The value of the definite integral intlimits_ | vedclass

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(B) Let $I = \int_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$.Substitute $e^{-z} = \sin 	heta$,then $-e^{-z} \, dz = \cos 	heta \, d	heta

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$\frac{\pi}{2} \ln 2$

Vedclass · Answer

(B) Let $I = \int_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$.Substitute $e^{-z} = \sin 	heta$,then $-e^{-z} \, dz = \cos 	heta \, d	heta$,which implies $dz = -\frac{\cos 	heta}{e^{-z}} \, d	heta = -\frac{\cos 	heta}{\sin 	heta} \, d	heta$.When $z = \infty$,$e^{-z} = 0$,so $\sin 	heta = 0 \implies 	heta = 0$.When $z = 0$,$e^{-z} = 1$,so $\sin 	heta = 1 \implies 	heta = \frac{\pi}{2}$.Also,$z = -\ln(\sin 	heta)$.Substituting these into the integral:$I = \int_{0}^{\pi/2} \frac{-\ln(\sin 	heta) \cdot \sin 	heta}{\sqrt{1-\sin^2 	heta}} \cdot \frac{\cos 	heta}{\sin 	heta} \, d	heta$$I = \int_{0}^{\pi/2} \frac{-\ln(\sin 	heta) \cdot \sin 	heta}{\cos 	heta} \cdot \frac{\cos 	heta}{\sin 	heta} \, d	heta$$I = -\int_{0}^{\pi/2} \ln(\sin 	heta) \, d	heta$.Using the standard integral result $\int_{0}^{\pi/2} \ln(\sin 	heta) \, d	heta = -\frac{\pi}{2} \ln 2$,we get:$I = -(-\frac{\pi}{2} \ln 2) = \frac{\pi}{2} \ln 2$.

The value of the definite integral $\int\limits_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$ is:

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