Show that $\int_0^\infty \frac{1}{x^\alpha(1+\log^2x)^\alpha}dx$ diverges for any $\alpha>0$

Question

How can I prove that $$\int_0^\infty \frac{1}{x^\alpha(1+\log^2x)^\alpha}dx$$ diverges for any $\alpha>0$?

Answer 1

Actually the integral converges if $\alpha=1$, but only then.

To prove this, decompose the integral dyadically to see

$$\int_0^\infty \frac1{x^\alpha (1+\log^2 x)^\alpha} dx = \sum_{k\in\mathbb{Z}} \int_{2^k}^{2^{k+1}} \frac1{x^\alpha (1+\log^2 x)^\alpha} dx $$

Note that $\int_{2^k}^{2^{k+1}} \frac1{x^\alpha (1+\log^2 x)^\alpha} dx $ is bounded from above and below (up to constants independent of $k$) by $$\frac{2^k }{2^{k\alpha} (1+k^2)^\alpha}$$

Thus, the integral converges if and only if the sum $$\sum_{k\in\mathbb{Z}} \frac{2^{k(1-\alpha)}}{(1+k^2)^{\alpha}}$$

converges.

Clearly, this is the case if and only if $\alpha=1$, since if $\alpha < 1$ the portion where $k$ tends to $+\infty$ diverges and if $\alpha > 1$ the portion where $k$ tends to $-\infty$ diverges.

Remark. This procedure is referred to as Cauchy condensation test.