級数•積分botの積分を示してみる②

\begin{aligned} I(a)&:=\displaystyle\int_0^\pi \ln\left(a+b\cos x\right) \mathrm{d}x\\ I’(a)&=\frac{\partial}{\partial a} \displaystyle\int_0^\pi \ln\left(a+b\cos x\right) \mathrm{d}x\\ &= \displaystyle\int_0^\pi \frac{1}{a+b\cos x} \mathrm{d}x\\ &= \dfrac{2}{a-b} \displaystyle\int_0^\infty \frac{1}{\frac{a+b}{a-b}+t^2}\mathrm{d}t \quad\left(\tan\frac x2\to t\right)\\ &= \frac\pi{\sqrt{a^2-b^2}} \end{aligned}
より
\begin{aligned} I(a)&=\pi\displaystyle\int \frac{1}{\sqrt{a^2-b^2}} \mathrm{d}a\\ &= \pi\ln(a+\sqrt{a^2-b^2})+C \end{aligned}
b=0ならば
\begin{aligned} I(a)&=\pi\ln a\\ \pi\ln(2a)+C&=\pi\ln a\\ C&=-\pi\ln2 \end{aligned}
よって
\begin{aligned} I(a)&=\pi\ln{\dfrac{a+\sqrt{a^2-b^2}}{2}} \end{aligned}

示す積分をIとします。
\begin{aligned} I&=\displaystyle\int_0^\frac\pi2 \ln\left(\frac{17}{16}+\frac{\cos 2x}2\right)\mathrm{d}x\\ &= \dfrac12 \displaystyle\int_0^\pi \ln\left(\frac{17}{16}+\frac{\cos t}2\right)\mathrm{d}t\quad(2x\to t)\\ &= \pi\ln1\quad\left(a=\frac{17}{16},b=\frac12\right)\\ &=0 \end{aligned}