\( \int_{0}^{\pi}\frac{d\theta}{1-2a\cos \theta+a^2} \)

[Tolaso J Kos]

For \(|a|<1 \) evaluate the integral: $$\int_{0}^{\pi}\frac{d\theta}{1-2a\cos \theta+a^2}$$

[Papapetros Vaggelis]

Let \(\displaystyle{a\in\left[0,1\right)}\) and \(\displaystyle{I(a)=\int_{0}^{\pi}\dfrac{1}{1-2\,a\,\cos\,\theta+a^2}\,\mathrm{d}\theta}\).

We have that

\(\displaystyle{I(a)=\dfrac{1}{2(1-a^2)}\,\int_{-\pi}^{\pi}\dfrac{1-a^2}{1-2\,a\,\cos\,\theta+a^2}\,\mathrm{d}\theta}\)

and the integrand function is the \(\displaystyle{a}\) - null of \(\displaystyle{\rm{Poisson}}\). So, if

\(\displaystyle{P_{a}(\theta)=\dfrac{1-a^2}{1-2\,\cos\,\theta+a^2}\,,\theta\in\left[-\pi,\pi\right]}\), then

\(\displaystyle{\int_{-\pi}^{\pi}P_{a}(\theta)\,\mathrm{d}\theta=2\,\pi}\) and

\(\displaystyle{I(a)=\dfrac{\pi}{1-a^2}}\).

Now, if \(\displaystyle{a\in\left(-1,0\right)}\), then

\(\displaystyle{\begin{aligned} I(a)&=\int_{0}^{\pi}\dfrac{1}{1-2\,a\,\cos\,\theta+a^2}\,\mathrm{d}\theta\\&=\int_{0}^{\pi}\dfrac{1}{1+2\,a\,\cos\,(\pi-\theta)+a^2}\,\mathrm{d}\theta\\&=\int_{0}^{\pi}\dfrac{1}{1-2\,(-a)\,\cos\,\theta+(-a^2)}\,\mathrm{d}\theta\\&\stackrel{0<-a<1}{=}I(-a)\\&=\dfrac{\pi}{1-(-a)^2}\\&=\dfrac{\pi}{1-a^2}\end{aligned}}\).

Finally, \(\displaystyle{I(a)=\int_{0}^{\pi}\dfrac{1}{1-2\,a\,\cos\,\theta+a^2}\,\mathrm{d}\theta=\dfrac{\pi}{1-a^2}\,,-1<a<1}\).

[Grigorios Kostakos]

A 2nd solution:

For \(|a|<1 \) holds
\begin{align}
1+\displaystyle2\mathop{\sum}\limits_{n=1}^{+\infty} a^n\cos(n\theta)&=\frac{1-a^2}{1-2\,a\,\cos\,\theta+a^2}\quad\Rightarrow\\
\frac{1}{1-a^2}+\frac{2}{1-a^2}\mathop{\sum}\limits_{n=1}^{+\infty} a^n\cos(n\theta)&=\frac{1}{1-2\,a\,\cos\,\theta+a^2}
\end{align}
Thus, integrating $(2)$ we get
\begin{align*}
\int_{0}^{\pi}\dfrac{1}{1-2\,a\,\cos\,\theta+a^2}\,\mathrm{d}\theta&=\frac{1}{1-a^2}\int_{0}^{\pi}\,\mathrm{d}\theta+\frac{2}{1-a^2}\int_{0}^{\pi}\bigg(\mathop{\sum}\limits_{n=1}^{+\infty} a^n\cos(n\theta)\bigg)\,\mathrm{d}\theta\\
&=\frac{\pi}{1-a^2}+\frac{2}{1-a^2}\mathop{\sum}\limits_{n=1}^{+\infty} a^n\int_{0}^{\pi}\cos(n\theta)\,\mathrm{d}\theta\\
&=\frac{\pi}{1-a^2}+\frac{2}{1-a^2}\mathop{\sum}\limits_{n=1}^{+\infty} a^n\,\frac{\cancelto{0}{\sin(n\pi)}}{n}\\
&=\frac{\pi}{1-a^2}\,.
\end{align*}


$(1):$ Proof left as an exercise.