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Đặt $u = {e^{{x^2}}}\,\,\, \Rightarrow \,\,du = 2x{e^{{x^2}}}dx$
$dv = \sin \,nx\,dx\,\,\, \Rightarrow \,\,v =
- \frac{1}{n}cosnx$
${I_n} = \int\limits_a^b {{e^{{x^2}}}\sin \,nx\,dx} =
[- \frac{1}{n}\left.
{{e^{{x^2}}}cos\,nx\,} \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 : $\mathop {\lim }\limits_{n \to + \infty } {I_n} = 0$
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