$f(x)$ is monotonic , positive, unbounded function then $\int_{0} ^{\infty } f(x)\sin(x)dx$ does not exist The question came from my try to show that:
$\int_{0} ^{\infty } (x^2+1)^n\sin(x)dx$ does not exist $\forall n \ge 1$.
Using Binomial method one gets, $\int_{0} ^{\infty } (b_1x^{2n} + b_2x^{2n-2} + ... + b_nx^{0})\sin(x)$ when $b_n$ are the binomial coefficients. 
I showed by induction that $\int_{0} ^{\infty } x^nsin(x)$ does not exists in the broad sense.  Yet I didn't succeed to prove that the sum of all those limits (using integral additivity) doesn't exists. (When in general sum of $\lim$ expressions which doesn't exist - might exist, for example $\lim_{x\rightarrow \infty} \sin^2(x)+\cos^2(x)$).
So the more particular question, how to prove that the sum of those integrals doesn't exist?
This question lead me to try and prove a more general claim as mentioned in the title , cause my intuition tells my that it is generally true, when $f(x)$ is unbounded , positive and monotone the the integral $\int_{0} ^{\infty } f(x)\sin(x)dx$ should not exist in the broad sense (neither finite nor infinite). 
Unfortunately I didn't succeed in proving it. Is there a counter example? 
otherwise if it's true can one use weaker conditions for $f(x)$ for the claim to hold?
(If needed I am familiar with Lebesgue integration and basic real analysis)
 A: The improper integral converges if and only if the Cauchy criterion is satisfied, that is for any $\epsilon> 0$ there exists $K > 0$ such that for all $c_2 > c_1 > K$
$$\left| \int_{c_1}^{c_2} f(x) \sin x \, dx \right| < \epsilon.$$
Taking $c_1 = \pi/4 +2n\pi$ and $c_2 = 3\pi/4 +2n\pi$, we get when $f$ is increasing
$$\left| \int_{\pi/4 +2n\pi}^{3 \pi/4 + 2n\pi} f(x) \sin x \, dx \right| \geqslant f(\pi/4) \frac{\pi}{2\sqrt{2}},$$
violating the Cauchy criterion.
Of course, there are weaker conditions for the improper integral to not exist. For example, $f$ could be bounded as $0 < m \leqslant f(x) \leqslant M$. If, however, $f$ is positive and monotonically decreasing with $\lim_{x \to \infty} f(x) = 0$, then the improper integral exists by the Dirichlet test.
A: Lemma. Let $f,g:(a,\infty)\to \mathbb{R}$ be Riemann-integrable on $(a,b)$ for all $b>a$. Suppose further that $g$ is non-integrable on $(a,\infty)$ in the improper Riemann sense and that $f\geq \varepsilon$ for some $\varepsilon>0$. Then $fg$ is non-integrable on $(a,\infty)$ as well.
Proof. Since $\lim_{b\to \infty}\int_{a}^bg(x)dx$ does not exist, there exists a $\varepsilon'>0$ and a sequence $\left\{b_k\right\}_{k\in \mathbb{N}}$ with $b_k\to \infty$ such that $\left|\int_{a}^{b_{k}}g(x)dx\right|\geq \varepsilon'$. Hence
$$\left|\int_{a}^{b_k}f(x)g(x)dx\right|\geq \varepsilon\left|\int_{a}^{b_k}g(x)dx\right|\geq \varepsilon \varepsilon',\qquad \forall k\in \mathbb{N} $$
Thus $\lim_{k\to \infty}\int_{a}^{b_k}f(x)g(x)dx$ does not exist, and hence neither does $\lim_{b\to \infty}\int_{a}^{b}f(x)g(x)dx$.
Apply the previous result to $g(x)=\sin(x)$, which is non-integrable on $(a,\infty)$ for all $a>0$. If $f$ is increasing and positive, then for all $\varepsilon >0$ there exists an $a>0$ such that $f(x)\geq \varepsilon$ on $(a,\infty)$. Unboundedness of $f$ is actually not needed.
EDIT: I adapted my proof to improper Riemann integrals.
A: For the improper integral to exist, we need in particular that
$$a_n:=\int_0^{n\pi}f(x)\sin x\,\mathrm dx $$
converges, in oparticular, is a Cauchy sequence. However,
$$ a_{2n+1}-a_{2n}=\int_{2n\pi}^{(2n+1)\pi}f(x)\sin x\,\mathrm dx\ge \int_{2n\pi}^{(2n+1)\pi}f(2n\pi)\sin x\,\mathrm dx=2f(2n\pi)$$
does not tend to $0$.
