Uniform convergence of $\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$ I am preparing for the exam. Please help me to solve the following problem:
Given a series 
$$\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$$
determine, if it is uniformly convergent on 1) $(0, \infty)$ ? 2) $(0, c)$, where $c < \infty$ ?
Thanks a lot for your help!
 A: First
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}
&=\frac14\int_0^\infty\frac{t^{-3/4}\,\mathrm{d}t}{(t+1)^{1/2}}\\
&=\frac14\frac{\Gamma\left(\frac14\right)\Gamma\left(\frac14\right)}{\Gamma\left(\frac12\right)}\\
&\le\int_0^1\mathrm{d}t+\int_1^\infty\frac{\mathrm{d}t}{t^2}\\[9pt]
&=1+1=2
\end{align}
$$
Furthermore,
$$
\begin{align}
\int_{nx}^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}
&\le\int_{nx}^\infty\frac{\mathrm{d}t}{t^2}\\
&=\frac1{nx}
\end{align}
$$
Therefore, using the harmonic mean of the bounds,
$$
\bbox[5px,border:2px solid #C0A000]{\int_{nx}^\infty\frac{\mathrm{d}t}{\sqrt{t^4+1}}\le\frac4{1+2nx}}
$$
Since
$$
\left|\sin((n+1)x)-\sin(nx)\right|\le2
$$
and
$$
\left|\sin((n+1)x)-\sin(nx)\right|\le(n+1)x-nx=x
$$
using the harmonic mean of the bounds, we have
$$
\bbox[5px,border:2px solid #C0A000]{\int_{\sin(nx)}^{\sin((n+1)x)}\sin\left(t^2\right)\,\mathrm{d}t\le\frac{4x}{2+x}}
$$
Thus, when $x\le2$, the sum is less than
$$
\begin{align}
\sum_{n=N}^\infty\frac{16x}{(1+2nx)(2+x)(1+n^3x^2)}
&\le\sum_{n=N}^\infty\frac{8x}{\left(1+2nx\right)\left(1+n^3x^2\right)}\\
&\le\sum_{n=N}^\infty\frac{8x}{\left(1+2nx\right)\left(1+Nn^2x^2\right)}\\
&\le\int_0^\infty\frac{8\mathrm{d}t}{\left(1+2t\right)\left(1+Nt^2\right)}\\
&\le\frac{4\pi}{\sqrt{N}}
\end{align}
$$
When $x\gt2$, the sum is less than
$$
\begin{align}
\sum_{n=N}^\infty\frac{16x}{(1+2nx)(2+x)(1+n^3x^2)}
&\le\sum_{n=N}^\infty\frac{16}{\left(1+4n\right)\left(1+4n^3\right)}\\
&\le\sum_{n=N}^\infty\frac1{n^4}\\
&\le\frac1{3(N-1)^3}
\end{align}
$$
Therefore, the convergence is uniform over $(0,\infty)$.
A: Let's single out the term at $n=0$ which is finite for all real numbers $x$.
Assuming we are asked to choose between the two answers where either 1) or 2) is true, then we may observe that
$$
\left|\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt\right|\leq 2,\quad \qquad n \in \{1,2,\cdots\},\,x \in \mathbb{R},
$$ and that
$$
\left|\int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}\right|< \left|\int_{nx}^{\infty}\frac{dt}{\sqrt{t^4}}\right|=\frac1{nx},\quad \qquad n \in \{1,2,\cdots\},\,x \in (0,\infty),
$$ giving, for any real number $\alpha$ such that $\alpha>0$,
$$
\sup_{x\in [\alpha, \infty)}\left| \dfrac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}\right|\leq \frac2{n\alpha(1+n^3\alpha^2)},
$$ the latter related series being convergent$\displaystyle \left(<\frac1{\alpha^3}\sum_{n\geq1}\frac1{n^4}\right)$, then we select answer $1)$.
