積分解説10

$$\begin{array}{rcl}I(a):& =& {\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2a}x-2{\sin }^{2}x}{\sin 2x\log \tan x}\mathrm{d}x\\ \frac{\partial I(a)}{\partial a}& =& 2{\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2a}x}{\sin 2x}\mathrm{d}x\\ & =& {\int }_0^1{x}^{2a-1}\mathrm{d}x(\tan x\mapsto x)\\ & =& \frac{1}{2a}\\ \therefore I(a)& =& \frac{1}{2}\log |a|+C\\ I(1)& =& C\\ \text{同時に、}\\ I(1)& =& {\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2}x-2{\sin }^{2}x}{\sin 2x\log \tan x}\mathrm{d}x\\ & =& \frac{1}{2}{\int }_0^1\frac{x^3-x}{\log x}\frac{\mathrm{d}x}{1+x^2}(\tan x\mapsto x)\\ & =& \frac{1}{2}{\int }_0^1{\int }_1^3{x}^t\mathrm{d}t\frac{\mathrm{d}x}{1+x^2}\\ & =& \frac{1}{2}{\int }_1^3{\int }_0^1\frac{x^t-x^{t+2}}{1-x^4}\mathrm{d}x\mathrm{d}t\\ & =& \frac{1}{8}{\int }_1^3{\int }_0^1\frac{x^{t/4}-x^{(t+2)/4}}{1-x}\frac{\mathrm{d}x}{x^{3/4}}\mathrm{d}t(x^4\mapsto x)\\ & =& \frac{1}{8}{\int }_1^3(\psi (\frac{t+2}{4}-\frac{3}{4}+1)-\psi (\frac{t}{4}-\frac{3}{4}+1))\mathrm{d}t\\ & =& \frac{1}{2}[\log \mathrm{\Gamma }(\frac{t+3}{4})-\log \mathrm{\Gamma }(\frac{t+1}{4}){]}_1^3\\ & =& \frac{1}{2}\log \frac{\pi }{2}\\ \text{ゆえに}\\ I(a)& =& \frac{1}{2}(\log |a|+\log \frac{\pi }{2})\\ \text{ここで}a& =& e\text{として結果が得られます。}\end{array}$$

$$\text{また、}$$
$${\int }_0^{\frac{\pi }{4}}\frac{{\tan }^{2e}x-2{\sin }^{2}x}{\sin 2x\log \tan x}\mathrm{d}x=\frac{1}{2}({\int }_0^1\frac{x^{2e-1}-x}{\log x}\frac{\mathrm{d}x}{1+x^2}+{\int }_0^1\frac{x^{2e+1}-x}{\log x}\frac{\mathrm{d}x}{1+x^2})$$
$$\text{という風に変形しても同様の手順で求まります。}$$