Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x

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Theorem

$\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$

where $a$ and $b$ are real numbers with $a > b > 0$.


Proof 1

Let $C$ be the unit open disk in the complex plane centred at $0$.

The boundary of $C$, $\partial C$, can be parameterized by:

$\map z x = e^{i x}$

for $0 \le x \le 2 \pi$.

We have:

\(\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\) \(=\) \(\ds \int_0^{2 \pi} \frac {\d x} {a + \frac b 2 \paren {e^{i x} + e^{-i x} } }\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \int_0^{2 \pi} \paren {\frac {2 e^{i x} } {2 e^{i x} } } \frac {\d x} {a + \frac b 2 \paren {e^{i x} + e^{-i x} } }\) multiplying by $1$
\(\ds \) \(=\) \(\ds 2 \int_0^{2 \pi} \frac {e^{i x} \rd x } {2 a e^{i x} + b e^{2 i x} + b}\)
\(\ds \) \(=\) \(\ds 2 \int_0^{2 \pi} \frac {e^{i x} \rd x } {b \paren {e^{2 i x} + \frac {2 a} b e^{i x} + 1} }\) rearranging
\(\ds \) \(=\) \(\ds \frac 2 {i b} \int_{\partial C} \frac {\d z} {z^2 + \frac {2 a} b z + 1}\) Definition of Complex Contour Integral, Derivative of Exponential Function, $\rd z = i e^{i x} \rd x$
\(\ds \) \(=\) \(\ds \frac 2 {i b} \int_{\partial C} \frac {\d z} {\paren {z + \frac a b}^2 - \frac {a^2} {b^2} + 1}\) Completing the Square

The integrand has poles where:

$\paren {z + \dfrac a b}^2 - \dfrac {a^2} {b^2} + 1 = 0$

That is, where:

$\size {z + \dfrac a b} = \dfrac {\sqrt {a^2 - b^2} } b$

So:

$z_1 = \dfrac {-a + \sqrt {a^2 - b^2} } b$

and:

$z_2 = \dfrac {-a - \sqrt {a^2 - b^2} } b$

are the poles of the integrand.

We have:

\(\ds \size {z_2}\) \(=\) \(\ds \size {\frac {-a - \sqrt {a^2 - b^2} } b}\)
\(\ds \) \(=\) \(\ds \frac {a + \sqrt {a^2 - b^2} } b\) as $a > b > 0$
\(\ds \) \(=\) \(\ds \frac a b \paren {1 + \sqrt {1 - \frac {b^2} {a^2} } }\)
\(\ds \) \(>\) \(\ds \frac a b\) as $\sqrt {1 - \dfrac {b^2} {a^2} } > 0$
\(\ds \) \(>\) \(\ds 1\)

So $z_2$ lies outside the closed disk $\size z \le 1$ for all real $a > b > 0$, so is of no concern.

We now establish a bound on $z_1$.

As $a > b > 0$, we have that:

$a^2 > a^2 - b^2 > 0$

So, from Square Root is Strictly Increasing:

$a > \sqrt {a^2 - b^2} > 0$

Multiplying through by $-2 \sqrt {a^2 - b^2}$:

$-2 a \sqrt{a^2 - b^2} < 2 \paren {b^2 - a^2} < 0$

This can be rewritten as:

\(\ds a^2 - 2 a \sqrt {a^2 - b^2} + a^2 - b^2\) \(=\) \(\ds a^2 - 2 a \sqrt {a^2 - b^2} + \paren {\sqrt {a^2 - b^2} }^2\)
\(\ds \) \(=\) \(\ds \paren {-a + \sqrt {a^2 - b^2} }^2\) Square of Sum
\(\ds \) \(<\) \(\ds b^2\)

giving:

$\size {-a + \sqrt {a^2 - b^2} } < b$

Therefore:

$\size {z_1} = \size {\dfrac {-a + \sqrt {a^2 - b^2} } b} < 1$

So $z_1$ is the only pole of the integrand lying within $C$.

Therefore:

\(\ds \frac 2 {i b} \int_C \frac {\d z} {z^2 + \frac {2 a} z + 1}\) \(=\) \(\ds \frac {4 \pi i} {i b} \Res {\frac 1 {z^2 + \frac {2 a} b z + 1} } {z_1}\) Cauchy's Residue Theorem
\(\ds \) \(=\) \(\ds \frac {4 \pi} b \paren {\lim_{z \mathop \to {z_1} } \paren {z - z_1} \frac 1 {z^2 + \frac {2 a} b z + 1} }\) Residue at Simple Pole
\(\ds \) \(=\) \(\ds \frac {4 \pi} b \paren {\lim_{z \mathop \to {\frac {-a + \sqrt {a^2 - b^2} } b} } \paren {z - \frac {-a + \sqrt {a^2 - b^2} } b} \frac 1 {\paren {z - \frac {-a + \sqrt {a^2 - b^2} } b} \paren {z - \frac {-a - \sqrt {a^2 - b^2} } b} } }\)
\(\ds \) \(=\) \(\ds \frac {4 \pi} b \sqbrk {\frac 1 {\paren {z - \frac {-a - \sqrt {a^2 - b^2} } b} } }_{z = \frac {-a + \sqrt {a^2 - b^2} } b}\) cancelling $\paren {z - \dfrac {-a + \sqrt {a^2 - b^2} } b} $
\(\ds \) \(=\) \(\ds \frac {4 \pi} b \paren { {\frac 1 {\paren {\frac {-a + \sqrt {a^2 - b^2} } b - \frac {-a - \sqrt {a^2 - b^2} } b} } } }\)
\(\ds \) \(=\) \(\ds \frac {4 \pi} b \paren { {\frac b {2 \paren {\sqrt {a^2 - b^2} } } } }\)
\(\ds \) \(=\) \(\ds \frac {2 \pi} {\sqrt {a^2 - b^2} }\)

$\blacksquare$


Proof 2

\(\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x}\) \(=\) \(\ds \int_0^\pi \frac {\d x} {a + b \cos x} + \int_\pi^{2 \pi} \frac {\d x} {a + b \cos x}\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 \pi + \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } \pi {2 \pi}\) Primitive of $\dfrac 1 {p + q \cos a x}$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} }\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \pi} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\)
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} }\) Tangent Function is Periodic on Reals
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren { \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan 0} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\)
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{x \mathop \to \pi^+} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} - \lim_{x \mathop \to \pi^-} \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} }\) simplfying
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\lim_{u \mathop \to \infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} - \lim_{u \mathop \to -\infty} \map \arctan {\sqrt {\frac {a - b} {a + b} } u} }\) letting $u = \tan \dfrac x 2$: Tangent Function $\to \pm \infty$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {a^2 - b^2} } \paren {\frac \pi 2 - \paren {-\frac \pi 2} }\) Limit to Positive and Negative Infinity of Arctangent Function
\(\ds \) \(=\) \(\ds \frac {2 \pi} {\sqrt {a^2 - b^2} }\)

$\blacksquare$


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