Find $\lambda_{M_G}(M_{G})$ for $\alpha >0$

Question

It is given that

$M_{g}:=\{(x,y,z) \in \mathbb R^{3}:z \in ]1,\infty[, x^{2}+y^{2}=g(z)^{2}\}$

Let $\alpha > 0$ and $g(z):=z^{-\alpha}$

I have been able to determine that $\lambda_{M_{g}}(M_{g})=2\pi\int_{]1,\infty[}g(z)\sqrt{1+(g^{'}(z))^{2}}dz$

Question: Dependending on $\alpha$, determine wheter $\lambda_{M_{g}}(M_{g})$ is finite or indeed infinite.

My starting point: $2\pi\int_{]1,\infty[}g(z)\sqrt{1+(g^{'}(z))^{2}}dz=2\pi\int_{]1,\infty[}z^{-\alpha}\sqrt{1+\alpha^{2}z^{-2(\alpha+1)}}dz$

I have attempted substitution to no avail, and I cannot find a clear trend in terms of the $\alpha$ select, for example if $\alpha=1$, we get:

$2\pi\int_{]1,\infty[}\frac{1}{z}\sqrt{1+\frac{1}{z^{4}}}dz\geq 2\pi\int_{]1,\infty[}\frac{1}{z}dz =\infty$

Answer 1

First of all, notice that $\lambda_{M_g} (M_g) \ge 0$ because the integrand is positive.

From now on, I shall drop the factor $2 \pi$ for convenience, since it doesn't change anything.

If $\alpha > 1$ then, assuming that your formula for $\lambda_{M_g} (M_g)$ is correct (I haven't checked it), you have that $z > 1$ implies $z^{-2(\alpha+1)} < 1$ (because $\alpha> -1$), therefore $\sqrt{1 + z^{-2(\alpha+1)}} < \sqrt 2$, so it follows that

$$\lambda_{M_g} (M_g) \le \sqrt 2 \int _1 ^\infty z^{-\alpha} \ \mathrm d z = \sqrt 2 \frac {z^{-\alpha + 1}} {-\alpha + 1} \Bigg|_1 ^\infty = \frac {\sqrt 2} {\alpha - 1}$$

which is finite, therefore in this case the integral is between $0$ and $\frac {\sqrt 2} {\alpha - 1}$, therefore finite.

If $\alpha = 1$ then

$$\lambda_{M_g} (M_g) = \int _1 ^\infty z^{-1}\sqrt{1+z^{-4}} \ \mathrm d z \ge \int _1 ^\infty z^{-1}\sqrt{1+0} \ \mathrm d z = \log z \big| _1 ^\infty = \infty \ ,$$

therefore the integral is infinite in this case.

Finally, if $\alpha \in (0,1)$, then

$$\lambda_{M_g} (M_g) = \int _1 ^\infty z^{-\alpha} \sqrt{1 + z^{-2(\alpha + 1)}} \ \mathrm d z \ge \int _1 ^\infty z^{-\alpha} \sqrt{1 + 0} \ \mathrm d z = \frac {z^{-\alpha + 1}} {-\alpha + 1} \Bigg| _1 ^\infty = \infty \ ,$$

therefore the integral is infinite in this case, too.

To conclude, the integral is finite for $\alpha > 1$ and infinite for $\alpha \in (0,1]$.

Answer 2

$$\forall z >1, f(z) = z^{-\alpha}\sqrt{1+z^{-2(\alpha +1)}} > 0$$

Let $\alpha > \beta > 1$: $$z^{\beta -\alpha}\sqrt{1+z^{-2(\alpha +1)}} \rightarrow_{z \rightarrow \infty} 0 $$

So $f(z) = \mathscr{o}_{z \rightarrow \infty}(z^{-\beta})$ but the integral between 1 and $\infty$ of $z \mapsto z^{-\beta}$ converges, thus the integral converges if $\alpha > 0 $.

Similarly, $ f(z) \sim_{z \rightarrow \infty} z^{-1}$ so the integral does not converge.

As for when $\alpha <1$, you can choose $1 > \beta > \alpha$ and you can show that $z^{-\beta} = \mathscr{o}_{z \rightarrow \infty}(f(z))$, thus the integral does not converge.

Answer 3

If $\alpha >0, z\ge 1,$ then $z^{-2(\alpha+1)}\le 1.$ Thus

$$ 1<\sqrt{1+\alpha^{2}z^{-2(\alpha+1)}}\le \sqrt {1 + \alpha^2}.$$

Thus the integral in question is finite iff $\int_1^\infty z^{-\alpha}\,dz <\infty.$ That, as you know, holds iff $\alpha >1.$