Contour Integral of $\int\limits_0^{2\pi}\frac{d\theta}{1+a\cos\theta}$ for $a^2<1$ (textbook wrong?)

Question

My book is telling me that the answer is $\frac{2\pi}{\sqrt{1-a^2}}$. I'm getting an extra a on the numerator. Could somebody verify if I'm wrong, or if it's my book (it has been wrong numerous times).

Answer 1

The textbook answer is correct. You maybe omitted some factor somewhere, can you do a re-check? Note that if you take $a=0$, your answer does not give the right value.

Answer 2

For small $a$,

$$\int_0^{2\pi}\frac{d\theta}{1+a\cos(\theta)}\approx\int_0^{2\pi}(1-a\cos(\theta)+a^2\cos^2(\theta)-a^3\cos^3(\theta))\,d\theta=2\pi\left(1-\frac{a^2}2\right),$$

which is quite compatible with $$\frac{2\pi}{\sqrt{1-a^2}}.$$

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Working backwards, the Taylor development of the exact function will give you the integrals of the even powers of the cosine.

Answer 3

Contour Integration $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)} &=\oint\frac{2\,\mathrm{d}z}{iz\left(2+az+\frac az\right)}\tag{1}\\ &=-\frac{2i}a\oint\frac{\mathrm{d}z}{z^2+\frac2az+1}\tag{2}\\ &=-\frac{2i}a\oint\frac{\mathrm{d}z}{\left(z+\frac1a+\sqrt{\frac1{a^2}-1}\right)\left(z+\frac1a-\sqrt{\frac1{a^2}-1}\right)}\tag{3}\\ &=\frac{2\pi}{\sqrt{1-a^2}}\tag{4} \end{align} $$ Explanation:
$(1)$: $z=e^{i\theta}$
$(2)$: algebra
$(3)$: factor the denominator
$(4)$: the residue of the integrand at $z=-\frac1a+\sqrt{\frac1{a^2}-1}$ is $\frac1{2\sqrt{\frac1{a^2}-1}}$
$\phantom{(4)\text{:}}$ the singularity at $z=-\frac1a-\sqrt{\frac1{a^2}-1}$ is outside the unit circle

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Weierstrass Substitution $$ \begin{align} \int_0^{2\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)} &=2\int_0^{\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)}\tag{5}\\ &=2\int_0^\infty\frac{\frac{2\,\mathrm{d}z}{1+z^2}}{1+a\frac{1-z^2}{1+z^2}}\tag{6}\\ &=4\int_0^\infty\frac{\mathrm{d}z}{(1+a)+(1-a)z^2}\tag{7}\\ &=\frac4{1+a}\sqrt{\frac{1+a}{1-a}}\int_0^\infty\frac{\mathrm{d}z}{1+z^2}\tag{8}\\ &=\frac{2\pi}{\sqrt{1-a^2}}\tag{9} \end{align} $$ Explanation:
$(5)$: periodicity and evenness of $\cos(x)$
$(6)$: Weierstrass substitution
$(7)$: algebra
$(8)$: substitute $z\mapsto\sqrt{\frac{1+a}{1-a}}\,z$