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.

$\blacksquare$


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