TITLE: Integral $\int_0^\infty\sin{(x^4)} dx$
QUESTION [7 upvotes]: Can anyone help with how to evaluate the following integral
$$
\int_0^\infty\sin{(x^4)} dx
$$
I know that I need to use the fact that $\int_0^\infty e^{-x^4}dx=\Gamma\left(\dfrac{5}{4}\right)$ and I know that I have to use Eulers formula and contour integration but I am really lost on how to start the problem.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\overbrace{\color{#66f}{\large%
\int_{0}^{\infty}\sin\pars{x^{4}}\,\dd x}}^{\ds{\dsc{x^{4}}\ \mapsto\ \dsc{x}}}\ =\
\frac{1}{4}\int_{0}^{\infty}x^{1/4}\,\frac{\sin\pars{x}}{x}\,\dd x
=\frac{1}{4}\int_{0}^{\infty}x^{1/4}\,\ \overbrace{%
\frac{1}{2\ic}\int_{-\ic}^{\ic}\expo{-kx}\,\dd k}
^{\dsc{\frac{\sin\pars{x}}{x}}}\ \,\dd x
\\[5mm]&=\frac{1}{8\ic}\int_{-\ic}^{\ic}\ \overbrace{%
\int_{0}^{\infty}x^{1/4}\expo{-kx}\,\dd x}
^{\ds{\dsc{kx}\ \mapsto\ \dsc{x}}}\ \,\dd k
=\frac{1}{8\ic}\int_{-\ic}^{\ic}k^{-5/4}\ \overbrace{%
\int_{0}^{\infty}x^{1/4}\expo{-x}\,\dd x}^{\dsc{\Gamma\pars{5/4}}}\ \,\dd k
\\[5mm]&=\Gamma\pars{\frac{5}{4}}\,\frac{1}{8\ic}
\bracks{-4i^{-1/4} + 4\pars{-\ic}^{-1/4}}
=\Gamma\pars{\frac{5}{4}}\,\frac{1}{2\ic}
\pars{-\expo{-\ic\pi/8} + \expo{\ic\pi/8}}
\\[5mm]&=\color{#66f}{\large%
\sin\pars{\frac{\pi}{8}}\Gamma\pars{\frac{5}{4}}}
\approx{\tt 0.3469}
\end{align}