Show that if |f(x)| converges in infinity, so is f(x).

I think that in a I should compare the function |f(x) - f(x) and 2|f(x)| but I am not sure how i would do that.

Also, I am not sure how i should duduce what i want to deduce in b after i find a.

• For part $b$, say that $\frac{\sin x}{x^2}$ converges if $\frac{| \sin x |}{x^2}$ converges, and that the latter converges because $\frac{| \sin x |}{x^2} \le \frac{1}{x^2}$ (implied that there are integrals in front of all these expressions). – anonymouse Apr 13 '16 at 22:06

For part (a), use $$\left| \int_c^{d} f(x) dx \right | \leq \int_c^{d} \left| f(x) \right | dx$$
For part (b), use the result of part (1) and $| \sin x| \leq 1$. and the fact that $$\int_1^{\infty} \frac{1}{x^2} dx$$ converges.
Let's prove (a). Set $$f_+(x)=\max\{f(x),0\},\quad f_-(x)=\max\{-f(x),0\}$$ Note that $f_+$ and $f_-$ are continuous as well; moreover $$f(x)=f_+(x)-f_-(x),\quad |f(x)|=f_+(x)+f_-(x)$$ Since $0\le f_+(x)\le|f(x)|$, we have, for every $b\ge a$, $$0\le \int_a^b f_+(x)\,dx\le\int_a^b |f(x)|\,dx\le\int_a^\infty|f(x)|\,dx$$ Since the function $$F_+(b)=\int_a^b f_+(x)\,dx$$ is non decreasing and bounded, its limit at $\infty$ exists and is finite. Similarly for $$F_-(b)=\int_a^b f_-(x)\,dx$$ Now \begin{align} \lim_{b\to\infty}\int_a^b f(x)\,dx &= \lim_{b\to\infty}\left(\int_a^b f_+(x)\,dx-\int_a^b f_-(x)\,dx\right) \\[6px] &= \lim_{b\to\infty}\int_a^b f_+(x)\,dx-\lim_{b\to\infty}\int_a^b f_-(x)\,dx \end{align}
For part (b), note that $$\left|\frac{\sin x}{x^2}\right|\le\frac{1}{x^2}$$ and that $$\int_{1}^\infty\frac{1}{x^2}\,dx=1$$