Show that $\int _0^{\infty }\:\frac{dz}{\sqrt{e^{2z}-1}}=\int _0^{1\:}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$

Question

The question is as follows:

By appropriate substitutions, show that $\int _0^{\infty }\:\frac{dz}{\sqrt{e^{2z}-1}}=\int _0^{1\:}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi }{2}$

Answer 1

Let $x = e^{-z} \iff z=-\ln(x)$.

Then $x\to0 \iff z\to\infty$ and $x\to1 \iff z=0$.

Differentiating, we also have $dx=-e^{-z}\,dz \iff dz=-\frac{dx}x$.

Thus

$$\int_0^\infty \frac{dz}{\sqrt{e^{2z}-1}} = \int_1^0 \frac{-\frac{dx}x}{\sqrt{\frac1{x^2} - 1}} = \int_0^1 \frac{dx}{x\sqrt{\frac1{x^2}-1}} = \int_0^1 \frac{dx}{\sqrt{1-x^2}}$$