Question

The value of definite integral ${\int }_{\infty }^0\frac{Z{e}^{-z}}{\sqrt{1-e^{-2z}}}dz$

• A. $-\frac{\pi }{2}ln2$
• B. $\frac{\pi }{2}ln2$
• C. $-\pi ln2$
• D. $\pi ln\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A), (D)

$\pi ln\frac{1}{\sqrt{2}}$
$-\frac{\pi }{2}ln2$

$l={\int }_{\infty }^0\frac{z{e}^{-z}}{\sqrt{1-e^{-2z}}}dz$ put $e^{-z}=\sin \theta$
$l=-{\int }_0^{\pi /2}\frac{ln\left(\sin \theta \right)(-\cos \theta )d\theta }{\sqrt{1-{\sin }^2\theta }}={\int }_0^{\pi /2}ln\sin \theta d\theta$
$\frac{-\pi }{2}ln2$