Question

`\rm \int_0^ \infty \frac{1}{(1 + {x}^( \phi) {)}^( \phi) } dx \\`​

Answer 1

Recall that
`\phi = 1 + \frac1\phi`.

In the integral, substitute
`y=1+x^\phi`, so that
`x=(y-1)^(1/\phi) = (y-1)^(\phi-1)` and
`dx = (\phi-1) (y-1)^(\phi-2) \, dy`. We have

`\displaystyle I = \int_0^\infty (dx)/((1+x^\phi)^\phi) \\\\ ~~~~ = \frac1\phi \int_1^\infty ((y-1)^(\phi-2))/(y^\phi) \, dy`

Substitute
`y=\frac1z` and
`dy=-(dz)/(z^2)` to transform
`I` to

`\displaystyle I = \frac1\phi \int_0^1 \left(\frac1z-1\right)^(\phi-2) z^(\phi-2) \, dz \\\\ ~~~~ = \frac1\phi \int_0^1 (1-z)^(\phi-2) \, dz`

Finally, substitute
`u=1-z`.

`\displaystyle I = \frac1\phi \int_0^1 u^(\phi-2) \, du \\\\ ~~~~ = \frac1\phi \cdot \frac1{\phi-1} = (\phi-1)/(\phi-1) = \boxed{1}`