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Question Number 52459 by Abdo msup. last updated on 08/Jan/19

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

1) 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 (α)

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

Commented by maxmathsup by imad last updated on 08/Jan/19

1) w e h a v e f^{^{'}} (α) = \int_{0}^{1} \frac{i x}{(1 + i α x) (1 + x^{2})} d x s o f o r α \neq 0

f^{^{'}} (α) = \frac{1}{α} \int_{0}^{1} \frac{1 + i α x - 1}{(1 + i α x) (x^{2} + 1)} d x = \frac{1}{α} - \frac{1}{α} \int_{0}^{1} \frac{d x}{(i α x + 1) (x^{2} + 1)} d x l e t

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

F (x) = \frac{a}{i α x + 1} + \frac{b x + c}{x^{2} + 1}

a = {l i m}_{x \to \frac{- 1}{i α}} (i α x + 1) F (x) = \frac{1}{({(\frac{1}{i α})}^{2} + 1)} = \frac{1}{\frac{1}{- α^{2}} + 1} = - \frac{1}{1 - α^{2}} α^{2} = \frac{α^{2}}{α^{2} - 1}

{l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{i α} + b \Rightarrow b = - \frac{a}{i α} = \frac{i a}{α} \Rightarrow

F (x) = \frac{α^{2}}{α^{2} - 1} \frac{1}{i α x + 1} + \frac{\frac{i a}{α} x + c}{x^{2} + 1} (w e s u p p o s e α \neq \bar{+} 1)

F (o) = 1 = \frac{α^{2}}{α^{2} - 1} + c \Rightarrow c = 1 - \frac{α^{2}}{α^{2} - 1} = \frac{- 1}{α^{2} - 1} \Rightarrow

F (x) = \frac{α^{2}}{(α^{2} - 1) (1 + i α x)} + \frac{\frac{i}{α} \frac{α^{2}}{α^{2} - 1} x - \frac{1}{α^{2} - 1}}{x^{2} + 1}

= \frac{α^{2}}{(α^{2} - 1) (1 + i α x)} + \frac{1}{α^{2} - 1} \frac{i α x - 1}{x^{2} + 1} \Rightarrow

f^{^{'}} (α) = \frac{1}{α} - \frac{1}{α} \int_{0}^{1} \frac{α^{2}}{(α^{2} - 1) (1 + i α x)} d x - \frac{1}{α (α^{2} - 1)} \int_{0}^{1} \frac{i α x - 1}{x^{2} + 1} d x

= \frac{1}{α} - \frac{α}{α^{2} - 1} \frac{1}{i α} \int_{0}^{1} \frac{i α d x}{1 + i α x} d x - \frac{i}{2 (α^{2} - 1)} {[l n (x^{2} + 1)]}_{0}^{1} + \frac{π}{4 α (α^{2} - 1)}

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

f (α) = l n ∣ α ∣ + i \int \frac{l n (1 + i α)}{α^{2} - 1} d α + i \frac{l n (2)}{2} \int \frac{d α}{α^{2} - 1} + \frac{π}{4} \int \frac{d α}{α (α^{2} - 1)} + c b u t

\int \frac{d α}{α^{2} - 1} = \frac{1}{2} \int (\frac{1}{α - 1} - \frac{1}{α + 1}) d α = \frac{1}{2} l n ∣ \frac{α - 1}{α + 1} ∣ l e t d e c o m p o s e

w (α) = \frac{1}{α (α^{2} - 1)} = \frac{1}{α (α - 1) (α + 1)}

w (α) = \frac{a}{α} + \frac{b}{α - 1} + \frac{c}{α + 1} \Rightarrow a = - 1

b = \frac{1}{2} a n d c = \frac{1}{2} \Rightarrow w (α) = - \frac{1}{α} + \frac{1}{2 (α - 1)} + \frac{1}{2 (α + 1)} \Rightarrow

\int \frac{d α}{α (α^{2} - 1)} = - l n ∣ α ∣ + \frac{1}{2} l n ∣ α^{2} - 1 ∣ \Rightarrow

f (α) = i \int \frac{l n (1 + i α)}{α^{2} - 1} d α + \frac{i l n (2)}{4} l n ∣ \frac{α - 1}{α + 1} ∣ + (1 - \frac{π}{4}) l n ∣ α ∣ + \frac{π}{8} l n ∣ α^{2} - 1 ∣ + c

Commented by Abdo msup. last updated on 08/Jan/19

2) w e h a v e 1 + i x = \sqrt{1 + x^{2}} (\frac{1}{\sqrt{1 + x^{2}}} + i \frac{x}{\sqrt{1 + x^{2}}}) = r e^{i θ} \Rightarrow

r = \sqrt{1 + x^{2}} a n d θ = a r c t a n x \Rightarrow

l n (1 + i x) = \frac{1}{2} l n (1 + x^{2}) + i a r c t a n x \Rightarrow

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

b y p a r t s I = \int_{0}^{1} \frac{a r c t a n x}{1 + x^{2}} d x = {[{a r c t a n}^{2} x]}_{0}^{1}

- \int_{0}^{1} \frac{a r c t a n x}{1 + x^{2}} = \frac{π^{2}}{16} - I \Rightarrow 2 I = \frac{π^{2}}{16} \Rightarrow I = \frac{π^{2}}{32}

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

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

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

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

= \frac{1}{\sqrt{t}} \int_{0}^{\sqrt{t}} \frac{u^{2}}{(u^{2} + 1) (u^{2} + t)} d u l e t d e c o m p o s e

F (u) = \frac{u^{2}}{(u^{2} + 1) (u^{2} + t)} \Rightarrow

F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + t}

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

= F (u) \Rightarrow a = 0 a n d c = 0 \Rightarrow

F (u) = \frac{b}{u^{2} + 1} + \frac{d}{u^{2} + t} w e s e e t h a t

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

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

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

\int_{0}^{\sqrt{t}} \frac{d u}{u^{2} + t} =_{u = \sqrt{t} α} \int_{0}^{1} \frac{\sqrt{t} d α}{t (1 + α^{2})} = \frac{1}{\sqrt{t}} \frac{π}{4} \Rightarrow

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

f (t) = \int \frac{a r c t a n (\sqrt{t})}{\sqrt{t} (t - 1)} d y - \frac{π}{4} \int \frac{d t}{t (t - 1)} + C

\int \frac{d t}{t (t - 1)} = \int (\frac{1}{t - 1} - \frac{1}{t}) d t = l n ∣ \frac{t - 1}{t} ∣ + c_{1}

\int \frac{a r c t a n (\sqrt{t})}{\sqrt{t} (t - 1)} d t =_{\sqrt{t} = u} \int \frac{a r c t a n (u)}{u (u^{2} - 1)} (2 u) d u

= 2 \int \frac{a r c t a n (u)}{u^{2} - 1} d u \dots . b e c o n t i n u e d \dots