Two questions about weakly convergent series related to sin(n2)\sin(n^2) and Weyl’s inequality

By using partial summation and Weyl’s inequality, it is not hard to show that the series $\sum_{n\geq 1}\frac{\sin(n^2)}{n}$ is convergent.

• Is is true that
• In the case of a positive answer to the previous question, what is

I recall a generalization of partial summation formula:

Suppose that $$\lambda_1,\lambda_2,\ldots\lambda_1,\lambda_2,\ldots$$ is a nondecreasing sequence of real numbers with limit infinity, that $$c_1,c_2,\ldotsc_1,c_2,\ldots$$ is an arbitrary sequence of real or complex numbers, and that $$f(x)f(x)$$ has a continuous derivative for $$x\geq \lambda_1x\geq \lambda_1$$. Put
$$C(x)=\sum_{\lambda_n\leq x}c_n, C(x)=\sum_{\lambda_n\leq x}c_n,$$
where the summation is over all $$nn$$ for which $$\lambda_n\leq x\lambda_n\leq x$$. Then for $$x\geq\lambda_1x\geq\lambda_1$$,
$$\sum_{\lambda_n\leq x}c_nf(\lambda_n)=C(x)f(x)-\int^{x}_{\lambda_1}C(t)f'(t)dt.\tag 1 \sum_{\lambda_n\leq x}c_nf(\lambda_n)=C(x)f(x)-\int^{x}_{\lambda_1}C(t)f'(t)dt.\tag 1$$

Now we can write if $$y=x^2y=x^2$$ and $$\lambda_n=n^2\lambda_n=n^2$$ and $$C(t)=[\sqrt{t}]C(t)=[\sqrt{t}]$$ (integer part of $$\sqrt{t}\sqrt{t}$$):
$$S=\sum_{1\leq n\leq x}\frac{\sin(n^2)}{n^a}=\sum_{\lambda_n\leq y}\frac{\sin(\lambda_n)}{\lambda_n^{a/2}}= S=\sum_{1\leq n\leq x}\frac{\sin(n^2)}{n^a}=\sum_{\lambda_n\leq y}\frac{\sin(\lambda_n)}{\lambda_n^{a/2}}=$$
$$=[\sqrt{y}]\frac{\sin(y)}{y^{a/2}}-\int^{y}_{1}[\sqrt{t}]\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt. =[\sqrt{y}]\frac{\sin(y)}{y^{a/2}}-\int^{y}_{1}[\sqrt{t}]\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt.$$
But it is $$[\sqrt{t}]=\sqrt{t}-\{\sqrt{t}\}[\sqrt{t}]=\sqrt{t}-\{\sqrt{t}\}$$, where $$\{\sqrt{t}\}\{\sqrt{t}\}$$ is the fractional part of $$\sqrt{t}\sqrt{t}$$. Hence
$$S=-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)-\{\sqrt{y}\}\frac{\sin(y)}{y^{a/2}}+ S=-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)-\{\sqrt{y}\}\frac{\sin(y)}{y^{a/2}}+$$
$$+\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt, +\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt,$$
where
$$E(a,z)=\int^{\infty}_{1}\frac{e^{-tz}}{t^a}dt E(a,z)=\int^{\infty}_{1}\frac{e^{-tz}}{t^a}dt$$
But when $$a>0a>0$$ and $$y\rightarrow+\inftyy\rightarrow+\infty$$ we have
$$\lim_{y\rightarrow+\infty}\left\{-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]\right\}+\sin(1)= \lim_{y\rightarrow+\infty}\left\{-\frac{1}{2}Re\left[iy^{1/2-a/2}E\left(\frac{1+a}{2},iy\right)\right]+\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]\right\}+\sin(1)=$$
$$=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1) =\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)$$
Also $$xx$$ is positive integer and $$\{\sqrt{y}\}=0\{\sqrt{y}\}=0$$.

Hence when $$a>0a>0$$, then
$$\lim_{x\rightarrow\infty}\sum^{x}_{n=1}\frac{\sin(n^2)}{n^a}=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)+\lim_{y\rightarrow\infty}\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt \lim_{x\rightarrow\infty}\sum^{x}_{n=1}\frac{\sin(n^2)}{n^a}=\frac{1}{2}Re\left[iE\left(\frac{1+a}{2},i\right)\right]+\sin(1)+\lim_{y\rightarrow\infty}\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt$$
But
$$\int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)t^{a/2}-a/2\sin(t)t^{a/2-1}}{t^a}dt= \int^{y}_{1}\{\sqrt{t}\}\frac{d}{dt}\left(\frac{\sin(t)}{t^{a/2}}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)t^{a/2}-a/2\sin(t)t^{a/2-1}}{t^a}dt=$$
$$\int^{y}_{1}\{\sqrt{t}\}\left(\cos(t)t^{-a/2}-a/2\sin(t)t^{-a/2-1}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt-\frac{a}{2}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt. \int^{y}_{1}\{\sqrt{t}\}\left(\cos(t)t^{-a/2}-a/2\sin(t)t^{-a/2-1}\right)dt=\int^{y}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt-\frac{a}{2}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt.$$
Clearly when $$aa$$ is positive and constant
$$\lim_{y\rightarrow+\infty}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt=2\lim_{x\rightarrow\infty}\int^{x}_{1}\frac{\sin(t^2)}{t^{a+1}}\{t\}dt<\infty, \lim_{y\rightarrow+\infty}\int^{y}_{1}\frac{\sin(t)}{t^{a/2+1}}\{\sqrt{t}\}dt=2\lim_{x\rightarrow\infty}\int^{x}_{1}\frac{\sin(t^2)}{t^{a+1}}\{t\}dt<\infty,$$
since $$0\leq\{t\}<10\leq\{t\}<1$$ and $$-1\leq\sin(t^2)\leq 1-1\leq\sin(t^2)\leq 1$$, for all $$t>0t>0$$.

Hence it remains to find under what condition on $$a>0a>0$$ we have
$$\int^{\infty}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt=2\int^{\infty}_{1}\cos(t^2)t^{1-a}\{t\}dt<\infty, \int^{\infty}_{1}\{\sqrt{t}\}\frac{\cos(t)}{t^{a/2}}dt=2\int^{\infty}_{1}\cos(t^2)t^{1-a}\{t\}dt<\infty,$$
knowinig already that for all $$0 we have
$$\int^{\infty}_{1}\cos(t^2)t^{1-a}dt<\infty. \int^{\infty}_{1}\cos(t^2)t^{1-a}dt<\infty.$$

However it is
$$F(x)=\int^{x}_{1}\cos(t)\{\sqrt{t}\}dt=\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)+ F(x)=\int^{x}_{1}\cos(t)\{\sqrt{t}\}dt=\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)+$$
$$+\sum_{2\leq k<\sqrt{x}}\sin(k^2)+\{\sqrt{x}\}\sin(x)=O\left(\sum_{2\leq k<\sqrt{x}}\sin(k^2)\right).\tag 2 +\sum_{2\leq k<\sqrt{x}}\sin(k^2)+\{\sqrt{x}\}\sin(x)=O\left(\sum_{2\leq k<\sqrt{x}}\sin(k^2)\right).\tag 2$$
The function $$\textrm{Fs}(x)\textrm{Fs}(x)$$ is the FresnelS function
$$\textrm{Fs}(z):=\int^{z}_{0}\sin\left(\pi t^2/2\right)dt \textrm{Fs}(z):=\int^{z}_{0}\sin\left(\pi t^2/2\right)dt$$
and it is known that
$$\lim_{x\rightarrow+\infty}\textrm{Fs}(x)=\frac{1}{2}. \lim_{x\rightarrow+\infty}\textrm{Fs}(x)=\frac{1}{2}.$$
Hence if we set
$$S(x):=\sum_{2\leq k
and assume that $$a=1/2-\epsilona=1/2-\epsilon$$, $$\epsilon>0\epsilon>0$$, then
$$I(x)=\int^{x}_{1}\frac{1}{t^{a/2}}\cos(t)\{\sqrt{t}\}dt=\int^{x}_{1}\frac{F'(t)}{t^{a/2}}dt=\frac{F(x)}{x^{a/2}}+\frac{a}{2}\int^{x}_{1}\frac{F(t)}{t^{a/2+1}}dt=S_1+S_2,\tag 3 I(x)=\int^{x}_{1}\frac{1}{t^{a/2}}\cos(t)\{\sqrt{t}\}dt=\int^{x}_{1}\frac{F'(t)}{t^{a/2}}dt=\frac{F(x)}{x^{a/2}}+\frac{a}{2}\int^{x}_{1}\frac{F(t)}{t^{a/2+1}}dt=S_1+S_2,\tag 3$$
where
$$S_1=\frac{1}{x^{a/2}}S(\sqrt{x})+\frac{\{\sqrt{x}\}\sin x}{x^{a/2}}+\frac{\sqrt{\pi/2}}{x^{a/2}}\left(\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)\right) S_1=\frac{1}{x^{a/2}}S(\sqrt{x})+\frac{\{\sqrt{x}\}\sin x}{x^{a/2}}+\frac{\sqrt{\pi/2}}{x^{a/2}}\left(\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\textrm{Fs}\left(\sqrt{\frac{2x}{\pi}}\right)\right)$$
and
$$S_2=\frac{a}{2}\int^{x}_{1}\frac{1}{t^{1/4-\epsilon/2+1}}\{\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2t}{\pi}}\right)+ S_2=\frac{a}{2}\int^{x}_{1}\frac{1}{t^{1/4-\epsilon/2+1}}\{\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2}{\pi}}\right)-\frac{\sqrt{2\pi}}{2}\textrm{Fs}\left(\sqrt{\frac{2t}{\pi}}\right)+$$
$$+\{\sqrt{t}\}\sin(t) +\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\}dt. +\{\sqrt{t}\}\sin(t) +\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\}dt.$$
But it is known that there exists constant $$CC$$ such that for infinite values of $$x\in\textbf{N}x\in\textbf{N}$$ holds
$$\sum_{2\leq k\leq x}\sin(k^2)>Cx^{1/2}.\tag 4 \sum_{2\leq k\leq x}\sin(k^2)>Cx^{1/2}.\tag 4$$
Hence for infinite values of $$xx$$ we will have (easily)
$$S_1>C_1x^{\epsilon/2}.\tag 5 S_1>C_1x^{\epsilon/2}.\tag 5$$
Moreover if we assume that
$$\left|\sum_{2\leq k\leq x}\sin(k^2)\right|=O\left(x^{c+\delta}\right)\textrm{, }\forall \delta>0\textrm{ and }x\rightarrow+\infty,\tag 6 \left|\sum_{2\leq k\leq x}\sin(k^2)\right|=O\left(x^{c+\delta}\right)\textrm{, }\forall \delta>0\textrm{ and }x\rightarrow+\infty,\tag 6$$
then in view of (4) it must be $$c\geq 1/2c\geq 1/2$$. Also
$$S_2=C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt. S_2=C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt.$$
Hence
$$\left|S_2\right|=\left|C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\leq \left|S_2\right|=\left|C_{\epsilon}(x)+\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\leq$$
$$\leq \left|\left|C_{\epsilon}(x)\right|+\left|\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\right|\leq \leq \left|\left|C_{\epsilon}(x)\right|+\left|\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)dt\right|\right|\leq$$
$$|C_{\epsilon}(x)|+\int^{x}_{1}\left|t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)\right|dt= |C_{\epsilon}(x)|+\int^{x}_{1}\left|t^{-1/4+\epsilon/2-1}\left(\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right)\right|dt=$$
$$=\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left|\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right|dt\leq =\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1/4+\epsilon/2-1}\left|\sum_{2\leq k\leq \sqrt{t}}\sin(k^2)\right|dt\leq$$
$$\leq\left|C_{\epsilon}(x)\right|+C_2\int^{x}_1t^{-1-1/4+\epsilon/2}t^{1/4+\delta/2}dt= \leq\left|C_{\epsilon}(x)\right|+C_2\int^{x}_1t^{-1-1/4+\epsilon/2}t^{1/4+\delta/2}dt=$$
$$=\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1+\epsilon/2+\delta/2}dt=\left|C_{\epsilon}(x)\right|+\frac{2}{\delta+\epsilon}\left(x^{(\delta+\epsilon)/2}-1\right)< =\left|C_{\epsilon}(x)\right|+C_2\int^{x}_{1}t^{-1+\epsilon/2+\delta/2}dt=\left|C_{\epsilon}(x)\right|+\frac{2}{\delta+\epsilon}\left(x^{(\delta+\epsilon)/2}-1\right)<$$
$$<|C_{0}|+\log x+C_3d\log^2 x,\tag 7 <|C_{0}|+\log x+C_3d\log^2 x,\tag 7$$
where $$\epsilon>0\epsilon>0$$ and $$\delta>0\delta>0$$ so small as we please and $$d=\frac{\epsilon+\delta}{2}>0d=\frac{\epsilon+\delta}{2}>0$$, $$C_3>0C_3>0$$ constant. It is also easy to see someone that $$\left|C_{\epsilon}(x)\right|\left|C_{\epsilon}(x)\right|$$ are bounded by a constant $$C_0>0C_0>0$$.

Hence from $$(3)(3)$$ and $$(5),(7)(5),(7)$$ we have if $$a=1/2-\epsilona=1/2-\epsilon$$, that
$$\left|\int^{x}_{1}\frac{\cos (t)\{\sqrt{t}\}}{t^{a/2}}dt\right|=|S_1+S_2|\geq |S_1|-|S_2|>C_1x^{\epsilon/2}-|C_0|-\log x-C_3d\log^2 x, \left|\int^{x}_{1}\frac{\cos (t)\{\sqrt{t}\}}{t^{a/2}}dt\right|=|S_1+S_2|\geq |S_1|-|S_2|>C_1x^{\epsilon/2}-|C_0|-\log x-C_3d\log^2 x,$$
For infinite values of $$x\in\textbf{N}x\in\textbf{N}$$.

Hence
$$\lim_{x\rightarrow+\infty}\int^{x}_{1}\frac{\cos(t)\{\sqrt{t}\}}{t^{a/2}}dt=+\infty \lim_{x\rightarrow+\infty}\int^{x}_{1}\frac{\cos(t)\{\sqrt{t}\}}{t^{a/2}}dt=+\infty$$
and we conclude that if (6) holds, then $$\textrm{inf}\geq1/2\textrm{inf}\geq1/2$$.

I will argue now about the the case $$a=\frac{1}{2}+2\epsilona=\frac{1}{2}+2\epsilon$$, $$\epsilon>0\epsilon>0$$ i.e
the case when $$aa$$ is not $$1/21/2$$ but rather a limiting case and doesnot cover the case $$a=1/2a=1/2$$. Both results $$\textrm{inf}\geq1/2\textrm{inf}\geq1/2$$ and $$\textrm{inf}=1/2+2\epsilon\textrm{inf}=1/2+2\epsilon$$, clearly show us that for $$1/2 the sum $$\sum^{\infty}_{n=1}\frac{\sin(n^2)}{n^a}\sum^{\infty}_{n=1}\frac{\sin(n^2)}{n^a}$$ converges and diverges for $$0, under the hypothesis $$(6)(6)$$. For $$a=1/2a=1/2$$, we dont know.

For $$a=1/2+2\epsilona=1/2+2\epsilon$$, $$\epsilon>0\epsilon>0$$ and for $$x>>1x>>1$$, we chose $$\delta>0\delta>0$$ such that
$$S\left(\sqrt{x}\right)\leq C_1x^{1/4+\delta}, S\left(\sqrt{x}\right)\leq C_1x^{1/4+\delta},$$
we get
$$\left|I(x)\right|=|S_1+S_2|\leq \left|C_1\frac{S\left(\sqrt{x}\right)}{x^{a/2}}+C_2\frac{a}{2}\int^{x}_{1}\frac{S\left(\sqrt{t}\right)}{t^{a/2+1}}dt\right|\leq \left|I(x)\right|=|S_1+S_2|\leq \left|C_1\frac{S\left(\sqrt{x}\right)}{x^{a/2}}+C_2\frac{a}{2}\int^{x}_{1}\frac{S\left(\sqrt{t}\right)}{t^{a/2+1}}dt\right|\leq$$
$$\leq\left|C_1'\frac{x^{1/4+\delta}}{x^{1/4+\epsilon}}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{t^{1/4+\delta}}{t^{1+1/4+\epsilon}}dt\right|= \leq\left|C_1'\frac{x^{1/4+\delta}}{x^{1/4+\epsilon}}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{t^{1/4+\delta}}{t^{1+1/4+\epsilon}}dt\right|=$$
$$=\left|C_1'x^{-(\epsilon-\delta)}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{dt}{t^{1+\epsilon-\delta}}\right|. =\left|C_1'x^{-(\epsilon-\delta)}+C_2'\left(\frac{1}{4}+\epsilon\right)\int^{x}_{1}\frac{dt}{t^{1+\epsilon-\delta}}\right|.$$
For $$\delta=\epsilon/2\delta=\epsilon/2$$ we get
$$|I(x)|\leq \left|C_1'x^{-\epsilon/2}-\frac{2C_2'}{\epsilon}\left(\frac{1}{4}+\epsilon\right)\left(x^{-\epsilon/2}-1\right)\right|= |I(x)|\leq \left|C_1'x^{-\epsilon/2}-\frac{2C_2'}{\epsilon}\left(\frac{1}{4}+\epsilon\right)\left(x^{-\epsilon/2}-1\right)\right|=$$
$$=\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H-H\right|\leq =\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H-H\right|\leq$$
$$\leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|.\tag 8 \leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|.\tag 8$$
Now we set
$$X=C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)>0 X=C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)>0$$
and
$$Y=2C_2'\left(1-x^{-\epsilon/2}\right)+H>0 Y=2C_2'\left(1-x^{-\epsilon/2}\right)+H>0$$
and I use the inequality
$$\left|X+Y\right|\leq \left|X-\frac{Y}{4\epsilon}\right|,\tag 9 \left|X+Y\right|\leq \left|X-\frac{Y}{4\epsilon}\right|,\tag 9$$
which is is true for small $$\epsilon\epsilon$$ and $$x>1x>1$$ since we can write equivalent
$$\left|X+Y\right|^2\leq \left|X-\frac{Y}{4\epsilon }\right|^2\Leftrightarrow X^2+Y^2+2XY\leq X^2+\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon }\Leftrightarrow \left|X+Y\right|^2\leq \left|X-\frac{Y}{4\epsilon }\right|^2\Leftrightarrow X^2+Y^2+2XY\leq X^2+\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon }\Leftrightarrow$$
$$Y^2+2XY\leq\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon}\Leftrightarrow Y+2X\leq \frac{Y}{16\epsilon^2}-\frac{X}{2\epsilon}\Leftrightarrow Y^2+2XY\leq\frac{Y^2}{16\epsilon^2}-\frac{XY}{2\epsilon}\Leftrightarrow Y+2X\leq \frac{Y}{16\epsilon^2}-\frac{X}{2\epsilon}\Leftrightarrow$$
$$\left(\frac{1}{16\epsilon^2}-1\right)Y\geq X\left(2+\frac{1}{2\epsilon}\right) \left(\frac{1}{16\epsilon^2}-1\right)Y\geq X\left(2+\frac{1}{2\epsilon}\right)$$
This last inequality holds for all small $$\epsilon>0\epsilon>0$$ and $$x>>1x>>1$$ since it can be writen equivalently as
$$(1-16\epsilon^2)Y-X(32\epsilon^2+8\epsilon)\geq0\Leftrightarrow (1-16\epsilon^2)Y-X(32\epsilon^2+8\epsilon)\geq0\Leftrightarrow$$
$$2\epsilon(1+2\epsilon)\left(C_2'-4\epsilon C_1'+4C_2'\epsilon\right)x^{-\epsilon/2}\geq 0, 2\epsilon(1+2\epsilon)\left(C_2'-4\epsilon C_1'+4C_2'\epsilon\right)x^{-\epsilon/2}\geq 0,$$
where we have used the value
$$H=\frac{2(C_2'+4C_2\epsilon)}{1-4\epsilon} H=\frac{2(C_2'+4C_2\epsilon)}{1-4\epsilon}$$
Hence (9) is true and we can extract from relation (8) the conclusion
$$\left|I(x)\right|\leq \left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|= \left|I(x)\right|\leq \left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)+2C_2'\left(1-x^{-\epsilon/2}\right)+H\right|+\left|H\right|=$$
$$=\left|X+Y\right|+\left|H\right|\leq =\left|X+Y\right|+\left|H\right|\leq$$
$$\leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{H}{4\epsilon}\right|+|H|= \leq\left|C_1'x^{-\epsilon/2}+\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{C_2'}{2\epsilon}\left(1-x^{-\epsilon/2}\right)-\frac{H}{4\epsilon}\right|+|H|=$$
$$=\left|C_1'x^{-\epsilon/2}-\frac{H}{4\epsilon}\right|+|H|. =\left|C_1'x^{-\epsilon/2}-\frac{H}{4\epsilon}\right|+|H|.$$
Hence
$$\epsilon |I(x)|\leq C_1'x^{-\epsilon/2}\epsilon+H/4+|H|\epsilon. \epsilon |I(x)|\leq C_1'x^{-\epsilon/2}\epsilon+H/4+|H|\epsilon.$$
Hence we conclude that
$$\epsilon \left|I(x)\right|=O(1) \epsilon \left|I(x)\right|=O(1)$$
is bounded. Hence for $$\epsilon>0\epsilon>0$$ small but constant the $$I(x)I(x)$$ are bounded.