Use-Residus-Theorem-to-explicit-f-a-n-1-1-n-sin-na-n-3-

Question Number 69238 by ~ À ® @ 237 ~ last updated on 21/Sep/19

U s e R e s i d u s T h e o r e m t o e x p l i c i t

f (a) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} s i n (n a)}{n^{3}}

Commented by mathmax by abdo last updated on 22/Sep/19

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

s (x) = I m (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} e^{i n x}}{n}) = I m (\sum_{n = 1}^{\infty} \frac{{(- e^{i x})}^{n}}{n})

l e t w (z) = \sum_{n = 1}^{\infty} \frac{z^{n}}{n} w i t h ∣ z ∣⩽ 1 \Rightarrow w^{^{'}} (z) = \sum_{n = 1}^{\infty} z^{n - 1} = \frac{1}{1 - z} \Rightarrow

w (z) = \int \frac{d z}{1 - z} + c = - l n (1 - z) + c (c = w (0) = 0) \Rightarrow w (z) = - l n (1 - z)

\Rightarrow \sum_{n = 1}^{\infty} \frac{{(- e^{i x})}^{n}}{n} = - l n (1 + e^{i x}) = - l n (1 + c o s x + i s i n x)

= - l n (2 {c o s}^{2} (\frac{x}{2}) + 2 i s i n (\frac{x}{2}) c o s (\frac{x}{2}))

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

\int s (x) d x = - \frac{x^{2}}{4} + c = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x)

x = 0 \Rightarrow c = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = - (2^{1 - 2} - 1) ξ (2) = \frac{1}{2} \frac{π^{2}}{6} = \frac{π^{2}}{12} \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} c o s (n x) = - \frac{x^{2}}{4} + \frac{π^{2}}{12} \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3}} s i n (n x) = - \frac{x^{3}}{12} + \frac{π^{2}}{12} x + c

x = 0 \Rightarrow c = 0 \Rightarrow \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3}} s i n (n x) = - \frac{x^{3}}{12} + \frac{π^{2} x}{12} \Rightarrow

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{3}} s i n (n x) = \frac{x}{12} (π^{2} - x^{2})

Answered by mind is power last updated on 22/Sep/19

W e h a v e

l i m x \to n \frac{π (x - n)}{s i n (π x)} = \frac{π}{(s i n {(π x)}_{x = n}^{^{'}}} = {(- 1)}^{n}

l e t f (z) = \frac{π s i n (z a)}{z^{3} s i n (n π)}

f h a s p o l e s a t z = n w i t h e n \in Z^{*} o f o r d e r 1 a n d i n z = 0 o f o r d e r 3

l e t C_{r} s q u a r o f s i d e r a n d c e n t e r i n 0

\Rightarrow \int_{C_{r}} f (z) d z = \sum_{Z_{k}} 2 i π (R e s (f ∣ z_{k})^{″} Z_{k} \in C_{r}^{″}

a n d w e h a v e l i m ∣ z f (z) ∣\Rightarrow 0

B y j o r d a n l e m m a w e g e t L i m r \to + \infty \int_{C_{r}} f (z) d z = 0

R e s i d u s t h \Rightarrow L i m r \to + \infty \int_{c r} f (z) d z = \sum_{n \hat{\in} I Z} 2 i π R e s (f (z) ∣∣ n)

\Rightarrow 0 = R e s (f ∣ 0) + \sum_{n \in {I Z}^{*}} R e s (f ∣ n)

R e s (f ∣ n) = l i m z \to n \frac{(z - n) π s i n (z a)}{z^{3} s i n (π z)} = \frac{π s i n (n a)}{n^{3} π c o s (n π)} = \frac{{(- 1)}^{n} s i n (n a)}{n^{3}}

R e s (f ∣ 0) w e u s e l a u r e n t s e r i e

\frac{s i n (z a)}{z^{3}} . \frac{π}{s i n (π z)} = \frac{(z a - \frac{{(z a)}^{3}}{6} \dots)}{z^{3}} . \frac{1}{π z - \frac{π^{3} z^{3}}{6} . .}

= \frac{1}{z^{3}} (a - \frac{z^{2} a^{3}}{6} \dots) (1 + \frac{π^{2} z^{2}}{6} \dots) = \frac{a}{z^{3}} + \frac{1}{z} (\frac{a π^{2}}{6} - \frac{a^{3}}{6}) \dots .

R e s (f ∣ 0) = \frac{a}{6} (- a^{2} + π^{2})

\Rightarrow \sum_{∣ n ∣⩾ 1} \frac{{(- 1)}^{n} s i n (n a)}{n^{3}} - \frac{a}{6} (a^{2} - π^{2}) = 0

\Rightarrow 2 \sum_{n = 1}^{+ \infty} \frac{{(- 1)}^{n} s i n (n a)}{n^{3}} = \frac{a}{6} (a^{2} - π^{2})

\Rightarrow \sum_{n = 1}^{+ \infty} \frac{{(- 1)}^{n} s i n (n a)}{n^{3}} = \frac{a}{12} (a^{2} - π^{2})