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The integral of 1 / (a + cos theta) over [0, 2 pi]

Example

For a real parameter $a$ with $∣ a ∣ > 1$ , $\int_{0}^{2 π} \frac{d θ}{a + \cos θ} = \frac{2 π sgn (a)}{\sqrt{a^{2} - 1}} .$ In particular, for $a > 1$ the value is $2 π / \sqrt{a^{2} - 1}$ .

Facts & Assumptions

Given: A real number $a$ with $∣ a ∣ > 1$ .

[L1]

The unit-circle substitution converts the trigonometric integral into a contour integral on $∣ z ∣ = 1$ (Trigonometric integrals become contour integrals by the unit-circle substitution).

Verification

technique · computation

1.1L1algebra

By [L1], $\int_{0}^{2 π} \frac{d θ}{a + \cos θ} = \int_{∣ z ∣ = 1} \frac{2 d z}{i (z^{2} + 2 a z + 1)} .$ The quadratic factors as $z^{2} + 2 a z + 1 = (z - z_{+}) (z - z_{-}), z_{\pm} = - a \pm \sqrt{a^{2} - 1} .$

2.1step 1.1algebra∎

Since $z_{+} z_{-} = 1$ , exactly one root lies inside the unit circle. It is $z_{+} = - a + sgn (a) \sqrt{a^{2} - 1}$ . The residue there is $\frac{2}{i (z_{+} - z_{-})} = \frac{sgn (a)}{i \sqrt{a^{2} - 1}} .$ Therefore $\int_{0}^{2 π} \frac{d θ}{a + \cos θ} = 2 π i \cdot \frac{sgn (a)}{i \sqrt{a^{2} - 1}} = \frac{2 π sgn (a)}{\sqrt{a^{2} - 1}} .$

Depends on

Trigonometric integrals become contour integrals by the unit-circle substitution

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. Howell and J. Mathews, Complex Analysis, Ch. 8 §8.2 (standard reference, not scraped)