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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

where:

$b$ is a real number
$a$ is a positive real number with $a \ge \size b$.


Proof

\(\ds \int_0^{2 \pi} \map \ln {a + b \cos x} \rd x\) \(=\) \(\ds \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_\pi^{2 \pi} \map \ln {a + b \cos x} \rd x\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \int_0^\pi \map \ln {a + b \cos x} \rd x - \int_\pi^0 \map \ln {a + b \map \cos {2 \pi - x} } \rd x\) substituting $x \mapsto 2 \pi - x$
\(\ds \) \(=\) \(\ds \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \map \cos {2 \pi - x} } \rd x\) Reversal of Limits of Definite Integral
\(\ds \) \(=\) \(\ds \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \cos x} \rd x\) Cosine of Conjugate Angle
\(\ds \) \(=\) \(\ds 2 \int_0^\pi \map \ln {a + b \cos x} \rd x\)
\(\ds \) \(=\) \(\ds 2 \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}\) Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Definite_Integral_from_0_to_2_Pi_of_Logarithm_of_a_plus_b_Cosine_x&oldid=524356"