便利さんの積分・級数botを解く①

\begin{align} I&=\int_{0}^{\frac{\pi}{2}}\log\left(\frac{9}{16}+\cos^2x\right)dx \\&=\int_{0}^{\frac{\pi}{2}}\log\left(\frac{9+16\cos^2x}{16}\right)dx \\&=\int_{0}^{\frac{\pi}{2}}\log\left(9+16\cos^2x\right)dx-\int_{0}^{\frac{\pi}{2}}\log16dx \\&=\int_{0}^{\frac{\pi}{2}}\log(9+16\cos^2x)dx-2\pi\log2 \end{align}
$$\int_{0}^{\frac{\pi}{2}}\log(9+16\cos^2x)dx=I_1$$とおいて計算していきます。
\begin{align} I_1&=\int_{0}^{\frac{\pi}{2}}\log(9+16\cos^2x)dx \\&=\int_{0}^{\frac{\pi}{2}}\log(9+8(1+\cos 2x))dx \\&=\frac{1}{2}\int_{0}^{\pi}\log(8(1+\cos x)+9)dx \\&=\int_{0}^{\infty}\log\left(8\left(\frac{1+t^2}{1+t^2}+\frac{1-t^2}{1+t^2}\right)+9\right) \frac{dt}{t^2+1} \\&=\int_{0}^{\infty}\log\left(\frac{9x^2+25}{x^2+1}\right)\frac{dx}{x^2+1} \\&=\int_{0}^{\infty}\frac{\log(9x^2+25)}{x^2+1}dx -\int_{0}^{\infty}\frac{\log(x^2+1)}{x^2+1}dx \end{align}
$$\int_{0}^{\infty}\frac{\log(9x^2+25)}{x^2+1}dx=I_2$$
$$\int_{0}^{\infty}\frac{\log(x^2+1)}{x^2+1}dx=I_3$$とおいて計算していきます。
\begin{align} I_3&=\int_{0}^{\infty}\frac{\log(x^2+1)}{x^2+1}dx \\&=\int_{0}^{\frac{\pi}{2}}\log\left(\frac{1}{\cos^2}\right)dx \\&=-2\int_{0}^{\frac{\pi}{2}}\log\cos xdx \\&=\pi\log2\qquad(わたしの記事で解説してます) \end{align}
\begin{align} I_2&=\int_{0}^{\infty}\frac{\log(9x^2+25)}{x^2+1}dx \end{align}
\begin{align} f(t)=\int_{0}^{\infty}\frac{\log(9x^2+t)}{x^2+1}dx \end{align}
\begin{align} f'(t)&=\int_{0}^{\infty}\frac{dx}{(9x^2+t)(x^2+1)}dx \\&=\frac{1}{t-9}\int_{0}^{\infty}\left(\frac{1}{x^2+1}-\frac{9}{9x^2+t}\right)dx \\&=\frac{1}{t-9}\left(\frac{\pi}{2}-\frac{3\pi}{2\sqrt{t}}\right) \\&=\frac{\pi}{2(t+3\sqrt t)} \end{align}
\begin{align} f(0)&=\int_{0}^{\infty}\frac{\log 9x^2}{x^2+1}dx \\&=2\log3\int_{0}^{\infty}\frac{dx}{x^2+1}+2\int_{0}^{\infty}\frac{\log x}{x^2+1}dx \\&=\pi\log3\qquad(わたしの記事で解説してます) \end{align}
\begin{align} f(25)-f(0)&=\int_{0}^{25}f'(t)dt \\&=\frac{\pi}{2}\int_{0}^{25}\frac{dt}{t+3\sqrt t} \\&=\pi\int_{0}^{5}\frac{dx}{x+3} \\&=\pi\left[\log(x+3)\right]^5_0 \\&=3\pi\log2-\pi\log3 \end{align}
$$f(25)=3\pi\log2$$
$$I_2=3\pi\log2$$
$$I_1=I_2-I_3$$
\begin{align} I&=I_1-2\pi\log2 \\&=(I_2-I_3)-2\pi\log2 \\&=(3\pi\log2-\pi\log2)-2\pi\log2 \\&=0 \end{align}