find-value-of-0-1-ln-1-ix-2-dx-and-0-1-ln-1-ix-2-dx-with-i-1-

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

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}