Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x/Proof 1
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Theorem
- $\ds \int_0^{2 \pi} \frac {\d x} {a + b \cos x} = \frac {2 \pi} {\sqrt {a^2 - b^2} }$
Proof
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} $ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \pi} b \paren { {\frac 1 {\paren {\frac {-a + \sqrt {a^2 - b^2} } b - \frac {-a - \sqrt {a^2 - b^2} } b} } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \pi} b \paren { {\frac b {2 \paren {\sqrt {a^2 - b^2} } } } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \pi} {\sqrt {a^2 - b^2} }\) |
$\blacksquare$