Let-J-0-f-x-x-1-2-dx-where-f-is-any-function-for-which-the-integral-exists-Show-that-J-0-x-2-f-x-x-1-2-dx-0-5-0-1-x-2-f-x-x-1-2-dx-0-f-u-2-du-

Question Number 2186 by Yozzi last updated on 07/Nov/15

L e t J = \int_{0}^{\infty} f ({(x - x^{- 1})}^{2}) d x w h e r e f i s

a n y f u n c t i o n f o r w h i c h t h e i n t e g r a l

e x i s t s . S h o w t h a t

J = \int_{0}^{\infty} x^{- 2} f ({(x - x^{- 1})}^{2}) d x = 0.5 \int_{0}^{\infty} (1 + x^{- 2}) f ({(x - x^{- 1})}^{2}) d x = \int_{0}^{\infty} f (u^{2}) d u .

Answered by prakash jain last updated on 07/Nov/15

\frac{1}{t} = x \Rightarrow x = 0, t = \infty; x = \infty, t = 0

- t^{- 2} d t = d x

J = \int_{0}^{\infty} f ({(x - x^{- 1})}^{2}) d x

= - \int_{\infty}^{0} t^{- 2} f ({(t^{- 1} - t)}^{2}) d t

= - \int_{\infty}^{0} t^{- 2} f ({(t - t^{- 1})}^{2}) d t

= \int_{0}^{\infty} t^{- 2} f ({(t - t^{- 1})}^{2}) d t

= \int_{0}^{\infty} x^{- 2} f ({(x - x^{- 1})}^{2}) d x (c h a n g e v a r i a b l e x = t)

J = \int_{0}^{\infty} f ({(x - x^{- 1})}^{2}) d x = \int_{0}^{\infty} x^{- 2} f ({(x - x^{- 1})}^{2}) d x

2 J = \int_{0}^{\infty} f ({(x - x^{- 1})}^{2}) d x + \int_{0}^{\infty} x^{- 2} f ({(x - x^{- 1})}^{2}) d x

2 J = \int_{0}^{\infty} (1 + x^{- 2}) f ({(x - x^{- 1})}^{2}) d

\Rightarrow J = 0.5 \int_{0}^{\infty} (1 + x^{- 2}) f ({(x - x^{- 1})}^{2}) d x

x - x^{- 1} = u \Rightarrow (1 + x^{- 2}) d x = d u

x = 0, u = - \infty

x = \infty, u = \infty

J = 0.5 \int_{0}^{\infty} (1 + x^{- 2}) f ({(x - x^{- 1})}^{2}) d x

= 0.5 \int_{- \infty}^{\infty} f (u^{2}) d u

f (u^{2}) i s e v e n, h e n c e

J = 0.5 \times 2 \int_{0}^{\infty} f (u^{2}) d u = \int_{0}^{\infty} f (u^{2}) d u

Commented by Yozzi last updated on 07/Nov/15

T h a n k s!

Facebook Tweet Pin