let-give-n-inyehr-natural-1-find-tbe-value-of-A-n-0-dx-x-2-1-x-2-2-x-2-n-

Question Number 37601 by prof Abdo imad last updated on 15/Jun/18

l e t g i v e n i n y e h r n a t u r a l ⩾ 1 f i n d t b e v a l u e o f

A_{n} = \int_{0}^{\infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) \dots . (x^{2} + n)}

Commented by math khazana by abdo last updated on 17/Jun/18

2 A_{n} = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 1) (x^{2} + 2) \dots . (x^{2} + n)}

l e t φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + 2) \dots . . (z^{2} + n)}

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

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

φ a r e i \sqrt{k} a n d - i \sqrt{k} w i t h k \in {1, 2, \dots, n}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{z_{k} \in w^{+}} R e s (φ, z_{k}) w i t h

w^{+} = {z \in C / I m (z) > 0}

\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}) = \frac{1}{p^{^{'}} (i \sqrt{k})} w i t h

p (x) = (x^{2} + 1) (x^{2} + 2) \dots (x^{2} + n)

l e t q (t) = (t + 1) (t + 2) \dots . (t + n) \Rightarrow

q^{^{'}} (t) = \sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (x + p)

p (x) = q (x^{2}) \Rightarrow p^{^{'}} (x) = 2 x q^{^{'}} (x^{2}) \Rightarrow

p^{^{'}} (x) = 2 x \sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (x^{2} + p) \Rightarrow

p^{^{'}} (i \sqrt{m}) = 2 i \sqrt{m} \sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (p - m)

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{m = 1}^{n} \frac{1}{2 i \sqrt{m}} {\sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (p - m)}

= \frac{π}{\sqrt{m}} \sum_{m = 1}^{n} {\sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (p - m)} .

Commented by math khazana by abdo last updated on 17/Jun/18

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{\sqrt{m}} \sum_{m = 1}^{n} {\sum_{m = 1}^{n} \prod_{p = 1}^{n} (p - m)}^{- 1}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \sum_{m = 1}^{n} \frac{1}{2 i \sqrt{m} \sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (p - m)}

= \sum_{m = 1}^{n} \frac{π}{\sqrt{m}} {\sum_{k = 1}^{n} \prod_{p = 1_{p \neq - k}}^{n} (p - m)}^{- 1}