級数•積分botの積分を示してみる②

$$\begin{array}{rl}I(a)& :={\displaystyle {\int }_0^{\pi }\ln (a+b\cos x)\mathrm{d}x}\\ I^{\prime }(a)& =\frac{\partial }{\partial a}{\displaystyle {\int }_0^{\pi }\ln (a+b\cos x)\mathrm{d}x}\\ & ={\displaystyle {\int }_0^{\pi }\frac{1}{a+b\cos x}\mathrm{d}x}\\ & ={\displaystyle \frac{2}{a-b}}{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{1}{\frac{a+b}{a-b}+t^2}\mathrm{d}t(\tan \frac{x}{2}\to t)}\\ & =\frac{\pi }{\sqrt{a^2-b^2}}\end{array}$$
より
$$\begin{array}{rl}I(a)& =\pi {\displaystyle \int \frac{1}{\sqrt{a^2-b^2}}\mathrm{d}a}\\ & =\pi \ln (a+\sqrt{a^2-b^2})+C\end{array}$$
b=0ならば
$$\begin{array}{rl}I(a)& =\pi \ln a\\ \pi \ln \left(2a\right)+C& =\pi \ln a\\ C& =-\pi \ln 2\end{array}$$
よって
$$\begin{array}{rl}I(a)& =\pi \ln {\displaystyle \frac{a+\sqrt{a^2-b^2}}{2}}\end{array}$$

示す積分をIとします。
$$\begin{array}{rl}I& ={\displaystyle {\int }_0^{\frac{\pi }{2}}\ln (\frac{17}{16}+\frac{\cos 2x}{2})\mathrm{d}x}\\ & ={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi }\ln (\frac{17}{16}+\frac{\cos t}{2})\mathrm{d}t(2x\to t)}\\ & =\pi \ln 1(a=\frac{17}{16},b=\frac{1}{2})\\ & =0\end{array}$$