For which values of m is the improper integral convergent

Question

For which values of m is the improper integral convergent:

$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$

I've been at this one for a while, here's what I've got so far:

$\int_0^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$ = $\int_0^1 \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$ + $\int_1^\infty \frac{1}{(x^2+x)^mln^2(1+2x)}tg(\frac{x^4}{x^7+x^4+1})dx$

$(x^2+x)^m$ ~ $x^m$ as $x->0$

$ln^2(1+2x)$ ~ $(2x)^2$ as $x->0$

$tg(\frac{x^4}{x^7+x^4+1})$ ~ $\frac{x^4}{x^7+x^4+1}$ as $x->0$

$(x^2+x)^m$ ~ $(x^2)^m$ as $x->\infty$

$ln^2(1+2x)$ ~ $ln^2x$ as $x->\infty$

$tg(\frac{x^4}{x^7+x^4+1})$ ~ $\frac{x^4}{x^7+x^4+1}$ as $x->\infty$

I get two integrals:

$\int_0^1 \frac{x^4}{x^m4x^2(x^7+x^4+1)}dx$ = $1/4\int_0^1 \frac{1}{\frac{x^m}{x^2}(x^7+x^4+1)}dx$

$\int_1^\infty \frac{x^4}{(x^2)^pln^2x(x^7+x^4+1)}dx$

And that's about as far as I've figured it out so far, I would be grateful if someone could help out.

Answer 1

Note that

$$\int_0^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx \\=\int_0^1 \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx\\+\\+\int_1^\infty \frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)dx$$

and for $x\to 0^+$

$$\frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)\sim \frac{x^4}{4x^{2+m}}\sim \frac{1}{4x^{m-2}}$$

and thus for $m<3$ the first integral converges by limit comparison test with $\int_0^1 \frac{1}{4x^{m-2}}$ and for $x\to \infty$

$$\frac{1}{(x^2+x)^m \ln^2(1+2x)}\tan\left(\frac{x^4}{x^7+x^4+1}\right)\sim \frac{1}{x^{2m+3}\ln^2 x}$$

and thus for $2m+3\ge1\implies m\ge -1$ the second integral converges by limit comparison test with $\int_1^\infty \frac{1}{x^{2m+3}\ln^2 x}$ .

Therefore the given integral converges for $-1\le m<3$.