Generalized Riemann Theorem Let $f$ be decreasing on $[0,\infty)$ with $\lim_{x\to\infty}f(x)=0$. Show that $$\lim_{n\to\infty}\int_0^\infty f(x)\sin nxdx=0.$$
What I have tried is as follows:
For general $f$ (not necessarily decreasing), I can prove that
$$\lim_{n\to\infty}\int_0^a f(x)\sin nxdx=0$$ for every $a>0$.
 A: You can view the following as an alternating series where the general term tends to $0$ because $f \ge 0$ is a decreasing function with $\lim_{x\rightarrow\infty}f(x)=0$:
$$
\begin{align}
    \int_{0}^{\infty}f(x)\sin(nx)\,dx & = \int_{0}^{\pi/n}f(x)|\sin(nx)|\,dx \\
        & -\int_{\pi/n}^{2\pi/n}f(x)|\sin(nx)|\,dx \\
        & +\int_{2\pi/n}^{3\pi/n}f(x)|\sin(nx)|\,dx - \cdots      
\end{align}
$$
Therefore, the sum converges, and the difference between the full alternating series and the truncated one is bounded by the absolute value of the first term which is neglected. That is,
$$
\begin{align}
      \left|\int_{0}^{\infty}f(x)\sin(nx)\,dx-\int_{0}^{k\pi/n}f(x)\sin(nx)\,dx\right|
       & \le \int_{k\pi/n}^{(k+1)\pi/n}f(x)|\sin(nx)|\,dx \\
       & \le f(k\pi/n)\int_{0}^{\pi/n}\sin(nx)\,dx \\
       & \le f(k\pi/n)\frac{2}{n}.
\end{align}
$$
Let $L$ be a given positive integer, and let $k=Ln$. Then
$$
     \left|\int_{0}^{\infty}f(x)\sin(nx)\,dx\right| \le \left|\int_{0}^{L\pi}f(x)\sin(nx)\,dx\right|+f(L\pi)\frac{2}{n}.
$$
Using what you know, which is $\lim_{n\rightarrow\infty}\int_{0}^{a}f(x)\sin(nx)\,dx=0$, then the above gives the desired result because both terms on the right tend to $0$ as $n\rightarrow\infty$.
