Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x
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Theorem
- $\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
where:
- $b$ is a real number
- $a$ is a positive real number with $a \ge \size b$.
Proof
Fix $b \in \R$ and define:
- $\ds \map I a = \int_0^\pi \map \ln {a + b \cos x} \rd x$
for $a \ge \size b$.
We have:
\(\ds \map {I'} a\) | \(=\) | \(\ds \frac \d {\d a} \int_0^\pi \map \ln {a + b \cos x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\pi \frac \partial {\partial a} \paren {\map \ln {a + b \cos x} } \rd x\) | Definite Integral of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\pi \frac 1 {a + b \cos x} \rd x\) | Derivative of Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \int_0^{2 \pi} \frac 1 {a + b \cos x} \rd x\) | Definite Integral of Even Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi {\sqrt {a^2 - b^2} }\) | Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$ |
So, by Primitive of $\sqrt {x^2 - a^2}$: Logarithm Form:
- $\map I a = \pi \map \ln {a + \sqrt {a^2 - b^2} } + C$
for all $a \ge \size b$, for some $C \in \R$.
We now investigate cases according to the sign of $b$.
Let $b = 0$.
We have:
\(\ds \map I a\) | \(=\) | \(\ds \pi \map \ln {a + \sqrt {a^2} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi \map \ln {a + \size a} + C\) | Definition 2 of Absolute Value | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \map \ln {2 a} + C\) | since $a \ge 0$ |
On the other hand:
\(\ds \map I a\) | \(=\) | \(\ds \int_0^\pi \ln a \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln a\) | Primitive of Constant |
So, by Sum of Logarithms:
- $\pi \ln 2 + \pi \ln a + C = \pi \ln a$
so:
- $C = -\pi \ln 2$
giving:
\(\ds \int_0^\pi \map \ln {a + b \cos x} \rd x\) | \(=\) | \(\ds \pi \map \ln {a + \sqrt {a^2 - b^2} } - \pi \ln 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}\) | Difference of Logarithms |
in the case $b = 0$.
$\Box$
Let $b > 0$.
Then:
\(\ds \map I b\) | \(=\) | \(\ds \pi \map \ln {b + \sqrt {b^2 - b^2} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b + C\) |
On the other hand:
\(\ds \map I b\) | \(=\) | \(\ds \int_0^\pi \map \ln {b + b \cos x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\pi \ln b \rd x + \int_0^\pi \map \ln {1 + \cos x} \rd x\) | Sum of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b + \int_0^\pi \map \ln {2 \cos^2 \frac x 2} \rd x\) | Primitive of Constant, Double Angle Formula for Cosine: Corollary $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b + \int_0^\pi \ln 2 \rd x + 2 \int_0^\pi \map \ln {\cos \frac x 2} \rd x\) | Sum of Logarithms, Logarithm of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b + \pi \ln 2 + 4 \int_0^{\frac \pi 2} \map \ln {\cos u} \rd u\) | substituting $u = \dfrac x 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b + \pi \ln 2 - 2 \pi \ln 2\) | Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\cos x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln b - \pi \ln 2\) |
giving:
- $C = -\pi \ln 2$
So we have:
- $\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b > 0$ too.
$\Box$
Let $b < 0$.
We have:
\(\ds \map I {\size b}\) | \(=\) | \(\ds \pi \map \ln {\size b + \sqrt {\size b^2 - b^2} } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi \map \ln {\size b} + C\) |
On the other hand:
\(\ds \map I {\size b}\) | \(=\) | \(\ds \int_0^\pi \map \ln {\size b + b \cos x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\pi \map \ln {\size b} \rd x + \int_0^\pi \map \ln {1 + \frac b {\size b} \cos x} \rd x\) | Sum of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln {\size b} + \int_0^\pi \map \ln {1 - \cos x} \rd x\) | since $\dfrac b {\size b} = -1$ for $b < 0$ and using Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln {\size b} + \int_0^\pi \map \ln {2 \sin^2 \frac x 2} \rd x\) | Double Angle Formula for Cosine: Corollary $2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln \size b + \int_0^\pi \ln 2 \rd x + \int_0^\pi \map \ln {\sin^2 \frac x 2} \rd x\) | Sum of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln \size b + \pi \ln 2 + 2 \int_0^\pi \map \ln {\sin \frac x 2} \rd x\) | Primitive of Constant, Logarithm of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln \size b + \pi \ln 2 + 4 \int_0^{\pi/2} \map \ln {\sin u} \rd u\) | substituting $u = \dfrac x 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln \size b + \pi \ln 2 - 2 \pi \ln 2\) | Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \ln \size b - \pi \ln 2\) |
We have in this case:
- $C = -\pi \ln 2$
So we have:
- $\ds \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
in the case $b < 0$ too.
$\Box$
The result follows.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.107$