lim-n-0-1-e-x-2-sin-nx-dx-

Question Number 174883 by infinityaction last updated on 13/Aug/22

\lim_{n \to \infty} \int_{0}^{1} e^{x^{2}} \sin (n x) d x

Answered by Mathspace last updated on 14/Aug/22

u_{n} = \int_{0}^{1} e^{x^{2}} s i n (n x) d x

\Rightarrow u_{n} =_{n x = t} \int_{0}^{n} e^{\frac{t^{2}}{n^{2}}} s i n t \frac{d t}{n}

= \int_{R} \frac{1}{n} e^{\frac{t^{2}}{n^{2}}} s i n t χ_{[0, n [} (t) d t

= \int_{R} f_{n} (t) d t

f_{n} \to 0 (c s) \Rightarrow l i m u_{n} = 0

Answered by Mathspace last updated on 14/Aug/22

a n o t h e r w a y

b y ρ a r t s

u_{n} = {[- \frac{1}{n} c o s (n x) e^{x^{2}}]}_{0}^{1}

+ \frac{1}{n} \int_{0}^{1} e^{x^{2}} c o s (n x) d x

\frac{1}{n} - e \frac{c o s n}{n} + \frac{1}{n} \int_{0}^{1} e^{x^{2}} c o s (n x) d x

∣ u_{n} ∣⩽ \frac{1}{n} + \frac{e}{n} + \frac{1}{n} \int_{0}^{1} e^{x^{2}} d x \to 0 (n \to + \infty)

\Rightarrow l i m u_{n} = 0

Commented by infinityaction last updated on 14/Aug/22

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