便利さんの積分・級数botを解く①

$$\begin{array}{rl}I& ={\int }_0^{\frac{\pi }{2}}\log (\frac{9}{16}+{\cos }^{2}x)dx\\ & ={\int }_0^{\frac{\pi }{2}}\log \left(\frac{9+16{\cos }^{2}x}{16}\right)dx\\ & ={\int }_0^{\frac{\pi }{2}}\log (9+16{\cos }^{2}x)dx-{\int }_0^{\frac{\pi }{2}}\log 16dx\\ & ={\int }_0^{\frac{\pi }{2}}\log (9+16{\cos }^{2}x)dx-2\pi \log 2\end{array}$$
$${\int }_0^{\frac{\pi }{2}}\log (9+16{\cos }^{2}x)dx=I_1$$とおいて計算していきます。
$$\begin{array}{rl}{I}_1& ={\int }_0^{\frac{\pi }{2}}\log (9+16{\cos }^{2}x)dx\\ & ={\int }_0^{\frac{\pi }{2}}\log (9+8(1+\cos 2x))dx\\ & =\frac{1}{2}{\int }_0^{\pi }\log (8(1+\cos x)+9)dx\\ & ={\int }_0^{\mathrm{\infty }}\log (8(\frac{1+t^2}{1+t^2}+\frac{1-t^2}{1+t^2})+9)\frac{dt}{t^2+1}\\ & ={\int }_0^{\mathrm{\infty }}\log \left(\frac{9x^2+25}{x^2+1}\right)\frac{dx}{x^2+1}\\ & ={\int }_0^{\mathrm{\infty }}\frac{\log (9x^2+25)}{x^2+1}dx-{\int }_0^{\mathrm{\infty }}\frac{\log (x^2+1)}{x^2+1}dx\end{array}$$
$${\int }_0^{\mathrm{\infty }}\frac{\log (9x^2+25)}{x^2+1}dx=I_2$$
$${\int }_0^{\mathrm{\infty }}\frac{\log (x^2+1)}{x^2+1}dx=I_3$$とおいて計算していきます。
$$\begin{array}{rl}{I}_3& ={\int }_0^{\mathrm{\infty }}\frac{\log (x^2+1)}{x^2+1}dx\\ & ={\int }_0^{\frac{\pi }{2}}\log \left(\frac{1}{{\cos }^2}\right)dx\\ & =-2{\int }_0^{\frac{\pi }{2}}\log \cos xdx\\ & =\pi \log 2(わたしの記事で解説してます)\end{array}$$
$$\begin{array}{rl}{I}_2& ={\int }_0^{\mathrm{\infty }}\frac{\log (9x^2+25)}{x^2+1}dx\end{array}$$
$$\begin{array}{r}f(t)={\int }_0^{\mathrm{\infty }}\frac{\log (9x^2+t)}{x^2+1}dx\end{array}$$
$$\begin{array}{rl}{f}^{\prime }(t)& ={\int }_0^{\mathrm{\infty }}\frac{dx}{(9x^2+t)(x^2+1)}dx\\ & =\frac{1}{t-9}{\int }_0^{\mathrm{\infty }}(\frac{1}{x^2+1}-\frac{9}{9x^2+t})dx\\ & =\frac{1}{t-9}(\frac{\pi }{2}-\frac{3\pi }{2\sqrt{t}})\\ & =\frac{\pi }{2(t+3\sqrt{t})}\end{array}$$
$$\begin{array}{rl}f(0)& ={\int }_0^{\mathrm{\infty }}\frac{\log 9x^2}{x^2+1}dx\\ & =2\log 3{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+1}+2{\int }_0^{\mathrm{\infty }}\frac{\log x}{x^2+1}dx\\ & =\pi \log 3(わたしの記事で解説してます)\end{array}$$
$$\begin{array}{rl}f(25)-f(0)& ={\int }_0^{25}{f}^{\prime }(t)dt\\ & =\frac{\pi }{2}{\int }_0^{25}\frac{dt}{t+3\sqrt{t}}\\ & =\pi {\int }_0^5\frac{dx}{x+3}\\ & =\pi {[\log (x+3)]}_0^5\\ & =3\pi \log 2-\pi \log 3\end{array}$$
$$f(25)=3\pi \log 2$$
$$I_2=3\pi \log 2$$
$$I_1=I_2-I_3$$
$$\begin{array}{rl}I& =I_1-2\pi \log 2\\ & =(I_2-I_3)-2\pi \log 2\\ & =(3\pi \log 2-\pi \log 2)-2\pi \log 2\\ & =0\end{array}$$