let z ∈C and ∣z∣<1 find f(x)=∫


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Question Number 60424 by Mr X pcx last updated on 20/May/19

$l e t z \in C a n d ∣ z ∣< 1 f i n d$ $f (x) = \int_{0}^{1} l n (1 + z x) d x .$
Commented bymaxmathsup by imad last updated on 26/May/19
$we have f(z) =∫_0 ^1 ln(1+zx)dx ⇒f^′ (z) =∫_0 ^1 (x/(1+zx)) dx =(1/z) ∫_0 ^1 ((1+zx−1)/(1+zx)) dx =(1/z) −(1/z) ∫_0 ^1 (dx/(1+zx)) =(1/z) −(1/z) ∫_0 ^1 (Σ_(n=0) ^∞ (−1)^n z^n x^n )dx =(1/z) −(1/z)Σ_(n=0) ^∞ (−1)^n z^n ∫_0 ^1 x^n dx =(1/z) −(1/z) Σ_(n=0) ^∞ (((−1)^n )/(n+1)) z^n =(1/z) −(1/z){1+Σ_(n=1) ^∞ (((−1)^n )/(n+1))z^n } =−Σ_(n=1) ^∞ (((−1)^n )/(n+1)) z^(n−1) ⇒f(z) =−∫ (Σ_(n=1) ^∞ (((−1)^n )/(n+1))z^(n−1) )dz +c =−Σ_(n=1) ^∞ (((−1)^n )/(n(n+1))) z^n +c we have f(0) =0=c ⇒ f(z) =−Σ_(n=1) ^∞ (−1)^n {(1/n) −(1/(n+1))}z^n =−Σ_(n=1) ^∞ (((−1)^n )/n) z^n +Σ_(n=1) ^∞ (((−1)^n )/(n+1)) z^n (((−1)^n )/n)z^n =Σ_(n=1) ^∞ (((−z)^n )/n) =−ln(1+z) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) z^n =(1/z) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) z^(n+1) =(1/z) Σ_(n=2) ^∞ (((−1)^(n−1) )/n) z^n =(1/z){ Σ_(n=1) ^∞ (((−1)^(n−1) )/n)z^n −z} =(1/z){ln(1+z) −z} =((ln(1+z))/z) −1 ⇒ f(z) = ln(1+z) +((ln(1+z))/z) −1 ⇒ f(z) =((z+1)/z)ln(1+z) −1.$
$w e h a v e f (z) = \int_{0}^{1} l n (1 + z x) d x \Rightarrow f^{^{'}} (z) = \int_{0}^{1} \frac{x}{1 + z x} d x$ $= \frac{1}{z} \int_{0}^{1} \frac{1 + z x - 1}{1 + z x} d x = \frac{1}{z} - \frac{1}{z} \int_{0}^{1} \frac{d x}{1 + z x}$ $= \frac{1}{z} - \frac{1}{z} \int_{0}^{1} (\sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} x^{n}) d x = \frac{1}{z} - \frac{1}{z} \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} \int_{0}^{1} x^{n} d x$ $= \frac{1}{z} - \frac{1}{z} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n} = \frac{1}{z} - \frac{1}{z} {1 + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n}}$ $= - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n - 1} \Rightarrow f (z) = - \int (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n - 1}) d z + c$ $= - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n (n + 1)} z^{n} + c w e h a v e f (0) = 0 = c \Rightarrow$ $f (z) = - \sum_{n = 1}^{\infty} {(- 1)}^{n} {\frac{1}{n} - \frac{1}{n + 1}} z^{n} = - \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} z^{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n}$ $\frac{{(- 1)}^{n}}{n} z^{n} = \sum_{n = 1}^{\infty} \frac{{(- z)}^{n}}{n} = - l n (1 + z)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n} = \frac{1}{z} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n + 1} z^{n + 1} = \frac{1}{z} \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n - 1}}{n} z^{n}$ $= \frac{1}{z} {\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} z^{n} - z} = \frac{1}{z} {l n (1 + z) - z} = \frac{l n (1 + z)}{z} - 1 \Rightarrow$ $f (z) = l n (1 + z) + \frac{l n (1 + z)}{z} - 1 \Rightarrow f (z) = \frac{z + 1}{z} l n (1 + z) - 1 .$
Commented bymaxmathsup by imad last updated on 26/May/19

$t h e Q i s f i n d f (z) = \int_{0}^{1} l n (1 + z x) d x .$
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