Question Number 219704 by Spillover last updated on 01/May/25
Answered by SdC355 last updated on 01/May/25
$$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\mathrm{ln}\left(\frac{\mathrm{1}}{{z}}−\mathrm{1}\right)\:\mathrm{d}{z}=\mathrm{0}\: \\ $$$$\int_{\mathrm{0}} ^{\:\rho} \:+\int_{\:\rho} ^{\:\mathrm{1}} =\mathrm{0}\:,\:\left(\rho<\mathrm{1}\right) \\ $$
Answered by vnm last updated on 01/May/25
![∫_0 ^1 ln(−1+(1/x))dx= [−1+(1/x)=e^u , x=(1/(e^u +1)), dx=−((e^u du)/((e^u +1)^2 )), u_1 =ln(−1+(1/0^+ ))=+∞, u_2 =ln(−1+(1/(1−0^+ )))=−∞]= ∫_(+∞) ^(−∞) u(−(e^u /((e^u +1)^2 )))du=∫_(−∞) ^(+∞) ((e^u u)/((e^u +1)^2 ))du= ∫_(−∞) ^(+∞) (u/(e^u +2+e^(−u) ))du=0 Integral of an odd function which obviously converges](https://www.tinkutara.com/question/Q219726.png)
$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(−\mathrm{1}+\frac{\mathrm{1}}{{x}}\right){dx}= \\ $$$$\left[−\mathrm{1}+\frac{\mathrm{1}}{{x}}={e}^{{u}} ,\:\:{x}=\frac{\mathrm{1}}{{e}^{{u}} +\mathrm{1}},\:\:{dx}=−\frac{{e}^{{u}} {du}}{\left({e}^{{u}} +\mathrm{1}\right)^{\mathrm{2}} },\:\:{u}_{\mathrm{1}} =\mathrm{ln}\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{0}^{+} }\right)=+\infty,\:\:{u}_{\mathrm{2}} =\mathrm{ln}\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{0}^{+} }\right)=−\infty\right]= \\ $$$$\int_{+\infty} ^{−\infty} {u}\left(−\frac{{e}^{{u}} }{\left({e}^{{u}} +\mathrm{1}\right)^{\mathrm{2}} }\right){du}=\int_{−\infty} ^{+\infty} \frac{{e}^{{u}} {u}}{\left({e}^{{u}} +\mathrm{1}\right)^{\mathrm{2}} }{du}= \\ $$$$\int_{−\infty} ^{+\infty} \frac{{u}}{{e}^{{u}} +\mathrm{2}+{e}^{−{u}} }{du}=\mathrm{0} \\ $$$$\mathrm{Integral}\:\mathrm{of}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function} \\ $$$$\mathrm{whi}{c}\mathrm{h}\:\mathrm{obviously}\:\mathrm{converges} \\ $$