If $f:(0, \infty) \rightarrow \mathbb{R}$ is decreasing and continuously differentiable with $\lim_{x \rightarrow \infty}f(x)= 0$. Then $$\int_1^{\infty}f(x)\sin(x)dx$$ converges.
MY attempt:
for $\epsilon >0$ choose $d \in \mathbb{R}$ so that $|f(x)| < \epsilon /2 \hspace{3mm}\forall x \geq d$ Using integration by parts we have $\int_1^df(x)\sin(x)dx = f(1)\cos(1) - f(d)\cos(d) + \int_1^df'(x)\cos(x)dx$
since $f'(x) <0$ we will have
$|\int_1^df'(x)\cos(x)dx| < \int_1^d (-f'(x))dx = f(1) - f(d)< f(1) +\epsilon/2$
Therefore we have $|\int_1^df(x)\sin(x)dx |\leq |f(1)\cos(1) - f(d)\cos(d)| + |f(1) +\epsilon/2| < 2 |f(1) +\epsilon/2| $
Here I have shown that the integral is bounded.
My question is does the boundedness implies convergence of integral? If not then what can be added.
Boundedness does not imply convergence. Take the simple example of $f\equiv 1$. Then, we have
$$\int_1^L f(x)\sin(x)\,dx=\cos(1)-\cos(L)$$
which fails to converge as $L\to \infty$.
To show convergence, we use the Cauchy Criterion for convergence to analyze the integral $\int_{u}^{v}f(x)\sin(x)\,dx$, where $1\le u<v$.
Using the Bonnet form of the second mean value theorem, we have for some number $\xi \in [u,v]$
$$\left|\int_u^vf(x)\sin(x)\,dx\right|=f(u) \left|\int_u^\xi \sin(x)\,dx\right|\le 2f(u)$$
Then, for any $\epsilon>0$, we choose a number $N$ so large that $|f(u)|<\epsilon/2$, for $N<u<v$. And for such an $N$ we see that
$$\left|\int_u^vf(x)\sin(x)\,dx\right|<\epsilon$$
Inasmuch as the conditions for the Cauchy Criterion are satisfied, the given integral converges.
NOTE:
There was nothing particularly special about the sine function in the development other than the fact that its integral, $\int_1^v \sin(x)\,dx$, is bounded for all $v\ge 1$.