Question

Вычислите несобственный интеграл I рода или установите его расходимость

Answer 1

Решение:

$\int\limits^\infty_0 {\frac{arctg2x}{\pi(1+4x^{2})}} \, dx=\lim_{a \to \infty}(\int\limits^a_0 {\frac{arctg2x}{\pi(1+4x^{2})}} \, dx)=\lim_{a \to \infty}(\int\limits {\frac{arctg2x}{\pi(1+4x^{2})}} \, dx)=\lim_{a \to \infty}(\frac{1}{\pi }*\int\limits {\frac{arctg2x}{1+4x^{2}}} \, dx)=|^{t=arctg2x}_{dt=dx}|=\lim_{a \to \infty}(\frac{1}{\pi}* \int\limits {\frac{t}{2}} \, dt)=\lim_{a \to \infty}(\frac{1}{\pi}*\frac{1}{2}*\int\limits {t \, dt)=lim_{a \to \infty}(\frac{1}{2\pi}*\frac{t^{2}}{2})=$$=lim_{a \to \infty}(\frac{1}{2\pi}*\frac{arctg^{2}2x}{2})=lim_{a \to \infty}(\frac{arctg^{2}2x}{4\pi}|^{a}_0)=lim_{a \to \infty}(\frac{arctg^{2}2a}{4\pi}-\frac{arctg^{2}2*0}{4\pi})=lim_{a \to \infty}(\frac{arctg^{2}2a}{4\pi}-0)=lim_{a \to \infty}(\frac{arctg^{2}2a}{4\pi})=\frac{\lim_{a \to \infty}(arctg^{2}2a)}{\lim_{a \to \infty}(4\pi)}=\frac{(\lim_{a \to \infty}(arctg2a))^{2}}{4\pi}=\frac{arctg( \lim_{a \to \infty}(2a))^{2}}{4\pi}=\frac{arctg(\infty)^{2}}{4\pi}=\frac{(\frac{\pi}{2})^{2}}{4\pi}$$=\frac{\frac{\pi^{2}}{4}}{4\pi}=\frac{\pi^{2}}{4}*\frac{1}{4\pi}=\frac{\pi}{16}$

Ответ: $\frac{\pi}{16}$