1-calculate-U-n-0-e-n-x-sin-pix-n-dx-nnatural-and-n-1-2-determine-nature-of-U-n-

Question Number 87530 by mathmax by abdo last updated on 04/Apr/20

1) c a l c u l a t e U_{n} = \int_{0}^{\infty} e^{- n [x]} s i n (\frac{π x}{n}) d x n n a t u r a l a n d n ⩾ 1

2) d e t e r m i n e n a t u r e o f Σ U_{n}

Commented by mathmax by abdo last updated on 05/Apr/20

1) U_{n} = \sum_{p = 0}^{\infty} \int_{p}^{p + 1} e^{- n p} s i n (\frac{π x}{n}) d x

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

= \sum_{p = 0}^{\infty} e^{- n p} {[- \frac{n}{π} c o s (\frac{π x}{n})]}_{p}^{p + 1}

= - \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} {c o s (\frac{π}{n} (p + 1)} - c o s (\frac{p π}{n})}

= - \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{π (p + 1)}{n}) + \frac{n}{π} \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{p π}{n})

w e h a v e \sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{p π}{n}) = R e (\sum_{p = 0}^{\infty} e^{- n p + i \frac{p π}{n}}) = R e (A_{n})

a n d A_{n} = \sum_{p = 0}^{\infty} {(e^{- n + \frac{i π}{n}})}^{p} = \frac{1}{1 - e^{- n + \frac{i π}{n}}}

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

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

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

\sum_{p = 0}^{\infty} e^{- n p} c o s (\frac{π (p + 1)}{n}) = \sum_{p = 1}^{\infty} e^{- n (p - 1)} c o s (\frac{π p}{n})

= e^{n} \sum_{p = 1}^{\infty} e^{- n p} c o s (\frac{π p}{n}) = e^{n} (\sum_{p = 1}^{\infty} e^{- n p} c o s (\frac{π p}{n}) - 1)

= e^{n} (A_{n} - 1) \Rightarrow U_{n} = - \frac{n}{π} e^{n} (A_{n} - 1) + \frac{n}{π} A_{n}

\Rightarrow U_{n} = \frac{n}{π} (1 - e^{n}) A_{n} + \frac{{n e}^{n}}{π}