$f\ge0$ is monotone such that $\int_0^\infty f(x)\sin xdx$ converges absolutely then the improper integral of $f$ converges

Question

$f$ is a monotonic non-negative function for which $\int_{a}^{\infty}f(x)\sin(x)dx$ converges absolutely. I proved in a previous section that: $$\lim_{x\to \infty}f(x)=0$$ I need to prove that the following improper integral exists and I'm a little stuck: ​ $$\int_{a}^{\infty}f(x)dx$$

would appreciate any hints/direction as to where to go from here, I tried to proof by contradiction that it doesn't exist and then $F(x)=\int_{a}^{\infty}f(x)dx$'s limit at $\infty$ is $\infty$ but got stuck there. Also tried to use the Cauchy condition but without success

Thanks a lot

Answer 1

Without loss of generality, assume that $a=0$. We have \begin{align*} &\int_{0}^{\infty}f(x)|\sin x|dx\\ &=\sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}f(x)|\sin x|dx\\ &=\sum_{n=0}^{\infty}\left(\int_{n\pi}^{(n+1/4)\pi}+\int_{(n+1/4)\pi}^{(n+3/4)\pi}+\int_{(n+3/4)\pi}^{(n+1)\pi}\right)f(x)|\sin x|dx\\ &\geq\dfrac{1}{\sqrt{2}}\sum_{n=0}^{\infty}\int_{(n+1/4)\pi}^{(n+3/4)\pi}f(x)+\sum_{n=0}^{\infty}f((n+1)\pi)\left(\int_{n\pi}^{(n+1/4)\pi}+\int_{(n+3/4)\pi}^{(n+1)\pi}\right)|\sin x|dx\\ &=\dfrac{1}{\sqrt{2}}\sum_{n=0}^{\infty}\int_{(n+1/4)\pi}^{(n+3/4)\pi}f(x)+C\sum_{n=0}^{\infty}f((n+1)\pi), \end{align*} where $C=2\int_{0}^{\pi/4}\sin xdx$.

Now we have \begin{align*} &\int_{0}^{\infty}f(x)dx\\ &=\sum_{n=0}^{\infty}\left(\int_{n\pi}^{(n+1/4)\pi}+\int_{(n+1/4)\pi}^{(n+3/4)\pi}+\int_{(n+3/4)\pi}^{(n+1)\pi}\right)f(x)dx\\ &\leq\sum_{n=0}^{\infty}\int_{(n+1/4)\pi}^{(n+3/4)\pi}f(x)+\sum_{n=0}^{\infty}f(n\pi)\left(\int_{n\pi}^{(n+1/4)\pi}+\int_{(n+3/4)\pi}^{(n+1)\pi}\right)1dx\\ &=\sum_{n=0}^{\infty}\int_{(n+1/4)\pi}^{(n+3/4)\pi}f(x)+\dfrac{\pi}{2}\left(f(0)+\sum_{n=0}^{\infty}f((n+1)\pi)\right)\\ &\leq\dfrac{\pi}{2}\cdot f(0)+\left(\sqrt{2}+\dfrac{\pi}{2C}\right)\cdot\int_{0}^{\infty}f(x)|\sin x|dx\\ &<\infty. \end{align*}

Answer 2

The result is false if $f$ is allowed to explode like $1/(x-a)$ at the left endpoint $a$ when $a$ is a zero of $\sin(x)$, so we assume this does not occur. For example, we can imagine $f$ is bounded.

Without loss of generality, $a=0$. $f\ge 0$ is decreasing to $0$. We therefore have $$ f((n+1)\pi )\int_{n\pi}^{(n+1)\pi}|\sin t|dt\le \int_{n\pi}^{(n+1)\pi}f(t)|\sin t|dt $$ Note $\int_{n\pi}^{(n+1)\pi}|\sin t|dt=C$ does not depend on $n$. This implies that $$ \sum_{n=1}^\infty f((n+1)\pi) \le\frac1C \int_0^\infty f(t)|\sin t|dt< \infty .$$ By integral comparison test (which works both ways), it follows that $$ \int_0^\infty f(t) dt < \infty.$$

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Directly without integral comparison or infinite series: similarly to the first line we have $$ \int_{(n+1)\pi}^{(n+2)\pi}f(t)dt \le \pi f((n+1)\pi)$$ which means $$ \int_{(n+1)\pi}^{(n+2)\pi}f(t)dt \le \frac\pi C \int_{n\pi}^{(n+1)\pi}f(t)|\sin t|dt$$ Summing from $n=0$ to $N$ we conclude $$ \int_1^{(N+2)\pi}f(t)dt \le \frac\pi C \int_{0}^{\infty}f(t)|\sin t|dt$$ Therefore $\{\int_0^{N\pi}f(t)dt\}_{N\ge0}$ is a bounded and monotonically increasing sequence; it therefore has a limit. As $\int_0^{s}f(t)dt$ is increasing in the continuous parameter $s$, the integral $\int_0^\infty f(t) dt$ converges in the sense of the improper Riemann integral.