calculate A(α) = ∫_0 ^1 ln(1+αix)dx 2) calculate ∫


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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
$we have 1+i(αx) =(√(1+α^2 x^2 )) ( (1/(√(1+α^2 x^2 ))) +i((αx)/(√(1+α^2 x^2 )))) =(√(1+α^2 x^2 )) e^(iθ) /cosθ =(1/(√(1+α^2 x^2 ))) and sinθ = ((αx)/(√(1+α^2 x^2 ))) ⇒ tanθ =αx ⇒θ =arctan(αx) ln(1 +iαx)=(1/2)ln(1+α^2 x^2 ) +i arctan(αx) ⇒ A(α) = (1/2) ∫_0 ^1 ln(1+α^2 x^2 )dx +i ∫_0 ^1 arctan(αx)dx =f(α) +i g(α) 2f^′ (α) = ∫_0 ^1 ((2αx^2 )/(1+α^2 x^2 ))dx = (2/α) ∫_0 ^1 ((1+α^2 x^2 −1)/(1+α^2 x^2 ))dx = (2/α) −(2/α) ∫_0 ^1 (dx/(1+α^2 x^2 )) =_(αx =u) (2/α) −(2/α) ∫_0 ^α (1/(1+u^2 )) (du/α) =(2/α) −(2/α^2 ) arctan(α) ⇒ f^′ (α) = (1/α) −((arctan(α))/α^2 ) ⇒ f(α)= ln∣α∣ − ∫^α ((arcrctant)/t^2 ) dt by parts ∫ ((arctant)/t^2 )dt = −(1/t) arctant −∫ ((−1)/t) (dt/(1+t^2 )) =−(1/t)arctant + ∫ (dt/(t(1+t^2 ))) =−((arctant)/t) + ∫ ((1/t) −(t/(1+t^2 )))dt =−((arctant)/t) +ln∣t∣ −(1/2) ln(1+t^2 ) ⇒ f(α)= ((arctan(α))/α) +(1/2)ln(1+α^2 ) +c c =lim_(α→.0) (f(α)−((arctan(α))/α) −(1/2)ln(1+α^2 ))=−1 f(α) =((arctan(α))/α) +(1/2)ln(1+α^2 ) −1 g(α) = ∫_0 ^1 arctan(αx)dx =_(αx =t) ∫_0 ^α arctant (dt/α) =(1/α) ∫_0 ^α arctant dt = (1/α){ [t arctant]_0 ^α −∫_0 ^α (t/(1+t^2 ))dt} = (1/α){ α arctan(α) −(1/2)ln(1+α^2 )} ⇒ A(α) = ((arctan(α))/α) +(1/2)ln(1+α^2 ) −1 i{ arctan(α) −(1/(2α))ln(1+α^2 )}$
$w e h a v e 1 + i (α x) = \sqrt{1 + α^{2} x^{2}} (\frac{1}{\sqrt{1 + α^{2} x^{2}}} + i \frac{α x}{\sqrt{1 + α^{2} x^{2}}})$ $= \sqrt{1 + α^{2} x^{2}} e^{i θ} / c o s θ = \frac{1}{\sqrt{1 + α^{2} x^{2}}} a n d$ $s i n θ = \frac{α x}{\sqrt{1 + α^{2} x^{2}}} \Rightarrow t a n θ = α x \Rightarrow θ = a r c t a n (α x)$ $l n (1 + i α x) = \frac{1}{2} l n (1 + α^{2} x^{2}) + i a r c t a n (α x) \Rightarrow$ $A (α) = \frac{1}{2} \int_{0}^{1} l n (1 + α^{2} x^{2}) d x + i \int_{0}^{1} a r c t a n (α x) d x$ $= f (α) + i g (α)$ $2 f^{^{'}} (α) = \int_{0}^{1} \frac{2 α x^{2}}{1 + α^{2} x^{2}} d x = \frac{2}{α} \int_{0}^{1} \frac{1 + α^{2} x^{2} - 1}{1 + α^{2} x^{2}} d x$ $= \frac{2}{α} - \frac{2}{α} \int_{0}^{1} \frac{d x}{1 + α^{2} x^{2}}$ $=_{α x = u} \frac{2}{α} - \frac{2}{α} \int_{0}^{α} \frac{1}{1 + u^{2}} \frac{d u}{α}$ $= \frac{2}{α} - \frac{2}{α^{2}} a r c t a n (α) \Rightarrow f^{^{'}} (α) = \frac{1}{α} - \frac{a r c t a n (α)}{α^{2}}$ $\Rightarrow f (α) = l n ∣ α ∣ - \int^{α} \frac{a r c r c t a n t}{t^{2}} d t b y p a r t s$ $\int \frac{a r c t a n t}{t^{2}} d t = - \frac{1}{t} a r c t a n t - \int \frac{- 1}{t} \frac{d t}{1 + t^{2}}$ $= - \frac{1}{t} a r c t a n t + \int \frac{d t}{t (1 + t^{2})}$ $= - \frac{a r c t a n t}{t} + \int (\frac{1}{t} - \frac{t}{1 + t^{2}}) d t$ $= - \frac{a r c t a n t}{t} + l n ∣ t ∣ - \frac{1}{2} l n (1 + t^{2}) \Rightarrow$ $f (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) + c$ $c = {l i m}_{α \to . 0} (f (α) - \frac{a r c t a n (α)}{α} - \frac{1}{2} l n (1 + α^{2})) = - 1$ $f (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) - 1$ $g (α) = \int_{0}^{1} a r c t a n (α x) d x$ $=_{α x = t} \int_{0}^{α} a r c t a n t \frac{d t}{α} = \frac{1}{α} \int_{0}^{α} a r c t a n t d t$ $= \frac{1}{α} {{[t a r c t a n t]}_{0}^{α} - \int_{0}^{α} \frac{t}{1 + t^{2}} d t}$ $= \frac{1}{α} {α a r c t a n (α) - \frac{1}{2} l n (1 + α^{2})} \Rightarrow$ $A (α) = \frac{a r c t a n (α)}{α} + \frac{1}{2} l n (1 + α^{2}) - 1$ $i {a r c t a n (α) - \frac{1}{2 α} l n (1 + α^{2})}$
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

	$\int_{0}^{1} l n (1 + i α x) = \int_{0}^{1} \frac{1}{2} l n (1^{2} + α^{2} x^{2}) + {i t a n}^{- 1} (\frac{α x}{1}) d x$ $= I_{1} + I_{2}$ $I_{1} = \frac{1}{2} \int_{0}^{1} (α^{2} x^{2} - \frac{α^{4} x^{4}}{2} + \frac{α^{6} x^{6}}{3} - \frac{α^{8} x^{8}}{4} + . . . . d x$ $u s i n g l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} -$ $= \frac{1}{2} ∣ \frac{α^{2} x^{3}}{3} - \frac{α^{4} x^{5}}{2 \times 5} + \frac{α^{6} x^{7}}{3 \times 7} - \frac{α^{8} x^{9}}{4 \times 9} . . . . . {\overset{1}{∣}}_{0}$ $= \frac{1}{2} (\frac{α^{2}}{3} - \frac{α^{4}}{10} + \frac{α^{6}}{21} - \frac{α^{8}}{36} . . .)$ $I_{2} = \int_{0}^{1} {i t a n}^{- 1} (α x) d x$
	Commented by tanmay.chaudhury50@gmail.com last updated on 09/May/18

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

	$I_{2} = \int_{0}^{1} {i t a n}^{- 1} α x d x$ $= i \int_{0}^{1} (α x - \frac{α^{3} x^{3}}{3} + \frac{α^{5} x^{5}}{5} . . . .) d x$ $= i ∣ (\frac{α x^{2}}{2} - \frac{α^{3} x^{4}}{3 \times 4} + \frac{α^{5} x^{6}}{5 \times 6} . . . .) ∣_{0}^{1}$ $= i (\frac{α}{2} - \frac{α^{3}}{12} + \frac{α^{5}}{30} . . .)$
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