4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ My friend was able to solve it, I will now post his solution $\endgroup$ Feb 2, 2021 at 12:10

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

This is a solution my friend came up with (credit doesn't go to me)

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .