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let-z-C-and-z-lt-1-find-f-x-0-1-ln-1-zx-dx-

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Question Number 60424 by Mr X pcx last updated on 20/May/19
let z ∈C and ∣z∣<1 find f(x)=∫_0 ^1 ln(1+zx)dx.
letz∈Cand∣z∣<1findf(x)=∫01ln(1+zx)dx.
Commented by maxmathsup 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.
wehavef(z)=∫01ln(1+zx)dx⇒f′(z)=∫01x1+zxdx=1z∫011+zx−11+zxdx=1z−1z∫01dx1+zx=1z−1z∫01(∑n=0∞(−1)nznxn)dx=1z−1z∑n=0∞(−1)nzn∫01xndx=1z−1z∑n=0∞(−1)nn+1zn=1z−1z{1+∑n=1∞(−1)nn+1zn}=−∑n=1∞(−1)nn+1zn−1⇒f(z)=−∫(∑n=1∞(−1)nn+1zn−1)dz+c=−∑n=1∞(−1)nn(n+1)zn+cwehavef(0)=0=c⇒f(z)=−∑n=1∞(−1)n{1n−1n+1}zn=−∑n=1∞(−1)nnzn+∑n=1∞(−1)nn+1zn(−1)nnzn=∑n=1∞(−z)nn=−ln(1+z)∑n=1∞(−1)nn+1zn=1z∑n=1∞(−1)nn+1zn+1=1z∑n=2∞(−1)n−1nzn=1z{∑n=1∞(−1)n−1nzn−z}=1z{ln(1+z)−z}=ln(1+z)z−1⇒f(z)=ln(1+z)+ln(1+z)z−1⇒f(z)=z+1zln(1+z)−1.
Commented by maxmathsup by imad last updated on 26/May/19
the Q is find f(z) =∫_0 ^1 ln(1+zx)dx .
theQisfindf(z)=∫01ln(1+zx)dx.

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