let-f-n-x-sin-nx-n-3-and-f-x-n-1-f-n-x-calculate-0-pi-f-x-dx-

Question Number 52680 by maxmathsup by imad last updated on 11/Jan/19

l e t f_{n} (x) = \frac{s i n (n x)}{n^{3}} a n d f (x) = \sum_{n = 1}^{\infty} f_{n} (x)

c a l c u l a t e \int_{0}^{π} f (x) d x .

Commented by maxmathsup by imad last updated on 11/Jan/19

i t s c l e a r t h a t t h e s e r i e Σ f_{n} (x) c o n v e r g e s i m p l . a n d u n i f . b e c a u s e

∣ f_{n} (x) ∣⩽ \frac{1}{n^{3}} a n d Σ \frac{1}{n^{3}} c o n v e r g e s w e h a v e

\int_{0}^{π} f (x) d x = \int_{0}^{π} \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n^{3}} = \sum_{n = 1}^{\infty} \frac{1}{n^{3}} \int_{0}^{π} s i n (n x) d x

= \sum_{n = 1}^{\infty} \frac{1}{n^{3}} {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} (1 - {(- 1)}^{n})

= 2 \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}} b u t w e h a v e p r o v e d t h a t \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{90} \Rightarrow

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}} + \frac{1}{16} \sum_{n = 1}^{\infty} \frac{1}{n^{4}} = \frac{π^{4}}{90} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}} = \frac{π^{4}}{90} - \frac{1}{16} \frac{π^{4}}{90}

= (1 - \frac{1}{16}) \frac{π^{4}}{90} = \frac{15}{16} \frac{π^{4}}{90} = \frac{3.5}{3.30 16} π^{4} = \frac{π^{4}}{6.16} = \frac{π^{4}}{96} \Rightarrow

\int_{0}^{π} f (x) d x = 2 . \frac{π^{4}}{96} \Rightarrow \int_{0}^{π} f (x) d x = \frac{π^{4}}{48} .