# prove if f is continuous and $\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$ then $\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$ diverges

## The Question

let $$f:\left[1,\infty\right)\to\mathbb{R}$$ be a continuous function with $$\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$$. then $$\int\limits _{1}^{\infty}f\left(x\right)\sin x\,dx$$ diverges.

I came across this question while preparing to my exam in calculus. I wasn't able to solve it: couldn't prove it nor give a counterexample.

## What have I tried?

I tried using tests such as the comparsion test, but I realised $$f\left(x\right)\sin x$$ isn't positive. Tried looking for counter examples but failed.

Here is my best attempt so far: since $$L\neq0$$ we can assume without loss of generality $$L>0$$, therefore $$f>0$$ almost always. for some $$\mathbb{N}\ni k,\, M=2k$$ we have $$\int\limits _{M\pi}^{k\pi}f\left(x\right)\sin x\,dx=\sum_{i=M}^{k}\left(-1\right)^{i}\int\limits _{0}^{\pi}f\left(x\right)\sin x=Q_{k}$$ and since $$2\left(L+\epsilon\right)\leq\int\limits _{0}^{\pi}f\left(x\right)\sin x\leq2\left(L+\epsilon\right)$$ for some $$\epsilon$$, we can write $$\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i+1}\epsilon\right)\leq Q_{k}\leq\sum_{i=M}^{k}\left(-1\right)^{i}\cdot2\cdot\left(L+\left(-1\right)^{i}\epsilon\right)$$

But here I got stuck, I'm not sure how to continue and I'm not sure if this is the right direction.

• My friend was able to solve it, I will now post his solution Feb 2, 2021 at 12:10

Without loss of genrality we may suppose $$L >0$$. If the intergal converges then $$\int_a^{b} f(x)\sin x dx$$ tends to $$0$$ as $$b >a \to \infty$$. Let us pove that this is not true. Consider $$\int_{2n\pi}^{2n\pi +\pi} f(x)\sin x dx$$. Note that $$\sin x$$ is positve in this interval Also, $$f(x) >\frac L 2$$ in this interval if $$n$$ is large enough. This gives $$\int_{2n\pi}^{2n\pi +\pi} f(x)\sin x dx >\frac L 2 \int_{2n\pi}^{2n\pi +\pi} \sin x dx=L$$. This is a contradiction.
we can assume without loss of generality $$L>0$$. let $$m:=\frac{L}{2}$$. since $$\lim\limits _{x\to\infty}f\left(x\right)=L\neq0$$ there exists $$x_0$$ such that $$\forall x\geq x_{0},\;f\left(x\right)\geq m$$. let $$\epsilon:=2m$$. for all $$N\in\mathbb{N}$$ exists $$N\in\mathbb{N}$$ such that $$2\pi n\geq N$$, then $$\left|\int\limits _{2\pi n}^{2\pi n+\pi}\sin xf\left(x\right)\,dx\right|=\int\limits _{2\pi n}^{2\pi n+\pi}\sin xf\left(x\right)\,dx\geq m\int\limits _{2\pi n}^{2\pi n+\pi}\sin x\,dx=2m=\epsilon$$ so be Cauchy's theorem the integral diverges.