calculate-A-0-1-ln-1-ix-dx-2-calculate-0-1-ln-1-ix-dx-i-2-1-

Question Number 34635 by abdo mathsup 649 cc last updated on 09/May/18

c a l c u l a t e A (α) = \int_{0}^{1} l n (1 + α i x) d x

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

Commented by tanmay.chaudhury50@gmail.com last updated on 09/May/18

Commented by abdo mathsup 649 cc last updated on 10/May/18

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

= \sqrt{1 + α^{2} x^{2}} e^{i θ} / c o s θ = \frac{1}{\sqrt{1 + α^{2} x^{2}}} a n d

s i n θ = \frac{α x}{\sqrt{1 + α^{2} x^{2}}} \Rightarrow t a n θ = α x \Rightarrow θ = a r c t a n (α x)

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

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

= f (α) + i g (α)

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

= \frac{2}{α} - \frac{2}{α} \int_{0}^{1} \frac{d x}{1 + α^{2} x^{2}}

=_{α x = u} \frac{2}{α} - \frac{2}{α} \int_{0}^{α} \frac{1}{1 + u^{2}} \frac{d u}{α}

= \frac{2}{α} - \frac{2}{α^{2}} a r c t a n (α) \Rightarrow f^{^{'}} (α) = \frac{1}{α} - \frac{a r c t a n (α)}{α^{2}}

\Rightarrow f (α) = l n ∣ α ∣ - \int^{α} \frac{a r c r c t a n t}{t^{2}} d t b y p a r t s

\int \frac{a r c t a n t}{t^{2}} d t = - \frac{1}{t} a r c t a n t - \int \frac{- 1}{t} \frac{d t}{1 + t^{2}}

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

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

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

f (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) + c

c = {l i m}_{α \to . 0} (f (α) - \frac{a r c t a n (α)}{α} - \frac{1}{2} l n (1 + α^{2})) = - 1

f (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) - 1

g (α) = \int_{0}^{1} a r c t a n (α x) d x

=_{α x = t} \int_{0}^{α} a r c t a n t \frac{d t}{α} = \frac{1}{α} \int_{0}^{α} a r c t a n t d t

= \frac{1}{α} {{[t a r c t a n t]}_{0}^{α} - \int_{0}^{α} \frac{t}{1 + t^{2}} d t}

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

A (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) - 1

i {a r c t a n (α) - \frac{1}{2 α} l n (1 + α^{2})}

Commented by abdo mathsup 649 cc last updated on 10/May/18

2) \int_{0}^{1} l n (1 + i x) d x = A (1)

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

Commented by abdo mathsup 649 cc last updated on 10/May/18

A (1) = \frac{π}{4} + \frac{1}{2} l n (2) - 1 + i (\frac{π}{4} - \frac{1}{2} l n (2)) .

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

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

= I_{1} + I_{2}

I_{1} = \frac{1}{2} \int_{0}^{1} (α^{2} x^{2} - \frac{α^{4} x^{4}}{2} + \frac{α^{6} x^{6}}{3} - \frac{α^{8} x^{8}}{4} + \dots . d x

u s i n g l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} -

= \frac{1}{2} ∣ \frac{α^{2} x^{3}}{3} - \frac{α^{4} x^{5}}{2 \times 5} + \frac{α^{6} x^{7}}{3 \times 7} - \frac{α^{8} x^{9}}{4 \times 9} \dots . . {\overset{1}{∣}}_{0}

= \frac{1}{2} (\frac{α^{2}}{3} - \frac{α^{4}}{10} + \frac{α^{6}}{21} - \frac{α^{8}}{36} \dots)

I_{2} = \int_{0}^{1} {i t a n}^{- 1} (α x) d x

Commented by tanmay.chaudhury50@gmail.com last updated on 09/May/18

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

I_{2} = \int_{0}^{1} {i t a n}^{- 1} α x d x

= i \int_{0}^{1} (α x - \frac{α^{3} x^{3}}{3} + \frac{α^{5} x^{5}}{5} \dots .) d x

= i ∣ (\frac{α x^{2}}{2} - \frac{α^{3} x^{4}}{3 \times 4} + \frac{α^{5} x^{6}}{5 \times 6} \dots .) ∣_{0}^{1}

= i (\frac{α}{2} - \frac{α^{3}}{12} + \frac{α^{5}}{30} \dots)