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calculste-0-1-e-zx-x-dx-with-z-from-C-and-Re-z-gt-0-




Question Number 90291 by mathmax by abdo last updated on 22/Apr/20
calculste ∫_0 ^∞  ((1−e^(zx) )/x)dx  with z from C and Re(z)>0
$${calculste}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{e}^{{zx}} }{{x}}{dx}\:\:{with}\:{z}\:{from}\:{C}\:{and}\:{Re}\left({z}\right)>\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 22/Apr/20
sorry Re(z)<0 .
$${sorry}\:{Re}\left({z}\right)<\mathrm{0}\:. \\ $$
Commented by mathmax by abdo last updated on 23/Apr/20
let f(z) =∫_0 ^∞  ((1−e^(zx) )/x)dx ⇒f^′ (z) =∫_0 ^∞ (∂/∂z)(((1−e^(zx) )/x))dx  =−∫_0 ^∞  e^(zx)  dx =−[(1/z)e^(zx) ]_0 ^(+∞)   =−(1/z)(−1)=(1/z)  (Rez<0) ⇒  f(z) =lnz +K  so if z =α+iβ     (α<0) ⇒z =(√(α^2  +β^2 ))e^(iarctan((β/α)))  ⇒  ln(z) =(1/2)ln(α^2  +β^2 ) +iarctan((β/α)) ⇒  f(z)=(1/2)ln(α^2  +β^2 )+iarctan((β/α))+K  rest to find  K
$${let}\:{f}\left({z}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{e}^{{zx}} }{{x}}{dx}\:\Rightarrow{f}^{'} \left({z}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\partial}{\partial{z}}\left(\frac{\mathrm{1}−{e}^{{zx}} }{{x}}\right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:{e}^{{zx}} \:{dx}\:=−\left[\frac{\mathrm{1}}{{z}}{e}^{{zx}} \right]_{\mathrm{0}} ^{+\infty} \:\:=−\frac{\mathrm{1}}{{z}}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{{z}}\:\:\left({Rez}<\mathrm{0}\right)\:\Rightarrow \\ $$$${f}\left({z}\right)\:={lnz}\:+{K}\:\:{so}\:{if}\:{z}\:=\alpha+{i}\beta\:\:\:\:\:\left(\alpha<\mathrm{0}\right)\:\Rightarrow{z}\:=\sqrt{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{e}^{{iarctan}\left(\frac{\beta}{\alpha}\right)} \:\Rightarrow \\ $$$${ln}\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+{iarctan}\left(\frac{\beta}{\alpha}\right)\:\Rightarrow \\ $$$${f}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)+{iarctan}\left(\frac{\beta}{\alpha}\right)+{K}\:\:{rest}\:{to}\:{find}\:\:{K} \\ $$$$ \\ $$

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