# Integral $\int_0^\infty\sin{(x^4)} dx$

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.

• Isn't it a divergent integral? – peterh Feb 17 '15 at 15:18
• hint: use contour integration to establish a relation between $\int e^{-x^4}$ and $\int e^{-ix^4}$. Be carfeul, you need to this in two different ways for the two exponetials in $\sin$. – tired Feb 17 '15 at 15:32
• the contour will be two wedges with angle $\pi/8$ – tired Feb 17 '15 at 15:34
• @peterh: No, it converges due to the Riemann-Lebesgue lemma. – Lucian Feb 17 '15 at 15:55
• @peterh I also thought that was divergent until I did it in Maple and was given a convergent result . – Vim Feb 17 '15 at 15:56

Consider the contour integral $\int_{\Gamma(R)} e^{-z^4}\, dz$, where $\Gamma(R)$ is the first quadrant sector of the quarter circle (centered at the origin), subtended by the angle $\pi/8$, with counterclockwise orientation. Then $\Gamma = \gamma_1 + \gamma_2 + \gamma_3$, where $\gamma_1$ is the line segment from $0$ to $R$, $\gamma_2$ is the arc of the sector, and $\Gamma_3$ is the line segment from $Re^{i\pi/8}$ to $0$. Along $\gamma_2$, $z = Re^{it}$, $0 \le t \le \pi/8$, so

$$|e^{-z^4}| = |e^{-R^4(\cos 4t+ i\sin 4t)}| = e^{-R^4\cos 4t}.$$

Since $\cos 4t \ge 1 - 8t/\pi$ for $0 \le t \le \pi/8$, $|e^{-z^4}| \le e^{-R^4}e^{8R^4t/\pi}$. Therefore

$$\left|\int_{\gamma_2} e^{-z^4}\, dz\right| \le e^{-R^4}\int_0^{\pi/8} e^{8R^4t/\pi} R\, dt = Re^{-R^4}\cdot \frac{\pi(e^{R^4} - 1)}{8R^4} = \frac{\pi}{8R^3}(1 - e^{-R^4})$$

The last expression tends to $0$ as $R\to \infty$. Since $\int_{\Gamma(R)} e^{-z^4}\, dz = 0$ by Cauchy's theorem, we have

$$0 = \lim_{R\to \infty} \left(\int_{\gamma_1} e^{-z^4}\, dz + \int_{\gamma_3} e^{-z^4}\, dz\right).$$

Along $\gamma_3$, $z = re^{i\pi/8}$, $0 \le r \le R$. Thus

$$\int_{\gamma_3} e^{-z^4}\, dz = -\int_0^R e^{-r^4e^{i\pi/2}}\, e^{i\pi/8}\, dr = -e^{i\pi/8}\int_0^R e^{-ir^4}\, dr.$$

Since

$$\int_{\gamma_1} e^{-z^4}\, dz + \int_{\gamma_3} e^{-z^4}\, dz = \int_0^R e^{-x^4}\, dx - e^{i\pi/8} \int_0^R e^{-ix^4}\, dx,$$

we deduce that

\begin{align}0 &= \int_0^\infty e^{-x^4}\, dx - \int_0^\infty e^{-i(x^4 - \pi/8)}\, dx\\ 0&= \Gamma(5/4) - \int_0^\infty [\cos(x^4 - \pi/8) + i \sin(x^4 - \pi/8)]\, dx\\ \Gamma(5/4)&= - \int_0^\infty (\cos(x^4)\cos(\pi/8) + \sin(x^4)\sin(\pi/8))\, dx\\& + \int_0^\infty (\sin(x^4)\cos(\pi/8) - \cos(x^4)\sin(\pi/8))\, dx\end{align}

Taking the real and imaginary parts, we obtain a system of equations

\begin{align} A\cos \pi/8 + B\sin \pi/8 &= \Gamma(5/4)\\ -A\sin \pi/8 + B\cos \pi/8 &= 0 \end{align}

Here $A = \int_0^\infty \cos(x^4)\, dx$ and $B = \int_0^\infty \sin(x^4)\, dx$. The solution is

\begin{align} A &= \cos(\pi/8)\Gamma(5/4)\\ B &= \sin(\pi/8)\Gamma(5/4) \end{align}

That is,

$$\int_0^\infty \sin x^4\, dx = \sin(\pi/8)\Gamma(5/4), \quad \int_0^\infty \cos x^4\, dx = \cos(\pi/8)\Gamma(5/4).$$

• Nice answer (+1), but i think it would be less work to work out $Exp[\pm i x^4]$ seperatly... – tired Feb 17 '15 at 16:32

Hint. Here is an approach.

Recall that, from the definition of the Euler $\Gamma$ function, we have \begin{align} \int_{0}^{\infty} e^{-bt} \, t^{s} \, dt = \frac{\Gamma(s+1)}{b^{s+1}}, \quad s>-1, \Re b>0. \tag1 \end{align} Put $b_\epsilon=\epsilon+i,\, \epsilon>0$, in $(1)$, then let $\epsilon \to 0^+$ to get \begin{align} \int_{0}^{\infty} t^{s} \sin t \, dt & = \cos \left(\frac{\pi s}{2}\right)\Gamma(s+1), \quad -1<s<0. \tag2 \end{align} Now, by the change of variable $t=x^4$, $x=t^{1/4}$, $dx=\frac14 t^{-3/4}dt$ we have \begin{align} \int_{0}^{\infty} \sin (x^4) \, dx & = \frac14\int_{0}^{\infty} t^{-3/4} \sin t \, dt =\frac14\cos \left(\frac{3\pi }{8}\right)\Gamma(1/4)=\frac{\sqrt{2-\sqrt{2}}}{8}\Gamma(1/4). \tag3 \end{align}

• Nice approach (+1) – tired Feb 17 '15 at 16:32

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}