1-expicite-f-x-0-1-ln-1-xt-2-1-t-2-dt-with-x-0-2-calculate-0-1-ln-1-t-2-1-t-2-dt-and-0-1-ln-1-2t-2-1-t-2-dt-

Question Number 79627 by mathmax by abdo last updated on 26/Jan/20

1) e x p i c i t e f (x) = \int_{0}^{1} \frac{l n (1 + {x t}^{2})}{1 + t^{2}} d t w i t h x ⩾ 0

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

Commented by mathmax by abdo last updated on 30/Jan/20

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

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

\int_{0}^{1} \frac{d t}{{x t}^{2} + 1} =_{t \sqrt{x} = u} \int_{0}^{\sqrt{x}} \frac{d u}{\sqrt{x} (1 + u^{2})} = \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) a n d

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

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

{l i m}_{t \to + \infty} t^{2} F (t) = 0 = b + \frac{d}{x} \Rightarrow d + b x = 0 \Rightarrow d = - b x

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

F (t) = \frac{1}{(1 - x) (t^{2} + 1)} - \frac{x}{(1 - x) ({x t}^{2} + 1)} (x \neq 1) \Rightarrow

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

= \frac{π}{4 (1 - x)} - \frac{x}{1 - x} \times \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) \Rightarrow

f^{^{'}} (x) = \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) - \frac{π}{4 (1 - x)} + \frac{\sqrt{x}}{1 - x} a r c t a n (\sqrt{x})

= (\frac{1}{\sqrt{x}} + \frac{\sqrt{x}}{1 - x}) a r c t a n (\sqrt{x}) - \frac{π}{4 (1 - x)}

= \frac{1}{(1 - x) \sqrt{x}} a r c t a n (\sqrt{x}) - \frac{π}{4 (1 - x)} \Rightarrow

f (x) = \int_{0}^{x} \frac{a r c t a n (\sqrt{u})}{(1 - u) \sqrt{u}} d u - \frac{π}{4} l n (∣ 1 - x ∣) + c

c = f (0) = 0 \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n (\sqrt{u})}{(1 - u) \sqrt{u}} d u - \frac{π}{4} l n ∣ 1 - x ∣

\sqrt{u} = z \Rightarrow \int_{0}^{x} \frac{a r c t a n (\sqrt{u})}{(1 - u) \sqrt{u}} d u = \int_{0}^{\sqrt{x}} \frac{a r c t a n (z)}{(1 - z^{2}) z} (2 z) d z

= 2 \int_{0}^{\sqrt{x}} \frac{a r c t a n (z)}{1 - z^{2}} d z \Rightarrow f (x) = 2 \int_{0}^{\sqrt{x}} \frac{a r c t a n (z)}{1 - z^{2}} d z - \frac{π}{4} l n ∣ 1 - x ∣ (x \neq 1)

Commented by mathmax by abdo last updated on 30/Jan/20

2) l e t A = \int_{0}^{1} \frac{l n (1 + t^{2})}{1 + t^{2}} d t \Rightarrow A = \int_{0}^{1} \frac{l n (1 + i t) + l n (1 - i t)}{1 + t^{2}} d t

= \int_{0}^{1} \frac{l n (1 + i t)}{1 + t^{2}} + c o n j (\int_{0}^{1} \frac{l n (1 + i t)}{1 + t^{2}} d t) = 2 R e (\int_{0}^{1} \frac{l n (1 + i t)}{1 + t^{2}} d t)

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

d e c o m p o s i t t i o n o f F (t) = \frac{t}{(1 + z t) (t^{2} + 1)}

= \frac{a}{z t + 1} + \frac{b t + c}{t^{2} + 1}

a = (z t + 1) F (t) ∣_{t = - \frac{1}{z}} = - \frac{1}{z (\frac{1}{z^{2}} + 1)} = - \frac{1}{z (\frac{1 + z^{2}}{z^{2}})} = - \frac{z}{1 + z^{2}}

{l i m}_{t \to + \infty} t F (t) = 0 = \frac{a}{z} + b \Rightarrow b = - \frac{a}{z} = \frac{1}{1 + z^{2}}

F (0) = 0 = a + c \Rightarrow c = - a = \frac{z}{1 + z^{2}} \Rightarrow

F (t) = - \frac{z}{(1 + z^{2}) (z t + 1)} + \frac{1}{1 + z^{2}} \times \frac{t + z}{t^{2} + 1} \Rightarrow

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

= - \frac{1}{1 + z^{2}} l n (z + 1) + \frac{1}{2 (1 + z^{2})} l n (2) + \frac{z}{1 + z^{2}} \times \frac{π}{4} \Rightarrow

f (z) = - \int_{0}^{z} \frac{l n (1 + u)}{u^{2} + 1} d u + \frac{l n (2)}{2} \int_{0}^{z} \frac{d u}{1 + u^{2}} + \frac{π}{8} \int_{0}^{z} \frac{2 u}{1 + u^{2}} d u + c

= \frac{l n (2)}{2} a r c t a n (z) + \frac{π}{8} l n (1 + z^{2}) - \int_{0}^{z} \frac{l n (1 + u)}{u^{2} + 1} d u (c = f (0) = 0)

f (1) = \int_{0}^{1} \frac{l n (1 + t)}{1 + t^{2}} d t = \frac{l n (2)}{2} \times \frac{π}{4} + \frac{π}{8} l n (2) - \int_{0}^{1} \frac{l n (1 + u)}{u^{2} + 1} d u \Rightarrow

2 \int_{0}^{1} \frac{l n (1 + u)}{1 + u^{2}} d u = \frac{π}{4} l n (2) \Rightarrow \int_{0}^{1} \frac{l n (1 + u)}{1 + u^{2}} d u = \frac{π}{8} l n (2)

t h e e x p l i c i t f o r m o f f (x) d o n t g i v e f (i) . . s o l e t t r y a n o t h e r w a y

\dots . b e c o n t i n u e d \dots