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The following integral arises in a problem in elasticity

\begin{equation*} \int_0^{\pi}\frac{\cos n\eta\,d\eta}{1+a^2+2a\cos m\eta} \end{equation*}

where $m$ and $n$ are both positive integers, and $0<a\le 1/(m-1)$. Mathematica can do this for special choices of $m$ and $n$, but I want a general result for arbitrary $a>0$, $m$ and $n$.

A similar but simpler problem is stated and answered at

Using residues to evaluate the integral $\int_{-\pi}^{\pi} \frac{\cos(n\theta)}{1-2a\cos(\theta)+a^2}d\theta$, $|a|<1$

Blue
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Jog
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1 Answers1

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Using techniques used in the linked question, we can show that

$$\begin{align}I=\int_{0}^{\pi}d\theta\frac{\cos n\theta}{1+a^2+2a\cos m\theta}&=\frac{1}{2}\int_{-\pi}^{\pi}d\theta\frac{\cos n\theta}{1+a^2+2a\cos m\theta}\\&=\frac{1}{2ia}\int_{|z|=1}dz\frac{z^{m+n-1}}{(z^m+a)(z^m+\frac{1}{a})}\end{align}$$

The latter integral is amenable to residue techniques. Assume $0\leq a <1$. Then only the poles attributed to $z^m+a$ are located within the unit circle and contribute to the sum. The poles occur at:

$$z_k=a^{1/m}e^{i\pi/m}e^{\frac{2k\pi i}{m}}$$

Performing the simple residue computation we obtain $(\omega=e^{2\pi i/m}$)

$$I=\sum_{k=0}^{m-1}\frac{z^{m+n-1}}{(z^m+a)'(z^m+\frac{1}{a})}\Bigg|_{z=z_k}=\frac{\pi}{m(1-a^2)}(ae^{i\pi})^{n/m}\sum_{k=0}^{m-1}\omega^{nk}$$

However we note that $$\sum_{k=0}^{m-1}(\omega^n)^k=\frac{\omega^{mn}-1}{\omega^n-1}=\begin{Bmatrix}m& ,m|n\\0&, \text{else}\end{Bmatrix}$$

and thus, denoting $d=\frac{n}{m}\in\mathbb{N}$, we finally conclude

$$I=(-1)^d\frac{\pi a^d}{1-a^2}\delta_{n,md}$$

DinosaurEgg
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  • Even the handbooks such as Gradshteyn give the result only for $m=1$ and $m=2$. So this should be very helpful. Brilliant! – Jog May 07 '20 at 05:50
  • From the elasticity solution alluded to before, and the above result, we get for $0\le \xi\le 1$, and $0\le a <1$, the following `Fourier series' \begin{equation} 1+\sum_{n=1}^{\infty} 2(-1)^na^n\xi^{nm}\cos(nm\eta)=\frac{1-a^2\xi^{2m}}{1+(a\xi^m)^2+2a\xi^m\cos m\eta}. \end{equation} – Jog May 07 '20 at 10:11
  • Using the series representation that I gave in my last comment we can now write for $a<1$, \begin{equation} \int_0^{\pi}\frac{f(\theta),d\theta}{1+a^2+2a\cos m\theta}=\frac{1}{1-a^2}\left[\int_0^{\pi} f(\theta),d\theta +2\sum_{p=1}^{\infty}(-1)^pa^p\int_0^{\pi}\cos(mp\theta)f(\theta),d\theta\right]. \end{equation} Mathematica can do the summation. As examples, for $f(\theta)=1$ and $f(\theta)=\cos n\theta$, $n\in\mathbb{N}$, we get the same expressions as with the contour integral. But now the method works for \emph{arbitrary} $f(\theta)$ without having to contour integrate. – Jog May 08 '20 at 06:26