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How to evaluate $\displaystyle\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$? I know that I can evaluate the integral using residue theorem, but which is the result with passages? Please

Pro
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    Similar integrals appeared on MSE many times. You can either split the integration range and apply Weierstrass' substitution, leading to the integral of a rational function, or directly apply the residue theorem. Another interesting technique is to notice that such integral depends on the area enclosed by an ellipse, due to the polar equation of the ellipse and the formula $A=\frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2,d\theta$. – Jack D'Aurizio Jun 21 '17 at 20:18
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    Your title and post are different. Is it $(A+B\cos\theta)^2$? –  Jun 21 '17 at 20:18
  • A taste of the last approach: https://math.stackexchange.com/a/2176755/44121 – Jack D'Aurizio Jun 21 '17 at 20:19
  • @JackD'Aurizio : Could I suggest calling it the tangent half-angle substitution rather than the Weierstrass substitution, because the latter name is justified only by a mistaken belief that Weierstrass had something to do with it? – Michael Hardy Jun 21 '17 at 20:23
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    @MichaelHardy: I am too late for performing an edit of my previous comment, but you are right, I should have called it that way. – Jack D'Aurizio Jun 21 '17 at 20:24
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    @tilper : The OP actually put {curly braces} around $A+B\cos\theta,$ perhaps not realizing that those would not appear in the result. – Michael Hardy Jun 21 '17 at 20:26
  • Judicious use of differentiating under the integral simplifies things a bit. I've posted a solution describing the details. – Mark Viola Jun 21 '17 at 20:33

2 Answers2

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With the substitution $z(\theta) = e^{i \theta}$ so that $d\theta = \frac{dz}{iz}$, we can rewrite the integral as the counterclockwise contour integral $$ \oint_{|z| = 1} \frac{1}{(A + \frac 12 B(z + z^{-1}))^2} \frac 1{iz}\,dz = \frac 4i \oint_{|z| = 1} \frac{z}{(Bz^2 + 2Az + B)^2}\,dz $$ Evaluate this integral using the residue theorem.

Assume that $|A| > |B|$ so that we have a nice formula. With the quadratic formula, we find that $$ Bz^2 + 2Az + B = 0 \implies z = \frac{-A \pm \sqrt{A^2 - B^2}}{B} = -\frac{A}{B} \pm \sqrt{\frac{A^2}{B^2} - 1} $$ Only the root $z_+ = \frac{-A + \sqrt{A^2 - B^2}}{B}$ satisfies $|z_+| < 1$. The residue theorem tells us that for $f(z) = \frac{z}{(Bz^2 + 2Az + B)^2}$, we have $$ \oint_{|z| = 1} \frac{z}{(Bz^2 + 2Az + B)^2} = 2 \pi i \operatorname{Res}(f;z_+) $$ Note that the pole at $z_+$ has order $2$. We calculate the residue $$ \operatorname{Res}(f;z_+) = \lim_{z \to z_+} \frac{d}{dz}\left[\frac{z}{B^2(z - z_-)^2}\right] = \lim_{z \to z_+} \frac{(z-z_-)^2 + 2z(z-z_-)}{B^2[(z - z_-)^2]^2} \\ = \frac{(z_+-z_-)^2 + 2z_+(z_+-z_-)}{B^2[(z_+ - z_-)^2]^2} $$ Simplify, noting that $z_+ - z_- = 2 \frac{\sqrt{A^2 - B^2}}{B}$.

Ben Grossmann
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As in THIS ANSWER, we let $I(A,B)$ be given by

$$I(A,B)=\int_0^{2\pi }\frac{1}{A+B\cos(\theta)}\,d\theta$$

where $|A|>|B|$.

Then, we have

$$-\frac{\partial I(A,B)}{\partial A}=\int_0^{2\pi }\frac{1}{(A+B\cos(\theta))^2}\,d\theta$$

Using either contour integration or the tangent half-angle substitution to evaluate $I(A,B)$, one finds $I(A,B)=\frac{2\pi}{\sqrt{A^2-B^2}}$.

Then differentiating yields

$$\int_0^{2\pi}\frac{1}{(A+B\cos(\theta))^2}\,d\theta=\frac{2\pi A}{(A^2-B^2)^{3/2}}$$

Mark Viola
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