Determine $ \int_0^\infty \frac{\sin x - e^{-x}-1}{x^{\frac{5}{2}}(x+1)}dx $ converge or not, using only Limit comparison test

Question

Given the following integral:

$$ \int_0^\infty \frac{\sin x - e^{-x}-1}{x^{\frac{5}{2}}(x+1)}dx $$

So I can tell that this integral diverges in $(0,1]$, however, I'm having trouble finding functions to compare with. I was trying to split into $3$ integrals using linearity of integration but I'm always getting $0$ or $\infty$ as the limit when comparing so obviously it's the wrong functions that I'm choosing.

Perhaps I can get to a more convenient form using substitution or integration by parts but I couldn't find something that works yet.

Answer 1

$$ \int_0^\infty \frac{\sin x - e^{-x}-1}{x^{\frac{5}{2}}(x+1)} dx $$

Since $\sin x\le 1, -e^{-x}<0$, we get: $\sin x - e^{-x}-1<0$, define:

$$\int_0^\infty \frac{\sin x - e^{-x}-1}{x^{\frac{5}{2}}(x+1)} dx=-\int_0^\infty \frac{1+ e^{-x}-\sin x}{x^{\frac{5}{2}}(x+1)} dx=-J$$

So $J\ge 0$, and we need to show $J$ is divergent for $x\in (0,1]$

$$J=\int_0^1 \frac{1+ e^{-x}-\sin x}{x^{\frac{5}{2}}(x+1)} dx\ge \int_0^1 \frac{ e^{-x}}{x^{\frac{5}{2}}(x+1)} dx\ge \int_0^1 \frac{ e^{-1}}{x^{\frac{5}{2}}(1+1)} dx=\frac{1}{2e}\int_0^1 \frac{1}{x^{\frac{5}{2}}} dx\to\infty$$

Update for Limit comparison test:

Take $g=\frac{1}{x^{5/2}}$

$$\lim_{x\to 0}\frac{f}{g}=\lim_{x\to 0} \frac{\frac{1+ e^{-x}-\sin x}{x^{5/2}(x+1)}}{\frac{1}{x^{5/2}}}=\lim_{x\to 0} \frac{1+ e^{-x}-\sin x}{(x+1)}=2$$