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

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:

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:

On the other hand:

So, by Sum of Logarithms:


 * $\pi \ln 2 + \pi \ln a + C = \pi \ln a$

so:


 * $C = -\pi \ln 2$

giving:

in the case $b = 0$.

Let $b > 0$.

Then:

On the other hand:

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.

Let $b < 0$.

We have:

On the other hand:

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.

The result follows.