Using comparison theorem

Question

So I think I have a grasp on this theorem: If a function f that is greater than another function g, and f is convergent, then g must also be convergent, and vice versa with divergence: if f is less than g and f is divergent, then g is also divergent.

That makes sense, but im having trouble applying this theorem to this integral:

$$\int_1^\infty \frac{|sin(x)| + {e}^{-x}}{x^2}dx$$

Ive tried going the route:

$$\frac{|sin(x)| + {e}^{-x}}{x^2} = \frac{e^x|sin(x)| + 1}{x^2e^x} > \frac{|sin(x)|}{x^2} > \frac{sin(x)}{x^2}$$

But I get stuck when I try to find $\int^\infty_1\frac{sin(x)}{x^2}dx$

Answer 1

The natural way to solve this problem is to note that$$x\geqslant1\implies\lvert\sin x\rvert+e^{-x}<2$$and to use the fact that the integral$$\int_1^\infty\frac1{x^2}\,\mathrm dx.$$converges.

Answer 2

• $|\sin x| \leq 1$ for $x \in [1,\infty)$.
• $\mathrm{e}^{-x} \leq 1$ for $x \in [1, \infty)$. (In fact, $\mathrm{e}^{-x} \leq 1/\mathrm{e}$ on this interval, but we don't need tight bounds.)
• So compare with $\displaystyle \int_1^\infty \; \frac{2}{x^2} \,\mathrm{d}x$.