calculate-A-n-0-1-sin-nx-e-2x-dx-with-n-integr-natural-

Question Number 47651 by prof Abdo imad last updated on 12/Nov/18

c a l c u l a t e A_{n} = \int_{0}^{1} s i n ([n x]) e^{- 2 x} d x w i t h n

i n t e g r n a t u r a l .

Commented by maxmathsup by imad last updated on 13/Nov/18

w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{\frac{k}{n}}^{\frac{k + 1}{n}} s i n ([n x]) e^{- 2 x} d x

= \sum_{k = 0}^{n - 1} \int_{\frac{k}{n}}^{\frac{k + 1}{n}} s i n (k) e^{- 2 x} d x = \sum_{k = 0}^{n} s i n (k) \int_{\frac{k}{n}}^{\frac{k + 1}{n}} e^{- 2 x} d x

= - \frac{1}{2} \sum_{k = 0}^{n - 1} s i n (k) (e^{- 2 \frac{k + 1}{n}} - e^{\frac{- 2 k}{n}})

= \frac{1}{2} \sum_{k = 0}^{n - 1} e^{\frac{- 2 k}{n}} - \frac{1}{2} \sum_{k = 0}^{n - 1} e^{- 2 \frac{k + 1}{n}} s i n (k)

= \frac{1 - e^{- \frac{2}{n}}}{2} \sum_{k = 0}^{n - 1} e^{\frac{- 2 k}{n}} s i n (k) b u t \sum_{k = 0}^{n -} e^{\frac{- 2 k}{n}} s i n k = I m (\sum_{k = 0}^{n - 1} e^{i k - \frac{2}{n} k}) a n d

\sum_{k = 0}^{n - 1} e^{i k - \frac{2}{n} k} = \sum_{k = 0}^{n - 1} e^{(i - \frac{2}{n}) k} = \frac{1 - {(e^{(i - \frac{2}{n})})}^{n}}{1 - e^{i - \frac{2}{n}}} = \frac{1 - e^{- 2} e^{i n}}{1 - e^{- \frac{2}{n}} e^{i}}

= \frac{1 - e^{- 2} c o s n - {i e}^{- 2} s i n (n)}{1 - e^{- \frac{2}{n}} c o s (1) - i e^{- \frac{2}{n}} s i n (1)}

= \frac{(1 - e^{- 2} c o s (n) - i e^{- 2} s i n n) (1 - e^{- \frac{2}{n}} c o s (1) + i e^{- \frac{2}{n}} s i n (1))}{{(1 - e^{- 2} c o s (1))}^{2} + e^{- \frac{4}{n}} {s i n}^{2} (1)}

= \frac{(1 - e^{- 2} c o s (n)) (1 - e^{- \frac{2}{n}} c o s (1)) + i (1 - e^{- 2} c o s (n)) e^{- \frac{2}{n}} s i n (1) - {i e}^{- 2} s i n (n) (1 - e^{- \frac{2}{n}} c o s (1)) + e^{\frac{4}{n}} s i n (n) s i n (1)}{{(1 - e^{- 2} c o s (1))}^{2} + e^{- \frac{4}{n}} {s i n}^{2} (1)}

s o t h e v a l u e o f A_{n} i s c l e a r d e t e r m i n e d a f t e r e x t r a c t i n g I m (Σ \dots .) .

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

\int_{0}^{\frac{1}{n}} s i n ([n x]) e^{- 2 x} d x + \int_{\frac{1}{n}}^{1} s i n ([n x]) e^{- 2 x} d x

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

s i n (1) \times ∣ \frac{e^{- 2 x}}{- 2} ∣_{\frac{1}{n}}^{1}

s i n (1) \times (\frac{e^{- 2} - e^{\frac{- 2}{n}}}{- 2})