let-f-x-0-ln-1-xt-2-2-t-2-dt-determine-a-explicit-form-of-f-x-2-calculate-0-ln-1-3x-2-2-t-2-dt-

Question Number 59278 by Mr X pcx last updated on 07/May/19

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

d e t e r m i n e a e x p l i c i t f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\infty} \frac{l n (1 + 3 x^{2})}{2 + t^{2}} d t

Commented by maxmathsup by imad last updated on 13/May/19

1) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{(2 + t^{2}) (1 + {x t}^{2})} d t

l e t d e c o m p o s e F (t) = \frac{t^{2}}{(t^{2} + 2) ({x t}^{2} + 1)} c a s e 1 x > 0

F (t) = \frac{a t + b}{t^{2} + 2} + \frac{c t + d}{{x t}^{2} + 1} w e h a v e F (- t) = F (t) \Rightarrow

\frac{- a t + b}{t^{2} + 2} + \frac{- c t + d}{{x t}^{2} + 1} = F (t) \Rightarrow a = c = 0 \Rightarrow F (t) = \frac{b}{t^{2} + 2} + \frac{d}{{x t}^{2} + 1}

{l i m}_{t \to + \infty} t^{2} F (t) = \frac{1}{x} = b + \frac{d}{x} \Rightarrow 1 = b x + d \Rightarrow d = 1 - b x

F (0) = 0 = \frac{b}{2} + d \Rightarrow b + 2 d = 0 \Rightarrow b = - 2 d \Rightarrow d = 1 - (- 2 d) x = 1 + 2 d x \Rightarrow

(1 - 2 x) d = 1 \Rightarrow d = \frac{1}{1 - 2 x} (w e s u p p o s e x \neq \frac{1}{2}) \Rightarrow b = \frac{- 2}{1 - 2 x} = \frac{2}{2 x - 1} \Rightarrow

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

= \frac{2}{2 x - 1} \int_{0}^{\infty} \frac{d t}{t^{2} + 2} - \frac{1}{2 x - 1} \int_{0}^{\infty} \frac{d t}{{x t}^{2} + 1} b u t

\int_{0}^{\infty} \frac{d t}{t^{2} + 2} =_{t = \sqrt{2} u} \int_{0}^{\infty} \frac{\sqrt{2} d u}{2 (1 + u^{2})} = \frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} = \frac{π}{2 \sqrt{2}}

\int_{0}^{\infty} \frac{d t}{{x t}^{2} + 1} =_{\sqrt{x} t = u} \int_{0}^{\infty} \frac{d u}{\sqrt{x} (1 + u^{2})} = \frac{π}{2 \sqrt{x}} \Rightarrow

f^{^{'}} (x) = \frac{2}{2 x - 1} \frac{π}{2 \sqrt{2}} - \frac{1}{2 x - 1} \frac{π}{2 \sqrt{x}} \Rightarrow f (x) = \frac{π}{\sqrt{2}} \int \frac{d x}{2 x - 1} - \frac{π}{2} \int \frac{d x}{\sqrt{x} (2 x - 1)} + c

= \frac{π}{2 \sqrt{2}} l n ∣ 2 x - 1 ∣ - \frac{π}{2} \int \frac{d x}{\sqrt{x} (2 x - 1)} + c

\int \frac{d x}{\sqrt{x} (2 x - 1)} =_{\sqrt{x} = u} \int \frac{2 u d u}{u (2 u^{2} - 1)} = 2 \int \frac{d u}{2 u^{2} - 1}

= 2 \int \frac{d u}{(\sqrt{2} u - 1) (\sqrt{2} u + 1)} = \int (\frac{1}{\sqrt{2} u - 1} - \frac{1}{\sqrt{2} u + 1}) d u

= \frac{1}{\sqrt{2}} l n ∣ \frac{\sqrt{2} u - 1}{\sqrt{2} u + 1} ∣ = \frac{1}{\sqrt{2}} l n ∣ \frac{\sqrt{2 x} - 1}{\sqrt{2 x} + 1} ∣ \Rightarrow

f (x) = \frac{π}{2 \sqrt{2}} l n ∣ 2 x - 1 ∣ - \frac{π}{2 \sqrt{2}} l n ∣ \frac{\sqrt{2 x} - 1}{\sqrt{2 x} + 1} ∣ + c

f (0) = 0 = c \Rightarrow f (x) = \frac{π}{2 \sqrt{2}} {l n ∣ \sqrt{2 x} - 1 ∣ + l n ∣ \sqrt{2 x} + 1 ∣ - l n ∣ \sqrt{2 x} - 1 ∣ + l n ∣ \sqrt{2 x} + 1 ∣}

= \frac{π}{2 \sqrt{2}} (2 l n ∣ \sqrt{2 x} + 1 ∣ \Rightarrow f (x) = \frac{π}{\sqrt{2}} l n ∣ \sqrt{2 x} + 1 ∣ w i t h x > 0 r e s t t o s t y d y t h e c a s e x < 0

Commented by maxmathsup by imad last updated on 13/May/19

d u o t o x > 0 w e h a v e f (x) = \frac{π}{\sqrt{2}} l n (\sqrt{2 x} + 1)

Commented by maxmathsup by imad last updated on 13/May/19

2) \int_{0}^{\infty} \frac{l n (1 + 3 t^{2})}{2 + t^{2}} d t = f (3) = \frac{π}{\sqrt{2}} l n (1 + \sqrt{6}) .