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.

  • $\begingroup$ Isn't it a divergent integral? $\endgroup$ – peterh Feb 17 '15 at 15:18
  • 1
    $\begingroup$ 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 $\sin$. $\endgroup$ – tired Feb 17 '15 at 15:32
  • 1
    $\begingroup$ the contour will be two wedges with angle $\pi/8$ $\endgroup$ – tired Feb 17 '15 at 15:34
  • 2
    $\begingroup$ @peterh: No, it converges due to the Riemann-Lebesgue lemma. $\endgroup$ – Lucian Feb 17 '15 at 15:55
  • $\begingroup$ @peterh I also thought that was divergent until I did it in Maple and was given a convergent result . $\endgroup$ – 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

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


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

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,

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

  • 1
    $\begingroup$ Nice answer (+1), but i think it would be less work to work out $Exp[\pm i x^4]$ seperatly... $\endgroup$ – 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 $$ \begin{align} \int_{0}^{\infty} \sin (x^4) \, dx & = \frac14\int_{0}^{\infty} t^{-3/4} \sin t \, dt =\frac14\cos \left(\frac{3\pi }{8}\right)\Gamma(1/4)=\frac{\sqrt{2-\sqrt{2}}}{8}\Gamma(1/4). \tag3 \end{align} $$

  • $\begingroup$ Nice approach (+1) $\endgroup$ – 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.