find-S-n-1-1-a-2-n-2-2-

Question Number 205173 by mr W last updated on 12/Mar/24

f i n d

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{{(a^{2} + n^{2})}^{2}}

Commented by lepuissantcedricjunior last updated on 12/Mar/24

S = \underset{n = 1}{\sum^{\infty}} \frac{1}{{(a^{2} + n^{2})}^{2}} = \int_{0}^{\infty} \frac{d x}{{(a^{2} + x^{2})}^{2}}

S = \int_{0}^{+ \infty} \frac{d x}{a^{4} {[1 + {(\frac{x}{a})}^{2}]}^{2}}

p o s o n s \frac{x}{a} = t a n β \Leftrightarrow d x = a (1 + t a \overset{2}{n β}) d β

q d : {\begin{cases} x \to 0 \\ x \to \infty \end{cases} \Rightarrow {\begin{cases} β \to 0 \\ β \to \frac{π}{2} \end{cases}

S = \int_{0}^{\frac{π}{2}} \frac{d β}{a^{3} (1 + t a \overset{2}{n β})} = \int_{0}^{\frac{π}{2}} \frac{c o \overset{2}{s β}}{a^{3}} d β

S = \frac{1}{a^{3}} {[\frac{1}{2} β + \frac{1}{4} s i n 2 β]}_{0}^{\frac{π}{2}}

S = \frac{π}{4 a^{3}}

Commented by mr W last updated on 12/Mar/24

h o w c a n y o u g e t

\underset{n = 1}{\sum^{\infty}} \frac{1}{{(a^{2} + n^{2})}^{2}} = \int_{0}^{\infty} \frac{d x}{{(a^{2} + x^{2})}^{2}} ?

Answered by Berbere last updated on 13/Mar/24

\sum_{n ⩾ 1} \frac{1}{(a^{2} + n^{2})} = f (a)

f (a) = \frac{1}{2 a i} (\sum_{n ⩾ 1} \frac{1}{n - i a} - \frac{1}{n + i a}) = \frac{1}{2 a i} (\sum_{n ⩾ 0} \int_{0}^{1} t^{n - i a} - t^{n + i a})

= \frac{1}{2 a i} \int_{0}^{1} \frac{t^{- i a} - t^{i a}}{1 - t} d t; Ψ (z) = - γ + \int_{0}^{1} \frac{1 - x^{z - 1}}{1 - x} d x

S = \frac{1}{2 a i} (Ψ (1 + i a) - Ψ (1 - i a))

= \frac{1}{2 a i} (\frac{1}{i a} + Ψ (i a) - Ψ (1 - i a))

Ψ (1 - z) - Ψ (z) = π c o t (π z)

= - \frac{1}{2 a^{2}} - \frac{1}{2 a i} π c o t (i π a) = - \frac{1}{2 a^{2}} + \frac{π}{2 a} c o t h (π a)

\sum_{n ⩾ 1} \frac{1}{(n^{2} + a^{2})} ⩽ ζ (2) s e r i e C v u n i f o m l y \Rightarrow

a \overset{f}{\to} \sum_{n ⩾ 1} \frac{1}{(n^{2} + a^{2})} i s d e r i v a b l e ? w e c a n c h a n g e Σ a n d d e r i v a t i o n

f^{'} (a) = \sum_{n ⩾ 1} \frac{- 2 a}{{(n^{2} + a^{2})}^{2}} = \frac{1}{a^{3}} - \frac{π}{2 a^{2}} c o t h (π a) + \frac{π^{2}}{2 a^{2}} . \frac{1}{{s i n h}^{2} (a)}

\sum_{n ⩾ 1} \frac{1}{{(n^{2} + a^{2})}^{2}} = - \frac{1}{2 a^{4}} + \frac{π}{4 a^{3}} c o t h (π a) + \frac{π^{2}}{4 a^{2} {s i n h}^{2} (π a)}

Commented by mr W last updated on 13/Mar/24

t h a n k s s i r!

n i c e s o l u t i o n!

Commented by Berbere last updated on 13/Mar/24

Y e s s i r W i t h e p l e a s u r

i l o s t m y A c c u n t W i t c h e r

G o d b l e s s y o u

Commented by mr W last updated on 13/Mar/24

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