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Question Number 121659 by Dwaipayan Shikari last updated on 10/Nov/20

\int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + a^{2} x^{2}) (1 + a^{4} x^{2}) (1 + a^{6} x^{2}) \dots . .}

Commented by Dwaipayan Shikari last updated on 10/Nov/20

\int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + a^{2} x^{2})} = \frac{π}{2 (a + 1)}

\int_{0}^{\infty} \frac{1}{(1 + x^{2}) (1 + a^{2} x^{2}) (1 + a^{4} x^{2})} = \frac{π}{2 (1 + a + a^{3})}

\int_{0}^{\infty} \frac{1}{(1 + x^{2}) (1 + a^{2} x^{2}) (1 + a^{4} x^{2}) (1 + a^{6} x^{2})} = \frac{π}{2 (1 + a + a^{3} + a^{6})}

\int_{0}^{\infty} \frac{1}{(1 + x^{2}) (1 + a^{2} x^{2}) (1 + a^{4} x^{2}) \dots .} = \frac{π}{2 (1 + a + a^{3} + a^{6} + a^{10} + C)}

Commented by TANMAY PANACEA last updated on 10/Nov/20

e x c e l l e n t

Commented by Dwaipayan Shikari last updated on 10/Nov/20

R a m a n u j u n s h o w e d t h i s r e s u l t w i t h o u t a n y p r o o f .

I h a v e t r i e d t o p r o v e t h e r e s u l t

Commented by Dwaipayan Shikari last updated on 10/Nov/20

Answered by mindispower last updated on 11/Nov/20

i f o u n d

\int_{0}^{\infty} \frac{d x}{\underset{k = 0}{\prod^{n}} (1 + a^{2 k} x^{2 k})} = \frac{π}{2} \underset{k = 0}{\sum^{n}} \frac{1}{a^{k} \underset{j = 0, j \neq k}{\prod^{n}} (1 - \frac{a^{2 j}}{a^{2 k}})}

Commented by Dwaipayan Shikari last updated on 11/Nov/20

x i s n o t v a r y i n g

Answered by mindispower last updated on 11/Nov/20

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

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

i n t e g r a t i n c o m p l e x u p e r h a l f p l a n I m (z) > 0

1 + a^{2 k} x^{2} = 0 \Rightarrow x = \underline{+} \frac{i}{a^{k}}

p o l e s {a r e t}_{k} = \frac{i}{a^{k}}, k \in [0, \dots n]

R e s (g, \frac{i}{a^{k}}) = \lim_{x \to \frac{i}{a^{k}}} (x - \frac{i}{a^{k}}) g (x) = \frac{1}{2 {i a}^{k}} . \underset{j = 0, j \neq k}{\prod^{n}} \frac{1}{(1 - \frac{a^{2 j}}{a^{2 k}})}

f (n) = \frac{1}{2} . 2 i π R e s (g, t_{k}, k \in [0, n])

= \frac{1}{2} . 2 i π \sum_{k ⩽ n} \frac{1}{2 {i a}_{k}} = \frac{π}{2} \sum_{k ⩽ n} \frac{1}{a^{k}} . \underset{j = 0, j \neq k}{\prod^{n}} \frac{1}{(1 - {(\frac{a^{j}}{a^{k}})}^{2})}