calculate-A-n-0-dx-x-2-1-x-2-2-x-2-n-with-n-integr-and-n-1-

Question Number 71663 by mathmax by abdo last updated on 18/Oct/19

c a l c u l a t e A_{n} = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) \dots . (x^{2} + n)}

w i t h n i n t e g r a n d n ⩾ 1

Commented by mathmax by abdo last updated on 18/Oct/19

w e h a v e 2 A_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) . . (x^{2} + n)} l e t

φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + 2) \dots (z^{2} + n)} = \frac{1}{\prod_{k = 1}^{n} (z^{2} + k)}

= \frac{1}{\prod_{k = 1}^{n} (z - i \sqrt{k}) (z + i \sqrt{k})} s o t h e p o l e s o f φ a r e \bar{+} i \sqrt{k}

a n d \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{k = 1}^{n} R e s (φ, i \sqrt{k})

R e s (φ, i \sqrt{k}) = {l i m}_{z \to i \sqrt{k}} (z - i \sqrt{k}) φ (z)

= \frac{1}{\prod_{k = 1}^{n} (2 i \sqrt{k}) \prod_{j = 1 a n d j \neq k}^{n} (i \sqrt{k} - i \sqrt{j})} = \frac{1}{{(2 i)}^{n} \sqrt{n!} \prod_{j = 1 a n d j \neq k}^{n} (i \sqrt{k} - i \sqrt{j})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{{(2 i)}^{n} \sqrt{n!} \prod_{j = 1}^{n} (i \sqrt{k} - i \sqrt{j})} \Rightarrow

A_{n} = \sum_{k = 1}^{n} \frac{i π}{{(2 i)}^{n} \sqrt{n!} \prod_{j = 1 a n d j \neq k}^{n} (i \sqrt{k} - i \sqrt{j})}

Answered by mind is power last updated on 18/Oct/19

\frac{1}{(x^{2} + 1) \dots . . (x^{2} + n)} = \underset{k = 1}{\sum^{n}} (\frac{a_{k}}{(x - i \sqrt{k})} + \frac{b_{k}}{(x + i \sqrt{k}})

a_{k} = \lim_{} (x - i \sqrt{k}) . \frac{1}{(x^{2} + 1) \dots \dots (x^{2} + n)}

= \underset{l = 1, l \neq k}{\prod^{n}} \frac{1}{2 i \sqrt{k} . (l - k)}

b_{k} = \underset{l = 1, l \neq k}{\prod^{n}} \frac{1}{- 2 i \sqrt{k} (l - k)}

\frac{1}{(x^{2} + 1) \dots . . (x^{2} + n)} = \underset{k = 1}{\sum^{n}} (\underset{l = 1, l \neq k}{\prod^{n}} \frac{1}{2 i \sqrt{k} (l - k)} (. \frac{1}{x - i \sqrt{k}} - \frac{1}{x + i \sqrt{k}})

= \underset{k = 1}{\sum^{n}} (\underset{l = 1, k \neq l}{\prod^{n}} \frac{1}{(l - k) (x^{2} + k)})

A_{n} = \underset{k = 1}{\sum^{n}} \underset{l = 1, l \neq k}{\prod^{n}} \frac{1}{(l - k)} \underset{0}{\int^{+ \infty}} \frac{dx}{x^{2} + k}

\int \frac{dx}{x^{2} + k} = \frac{1}{\sqrt{k}} . \tan^{- 1} (\frac{x}{\sqrt{k}}) + c

\int_{0}^{+ \infty} \frac{dx}{x^{2} + k} = \frac{π}{2 \sqrt{k}}

A_{n} = \underset{k = 1}{\sum^{n}} \underset{l = 1, l \neq k}{\prod^{n}} \frac{π}{2 \sqrt{k} . (l - k)}

Commented by mathmax by abdo last updated on 18/Oct/19

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