Uniform convergence of $\sum_{n=0}^{\infty} \frac{\int_{\sin nx}^{\sin(n+1)x}\sin t^2dt \int_{nx}^{\infty}\frac{dt}{\sqrt{t^4+1}}}{1+n^3x^2}$

Question

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!

Answer 1

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 $1)$.

Answer 2

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 $(0,\infty)$.