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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.
$${let}\:{z}\:\in{C}\:{and}\:\:\mid{z}\mid<\mathrm{1}\:\:{find} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{zx}\right){dx}. \\ $$
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.
$${we}\:{have}\:{f}\left({z}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{zx}\right){dx}\:\:\:\:\Rightarrow{f}^{'} \left({z}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{\mathrm{1}+{zx}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{{z}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{zx}−\mathrm{1}}{\mathrm{1}+{zx}}\:{dx}\:=\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{z}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{zx}} \\ $$$$=\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{z}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {z}^{{n}} {x}^{{n}} \right){dx}\:=\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{z}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {z}^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} {dx} \\ $$$$=\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{z}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{z}^{{n}} \:\:=\frac{\mathrm{1}}{{z}}\:−\frac{\mathrm{1}}{{z}}\left\{\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{z}^{{n}} \right\} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{z}^{{n}−\mathrm{1}} \:\Rightarrow{f}\left({z}\right)\:=−\int\:\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{z}^{{n}−\mathrm{1}} \right){dz}\:+{c} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:{z}^{{n}} \:+{c}\:\:\:{we}\:{have}\:\:{f}\left(\mathrm{0}\right)\:=\mathrm{0}={c}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}}{{n}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\}{z}^{{n}} \:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{z}^{{n}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{z}^{{n}} \\ $$$$\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{z}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−{z}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}+{z}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{z}^{{n}} \:=\frac{\mathrm{1}}{{z}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{z}^{{n}+\mathrm{1}} \:=\frac{\mathrm{1}}{{z}}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{z}^{{n}} \\ $$$$=\frac{\mathrm{1}}{{z}}\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{z}^{{n}} \:\:−{z}\right\}\:=\frac{\mathrm{1}}{{z}}\left\{{ln}\left(\mathrm{1}+{z}\right)\:−{z}\right\}\:=\frac{{ln}\left(\mathrm{1}+{z}\right)}{{z}}\:−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({z}\right)\:=\:{ln}\left(\mathrm{1}+{z}\right)\:+\frac{{ln}\left(\mathrm{1}+{z}\right)}{{z}}\:−\mathrm{1}\:\Rightarrow\:{f}\left({z}\right)\:=\frac{{z}+\mathrm{1}}{{z}}{ln}\left(\mathrm{1}+{z}\right)\:−\mathrm{1}. \\ $$
Commented by maxmathsup by imad last updated on 26/May/19
the Q is find f(z) =∫_0 ^1 ln(1+zx)dx .
$${the}\:{Q}\:{is}\:{find}\:{f}\left({z}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{zx}\right){dx}\:. \\ $$

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