積分解説10

\begin{eqnarray} 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 \quad(\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} \quad(\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 \quad(x^4\mapsto x)\\ &=&\frac{1}{8}\int_1^3\Big(\psi\big(\frac{t+2}{4}-\frac{3}{4}+1\big)-\psi\big(\frac{t}{4}-\frac{3}{4}+1\big)\Big)\mathrm{d}t \\ &=&\frac{1}{2}\Big[\log{\Gamma\big(\frac{t+3}{4}\big)}-\log{\Gamma\big(\frac{t+1}{4}\big)}\Big]_1^3 \\ &=&\frac{1}{2}\log{\frac{\pi}{2}} \\ \text{ゆえに}\\ I(a)&=&\frac{1}{2}\Big(\log{|a|}+\log{\frac{\pi}{2}}\Big) \\ \text{ここで}a&=&e\text{として結果が得られます。}\\ \end{eqnarray}

$$ \text{また、} $$
$$ \int_0^\frac{\pi}{4}\frac{\tan^{2e}{x}-2\sin^2{x}}{\sin{2x} \log{\tan{x}}}\mathrm{d}x=\frac{1}{2}\Big(\int_0^1\frac{x^{2e-1}-x}{\log{x}}\frac{\mathrm{d}x}{1+x^2}+\int_0^1\frac{x^{2e+1}-x}{\log{x}}\frac{\mathrm{d}x}{1+x^2}\Big) $$
$$ \text{という風に変形しても同様の手順で求まります。} $$