + n}{2}}$$

$$I_n(\alpha) = \sum_{k=0}^n\frac{1+(-1)^{k+n}}{2}2^{-n-k}\binom{n}{k}\binom{k+n}{\frac{k + n}{2}} \alpha^k$$

I am not entirely sure the binomial theorem can be avoided, but also the Weierstrass substitution ($(t = \tan x/2)$ transforms your integral into an integral of a rational function from $0$ to infinity, at which point the residue theorem is your friend.

EDIT A similar method is the following: first note that your integrand is even, so the integral is half of the integral from $-\pi$ to $\pi.$ Now, make the substitution $z = \exp(i x),$ so that $\cos x = \frac12\left(z + \frac1z\right),$ and so your integral is (half of) the integral over the unit circle:

$$\begin{multline}-i \int_C ((1+ a (1+z^2)/2z)(1+z^2)/2z)^n \frac{dz}z\\ = \frac{-i}{2^{2n}} \int_C((az^2 + 2 z + a)(1+z^2))^n/ z^{2n+1} dz \\= \frac{-i}{2^{2n}}\int_C (a + 2 z + + 2az^2 + 2 z^3 + az^4)^n/z^{2n+1} d z.\end{multline}$$

So, your goal in life is to find the coefficient of $z^{2 n}$ in $(a + 2 z + + 2az^2 + 2 z^3 + az^4)^n,$ since that will give you the residue at $0.$

• Absolutely the way to go for solving problems like this in general! – Brevan Ellefsen May 3 '17 at 3:24
• Nice suggestion, thanks! Above, I've added the result I obtained using this. It looks tricky to do the derivative though. – z.v. May 3 '17 at 16:00
• @z.v. See a variation after the edit... – Igor Rivin May 3 '17 at 16:22
• @ Igor Rivin Ok, this is quite nice and elegant. Although, to extract the coefficient I was trying to use the multinomial theorem, which also doesn't give me something very elegant. I guess there most likely does not exist some very simple closed form of this integral, so maybe my best bet would be to try and get some solution in a form of some recursion. – z.v. May 3 '17 at 18:58
• @z.v. My answer has a simple closed form. Given $n$ you can immediately compute the coefficient for $a^k$ in the polynomial for $I_n$. – orlp May 4 '17 at 10:56