find value of ∫_0 ^1 ln(1+ix^2 )dx and ∫_0 ^1 ln(1−ix^2 )dx with i=(√(−1))


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Question Number 79758 by mathmax by abdo last updated on 27/Jan/20

$f i n d v a l u e o f \int_{0}^{1} l n (1 + {i x}^{2}) d x a n d \int_{0}^{1} l n (1 - {i x}^{2}) d x w i t h i = \sqrt{- 1}$
Commented by mathmax by abdo last updated on 29/Jan/20
$let z from C and f(z)=∫_0 ^1 ln(1+zx^2 )dx ⇒ f^′ (z)=∫_0 ^1 (x^2 /(1+zx^2 ))dx =(1/z) ∫_0 ^1 ((zx^2 +1−1)/(1+zx^2 ))dx =(1/z) −(1/z) ∫_0 ^1 (dx/(1+zx^2 )) and ∫_0 ^1 (dx/(1+zx^2 )) =_(x(√z)=t) ∫_0 ^(√z) (dt/((√z)(1+t^2 ))) =(1/(√z)) arctan((√z)) ⇒f^′ (z)=(1/z)−((arctan((√z)))/(z(√z))) ⇒ f(z)=lnz +∫_1 ^z ((arctan((√u)))/(u(√u))) du +c (u=t^2 ) =lnz +∫_1 ^(√z) ((arctan(t))/t^3 )(2t)dt +c =ln(z)+2∫_1 ^(√z) ((arctan(t))/t^2 )dt +c f(1) =∫_0 ^1 ln(1+x^2 )dx =c ⇒ f(z) =lnz +2∫_1 ^(√z) ((arctan(t))/t^2 )dt +∫_0 ^1 ln(1+x^2 )dx we have by parts ∫_1 ^(√z) ((arctan(t))/t^2 )dt =[−((arctant)/t)]_1 ^(√z) +∫_1 ^(√z) (1/(t(1+t^2 )))dt =(π/4)−((arctan((√z)))/(√z)) +∫_1 ^(√z) ((1/t)−(t/(t^2 +1)))dt =(π/4)−((arctan((√z)))/(√z)) +ln((√z))−[(1/2)ln(t^2 +1)]_1 ^(√z) =(π/4)−((arctan((√z)))/(√z)) +(1/2)ln(z)−(1/2){ln(z+1)−ln(2)} ⇒ f(z)=ln(z)+(π/2)−((2arctan((√z)))/(√z)) +ln(z)−ln(z+1)+ln(2)+∫_0 ^1 ln(1+x^2 )dx f(z)=2ln(z)+(π/2) −2((arctan((√z)))/(√z)) −ln(z+1)+ln(2)+∫_0 ^1 ln(1+x^2 )dx by parts ∫_0 ^1 ln(1+x^2 )dx =[xln(1+x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 )/(1+x^2 ))dx =ln(2)−2 ∫_0 ^1 ((1+x^2 −1)/(1+x^2 ))dx =ln(2)−2 +2 ×(π/4) =ln(2)−2+(π/2) ⇒ f(z) =2ln(z) +2ln(2) +π −2 −2((arctan((√z)))/(√z)) −ln(z+1) =∫_0 ^1 ln(1+zx^2 )dx ⇒ ∫_0 ^1 ln(1+ix^2 )dx =f(i) =2ln(i)+2ln(2)+π−2−2 ×((arctan((√i)))/(√i)) −ln(i+1) =2×((iπ)/2) +2ln(2)+π−2 −2 ×((arctan(e^((iπ)/4) ))/e^((iπ)/4) )−ln((√2)e^((iπ)/4) ) =iπ +2ln(2)+π−2 −2e^(−((iπ)/4)) arctan(e^((iπ)/4) )−(1/2)ln(2)−((iπ)/4)$
$l e t z f r o m C a n d f (z) = \int_{0}^{1} l n (1 + {z x}^{2}) d x \Rightarrow$ $f^{^{'}} (z) = \int_{0}^{1} \frac{x^{2}}{1 + {z x}^{2}} d x = \frac{1}{z} \int_{0}^{1} \frac{{z x}^{2} + 1 - 1}{1 + {z x}^{2}} d x$ $= \frac{1}{z} - \frac{1}{z} \int_{0}^{1} \frac{d x}{1 + {z x}^{2}} a n d \int_{0}^{1} \frac{d x}{1 + {z x}^{2}} =_{x \sqrt{z} = t} \int_{0}^{\sqrt{z}} \frac{d t}{\sqrt{z} (1 + t^{2})}$ $= \frac{1}{\sqrt{z}} a r c t a n (\sqrt{z}) \Rightarrow f^{^{'}} (z) = \frac{1}{z} - \frac{a r c t a n (\sqrt{z})}{z \sqrt{z}} \Rightarrow$ $f (z) = l n z + \int_{1}^{z} \frac{a r c t a n (\sqrt{u})}{u \sqrt{u}} d u + c (u = t^{2})$ $= l n z + \int_{1}^{\sqrt{z}} \frac{a r c t a n (t)}{t^{3}} (2 t) d t + c = l n (z) + 2 \int_{1}^{\sqrt{z}} \frac{a r c t a n (t)}{t^{2}} d t + c$ $f (1) = \int_{0}^{1} l n (1 + x^{2}) d x = c \Rightarrow$ $f (z) = l n z + 2 \int_{1}^{\sqrt{z}} \frac{a r c t a n (t)}{t^{2}} d t + \int_{0}^{1} l n (1 + x^{2}) d x w e h a v e b y p a r t s$ $\int_{1}^{\sqrt{z}} \frac{a r c t a n (t)}{t^{2}} d t = {[- \frac{a r c t a n t}{t}]}_{1}^{\sqrt{z}} + \int_{1}^{\sqrt{z}} \frac{1}{t (1 + t^{2})} d t$ $= \frac{π}{4} - \frac{a r c t a n (\sqrt{z})}{\sqrt{z}} + \int_{1}^{\sqrt{z}} (\frac{1}{t} - \frac{t}{t^{2} + 1}) d t$ $= \frac{π}{4} - \frac{a r c t a n (\sqrt{z})}{\sqrt{z}} + l n (\sqrt{z}) - {[\frac{1}{2} l n (t^{2} + 1)]}_{1}^{\sqrt{z}}$ $= \frac{π}{4} - \frac{a r c t a n (\sqrt{z})}{\sqrt{z}} + \frac{1}{2} l n (z) - \frac{1}{2} {l n (z + 1) - l n (2)} \Rightarrow$ $f (z) = l n (z) + \frac{π}{2} - \frac{2 a r c t a n (\sqrt{z})}{\sqrt{z}} + l n (z) - l n (z + 1) + l n (2) + \int_{0}^{1} l n (1 + x^{2}) d x$ $f (z) = 2 l n (z) + \frac{π}{2} - 2 \frac{a r c t a n (\sqrt{z})}{\sqrt{z}} - l n (z + 1) + l n (2) + \int_{0}^{1} l n (1 + x^{2}) d x$ $b y p a r t s \int_{0}^{1} l n (1 + x^{2}) d x = {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{2}} d x$ $= l n (2) - 2 \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} d x = l n (2) - 2 + 2 \times \frac{π}{4} = l n (2) - 2 + \frac{π}{2} \Rightarrow$ $f (z) = 2 l n (z) + 2 l n (2) + π - 2 - 2 \frac{a r c t a n (\sqrt{z})}{\sqrt{z}} - l n (z + 1)$ $= \int_{0}^{1} l n (1 + {z x}^{2}) d x \Rightarrow$ $\int_{0}^{1} l n (1 + {i x}^{2}) d x = f (i) = 2 l n (i) + 2 l n (2) + π - 2 - 2 \times \frac{a r c t a n (\sqrt{i})}{\sqrt{i}}$ $- l n (i + 1)$ $= 2 \times \frac{i π}{2} + 2 l n (2) + π - 2 - 2 \times \frac{a r c t a n (e^{\frac{i π}{4}})}{e^{\frac{i π}{4}}} - l n (\sqrt{2} e^{\frac{i π}{4}})$ $= i π + 2 l n (2) + π - 2 - 2 e^{- \frac{i π}{4}} a r c t a n (e^{\frac{i π}{4}}) - \frac{1}{2} l n (2) - \frac{i π}{4}$
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