Show that $\int_{0}^{+\infty}x\sin({x}^{4})\,\mathrm dx$ converges, although the function is unlimited 
Show that $\int_{0}^{+\infty}x\sin({x}^{4})\,\mathrm dx$ converges, although the function is unlimited.

Summary proof: writing $a=\sqrt[4]{n\pi}$ and $b =\sqrt[4]{(n+1)\pi}$ we have to $$\int_{a}^{b}\left|x\sin({x}^{4}) \right|\,\mathrm dx < b(b-a)$$ where $b (b-a)$ is the area of the rectangle. Thus $$\pi={b}^{4}-{a}^{4}=\left(b-a \right)\left({a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3} \right)$$ this area is worth $$\left(b-a \right)=\frac{b\pi}{{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3} }$$ goes to zero when $n\rightarrow\infty$.
Let ${c}^{4}=\sqrt[4]{(n+2)\pi}$, the change of variable $x=\sqrt[4]{{u}^{4}+\pi}$ we have to $$\int_{a}^{b}\left|x\sin({x}^{4}) \right|\,\mathrm dx=\int_{a}^{b}\left|{u}^{2}\sin({u}^{4}) \right|\frac{{u}^{2}}{\sqrt[4]{{({u}^{4}+\pi)}^{2}}}\,\mathrm du$$  such that  ${a}_{n+1}<{a}_{n}$.  By Leibniz's theorem $\sum_{1}^{\infty}\left({-1}^{n} \right){a}_{n}$ converges for the value of integral.
OBS:
$\sqrt[4]{n\pi} < x < \sqrt[4]{(n+1)\pi} \Rightarrow n\pi< {x}^{4}-\pi< (n+1)\pi$ making a change of variable $u={x}^{4}-\pi \Rightarrow x=\sqrt[4]{{u}^{4}+\pi}$
Why $\int_{a}^{b}\left|x\sin({x}^{4}) \right|\mathrm dx < b(b-a)$? and how did you conclude ${a}_{n+1}<{a}_{n}$?
 A: Your idea is fine, but it is simpler to substitute $x:=u^{1/4}$. In this way you obtain
$$\int_0^b x\sin(x^4)\>dx={1\over4}\int_0^{b^4}{1\over u^{1/2}}\sin u\>du={1\over4}\sum_{k=1}^N(-1)^{k-1}\int_{(k-1)\pi}^{k\pi}{|\sin u|\over u^{1/2}}\>du\ ,$$
where the last bump is not counted in full, and $N\to\infty$ at the end.
A: I substituted $x=\dfrac{1}{t}$ 
The integral becomes
$$\int_0^{\infty } \frac{\sin \left(\frac{1}{t^4}\right)}{ t^3} \, dt$$
the numerator is bounded by 1 so the integrand is less than $\dfrac{1}{t^3}$ which converges 
A: The given integral is not absolutely convergent, but
$$ I=\lim_{R\to +\infty}\int_{0}^{R}x\sin(x^4)\,dx $$
is finite. Indeed, by a substitution and the Laplace transform we get:
$$ I \stackrel{x\mapsto w^{1/4}}{=}\underbrace{ \frac{1}{4}\int_{0}^{+\infty}\frac{\sin w}{\sqrt{w}}\,dw}_{\text{Convergent by Dirichlet's test}} \stackrel{\mathcal{L}}{=}\frac{1}{4\sqrt{\pi}}\int_{0}^{+\infty}\frac{ds}{(1+s^2)\sqrt{s}} $$
leading to:
$$ I = \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{1+t^4} = \color{red}{\frac{1}{4}\sqrt{\frac{\pi}{2}}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}x\sin\pars{x^{4}}\,\dd x &
\,\,\,\stackrel{x^{4}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}x^{-1/2}\sin\pars{x}\,\dd x
\\[5mm] & =
{1 \over 4}\,\Im\int_{0}^{\infty}x^{-1/2}\expo{\ic x}\,\dd x\qquad
\pars{~z^{-1/2}\ \mbox{is its}\ Principal\ Branch ~}
\\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\,&
-\,{1 \over 4}\,\Im\int_{0}^{\pi/2}\!\!\!R^{-1/2}\expo{-\ic\theta/2}
\exp\pars{\ic R\expo{\ic\theta}}\,R\expo{\ic\theta}\ic\,\dd\theta -
{1 \over 4}\,\Im\int_{\infty}^{0}\!\!\!y^{-1/2}\expo{-\ic\pi/4}
\expo{\ic\pars{\ic y}}\ic\,\dd y
\\[5mm] & =
-\,{1 \over 4}\ \overbrace{R^{1/2}\,\Re\int_{0}^{\pi/2}
\exp\pars{-R\sin\pars{\theta}}\exp\pars{\ic\bracks{R\cos\pars{\theta} + {1 \over 2}\,\theta}}\dd\theta}^{\ds{\to\ 0\ \mrm{as}\ R\ \to\ \infty}}
\\[2mm] & +
{1 \over 4}\,{\root{2} \over 2}\
\underbrace{\int_{0}^{\infty}y^{-1/2}\expo{-y}\,\dd y}
_{\ds{\Gamma\pars{1 \over 2} = \root{\pi}}}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & = \bbx{\root{2\pi} \over 8} \approx 0.3133
\end{align}
