let-give-f-x-0-ln-1-x-t-2-dt-with-x-lt-1-find-a-simple-form-of-f-x-

Question Number 32705 by caravan msup abdo. last updated on 31/Mar/18

l e t g i v e f (x) = \int_{0}^{\infty} l n (1 + \frac{x}{t^{2}}) d t w i t h ∣ x ∣< 1

f i n d a s i m p l e f o r m o f f (x) .

Commented by abdo imad last updated on 03/Apr/18

f^{^{'}} (x) = \int_{0}^{\infty} \frac{1}{t^{2} (1 + \frac{x}{t^{2}})} d t = \int_{0}^{\infty} \frac{d t}{x + t^{2}}

c a s e 1 0 < x < 1 f^{^{'}} (x) =_{t = \sqrt{x} u} \int_{0}^{\infty} \frac{1}{x (1 + t^{2})} \sqrt{x} d u

= \frac{π}{2 \sqrt{x}} \Rightarrow f (x) = π \sqrt{x} + λ b u t λ = f (0) = 0 \Rightarrow f (x) = π \sqrt{x}

- 1 < x < 0 \Rightarrow f^{^{'}} (x) = \int_{0}^{\infty} \frac{d t}{t^{2} - {(\sqrt{- x})}^{2}}

= \frac{1}{2 \sqrt{- x}} \int_{0}^{\infty} (\frac{1}{t - \sqrt{x}} - \frac{1}{t + \sqrt{- x}}) d t

= \frac{1}{2 \sqrt{- x}} {[l n ∣ \frac{t - \sqrt{- x}}{t + \sqrt{- x}} ∣]}_{0}^{\infty} = 0