0-1-log-1-x-3-dx-and-0-1-log-1-x-4-dx-and-if-possible-then-find-the-value-of-p-p-0-1-log-1-x-n-dx-n-N-

Question Number 207652 by universe last updated on 22/May/24

\int_{0}^{1} \log (1 + x^{3}) d x = ? a n d \int_{0}^{1} \log (1 + x^{4}) d x = ?

and if possible then find the value of p

p = \int_{0}^{1} \log (1 + x^{n}) d x = ? n \in N

Answered by Berbere last updated on 22/May/24

u s i n g x^{n} + 1 = \underset{k = 0}{\prod^{n - 1}} (x - e^{i (2 k + 1) \frac{π}{n}}) = \underset{k = 0}{\prod^{n - 1}} l n (x - e^{\frac{i (2 k + 1) π}{n}})

\int_{0}^{1} l n (x - e^{i (\frac{2 k + 1) π}{n})}) d x = l n (e^{i (π + \frac{2 k + 1}{n} π)} (1 - {x e}^{\frac{i (2 k + 1) π}{n}}))

u s i n g p r i n c i p a l l o g r e p r e s e n t a t i o n

l o g (z) = l n ∣ z ∣ + i a r g (z); a r g (z) \in [- \frac{π}{2}, \frac{3 π}{2} [

e^{i (π + \frac{2 k + 1}{n} π)} = e^{i (- π + \frac{2 k + 1}{n} π)}; l n (e^{i (- π + \frac{2 k + 1}{n} π)}) = i (- π + \frac{2 k + 1}{n} π)

p = \underset{k = 0}{\sum^{n - 1}} i (- π + \frac{2 k + 1}{n} π) + \underset{k = 0}{\sum^{n - 1}} \int_{0}^{1} l n (1 - {x e}^{- i (\frac{2 k + 1}{n}) π}) d x

= \underset{k = 0}{\sum^{n - 1}} \int_{0}^{1} l n (1 - {x e}^{- i (\frac{2 k + 1}{n}) π}) d x

{l {n (1 - x a)]}_{0}^{1} = \frac{1}{a} ((a x - 1) l n (1 - x a) - a x)]}_{0}^{1}

= \frac{1}{a} (a - 1) l n (1 - a) - 1

= \underset{k = 0}{\sum^{n - 1}} e^{i (\frac{2 k + 1}{n}) π} [(e^{- i (\frac{2 k + 1}{n}) π} - 1) l n (1 - e^{- \frac{i π}{n} (2 k + 1)}) - 1]

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