|
|
Đặt $u = {e^{{x^2}}}\,\,\, \Rightarrow \,\,du = 2x{e^{{x^2}}}dx$
$dv = \sin \,nx\,dx\,\,\, \Rightarrow \,\,v = - \frac{1}{n}\cos nx$
${I_n} = \int\limits_a^b {{e^{{x^2}}}\sin \,nx\,dx} = - \frac{1}{n}\left. {{e^{{x^2}}}\cos\,nx\,dx} \right|_a^b + \frac{2}{n}\int\limits_a^b {x{e^{{x^2}}}\,\cos\,nx\,dx} $
Ta có: $\mathop {\lim }\limits_{n \to + \infty } \left [{\frac{{ - {e^{{x^2}}}cos\,nx}}{n}} \right] |_a^b \le \mathop {\lim }\limits_{n \to + \infty } -\frac{1}{n}\left( {{e^{{b^2}}} - {e^{{a^2}}}} \right) = 0$
$\frac{2}{n}\int\limits_a^b {x{e^{{x^2}}}\,\cos\,nx\,dx} \le \frac{2}{n}\int\limits_a^b {\left| x \right|{e^{{x^2}}}\,dx} \le \frac{2}{n}\int\limits_a^b {M\,dx} = \frac{2}{n}M\left( {b - a} \right)$
Trong đó $M = \mathop {Max}\limits_{a \le x \le b} \left| x \right|{e^{{x^2}}}\,$
$\mathop {\lim }\limits_{n \to + \infty } \frac{2}{n}M\left( {b - a} \right) = 0$ Từ đó : $\mathop {\lim }\limits_{n \to + \infty } {I_n} = 0$
|