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

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Theorem

$\ds \int_0^\pi \map \ln {a^2 - 2 a b \cos x + b^2} \rd x = \begin{cases}2 \pi \ln a & a \ge b > 0 \\ 2 \pi \ln b & b \ge a > 0\end{cases}$


Proof

Note that:

$\paren {a - b}^2 \ge 0$

so by Square of Sum:

$a^2 - 2 a b + b^2 \ge 0$

So:

$a^2 + b^2 \ge 2 a b = \size {-2 a b}$

so we may apply Definite Integral from $0$ to $\pi$ of $\map \ln {a + b \cos x}$.

We then have:

\(\ds \int_0^\pi \map \ln {a^2 - 2 a b \cos x + b^2} \rd x\) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + \sqrt {\paren {a^2 + b^2}^2 - \paren {2 a b}^2} } 2}\)
\(\ds \) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + \sqrt {a^4 + 2 a^2 b^2 + b^4 - 4 a^2 b^2} } 2}\)
\(\ds \) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + \sqrt {a^4 - 2 a^2 b^2 + b^4} } 2}\)
\(\ds \) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + \sqrt {\paren {a^2 - b^2}^2} } 2}\)
\(\ds \) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + \size {a^2 - b^2} } 2}\) Definition of Absolute Value

Note that if $a \ge b > 0$ we have:

\(\ds \pi \map \ln {\frac {a^2 + b^2 + \size {a^2 - b^2} } 2}\) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + a^2 - b^2} 2}\) since $a \ge b > 0$, we have $a^2 \ge b^2$ and $\size {a^2 - b^2} = a^2 - b^2$
\(\ds \) \(=\) \(\ds \pi \map \ln {a^2}\)
\(\ds \) \(=\) \(\ds 2 \pi \ln a\) Logarithm of Power

Note that if $b \ge a > 0$ we have:

\(\ds \pi \map \ln {\frac {a^2 + b^2 + \size {a^2 - b^2} } 2}\) \(=\) \(\ds \pi \map \ln {\frac {a^2 + b^2 + b^2 - a^2} 2}\) since $a \ge b > 0$, we have $a^2 \ge b^2$ and $\size {a^2 - b^2} = b^2 - a^2$
\(\ds \) \(=\) \(\ds \pi \map \ln {b^2}\)
\(\ds \) \(=\) \(\ds 2 \pi \ln b\) Logarithm of Power

$\blacksquare$


Sources

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