Generalized Fresnel integral $\int_0^\infty \sin x^p \, {\rm d}x$

Question

I am stuck at this question. Find a closed form (that may actually contain the Gamma function) of the integral

$$\int_0^\infty \sin (x^p)\, {\rm d}x$$

I am interested in a Laplace approach, double integral etc. For some weird reason I cannot get it to work.

I am confident that a closed form may actually exist since for the integral:

$$\int_0^\infty \cos x^a \, {\rm d}x = \frac{\pi \csc \frac{\pi}{2a}}{2a \Gamma(1-a)}$$

there exists a closed form with $\Gamma$ and can be actually be reduced further down till it no contains no $\Gamma$. But trying to apply the method of Laplace transform that I have seen for this one , I cannot get it to work for the above integral that I am interested in.

May I have a help?

Answer 1

If $p>1$, $$ I(p)=\int_{0}^{+\infty}\sin(x^p)\,dx = \frac{1}{p}\int_{0}^{+\infty}x^{\frac{1}{p}-1}\sin(x)\,dx \tag{1}$$ but since $\mathcal{L}(\sin(x))=\frac{1}{s^2+1}$ and $\mathcal{L}^{-1}\left(x^{1/p-1}\right)=\frac{s^{-1/p}}{\Gamma\left(1-\frac{1}{p}\right)}$ we have: $$ I(p)=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_{0}^{+\infty}\frac{s^{-1/p}}{1+s^2}\,ds = \color{red}{\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\,\sec\left(\frac{\pi}{2p}\right)}.\tag{2}$$

Answer 2

The integral evaluates to $$ \int_0^{\infty}\sin x^a\ dx=\Gamma\left(1+\frac{1}{a}\right)\sin\frac{\pi}{2a}, $$ but the way I know uses complex analysis.

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Added by request:

We will integrate the function $\exp(-x^a)$ around the circular wedge of radius $R$ and opening angle $\pi/(2a)$, for $a>1$. By the Residue Theorem, $$ 0=\int_0^R\exp(-x^a)\ dx+\int_0^{\pi/(2a)}\exp(-R^ae^{i\theta})iRe^{i\theta}\ d\theta-e^{i\pi/(2a)}\int_0^R\exp(-ix^a)\ dx. $$ The middle integral is $O(Re^{-R^a})$, so sending $R\to\infty$ yields $$ e^{i\pi/(2a)}\int_0^{\infty}\exp(-ix^a)\ dx=\int_0^{\infty}\exp(-x^a)\ dx. $$ To compute the latter integral, let $t=x^a$ so that $dt/t=a\ dx/x$. This yields $$ \int_0^{\infty}\exp(-x^a)\ dx=\frac{\Gamma(1/a)}{a}. $$ Putting everything together, $$ \int_0^{\infty}\exp(-ix^a)\ dx=e^{-i\pi/(2a)}\Gamma\left(1+1/a\right). $$ Taking imaginary parts yields the result.

Answer 3

Hint: Use $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx~=~n!~\iff~\int_0^\infty e^{-x^n}~dx~=~\Gamma\bigg(1+\frac1n\bigg)~$ in conjunction with Euler's formula.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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With $\ds{\pars{~x \equiv t^{1/\pars{2p}} \implies t = x^{2p}~}}$: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{p}}\,\dd x} = {1 \over 2p}\int_{0}^{\infty} t^{\bracks{\color{red}{1/\pars{2p} + 1/2}} - 1} \,\,{\sin\pars{\root{t}} \over \root{t}}\,\dd y \end{align} Note that $\ds{{\sin\pars{\root{t}} \over \root{t}} = \sum_{k = 0}^{\infty} {\pars{-1}^{k} \over \pars{2k + 1}!}\,t^{k} = \sum_{k = 0}^{\infty} \color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{2 + 2k}}\,{\pars{-t}^{k} \over k!}}$.

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With Ramanujan-MT: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{p}}\,\dd x} \\[5mm] = &\ {1 \over 2p}\, \Gamma\pars{{1 \over 2p} + {1 \over 2}}\, {\Gamma\pars{1 - \bracks{1/\pars{2p} + 1/2}} \over \Gamma\pars{2 - 2\bracks{1/\pars{2p} + 1/2}}} \\[5mm] = &\ {1 \over 2p}\,{\Gamma\pars{1/\bracks{2p} + 1/2} \Gamma\pars{1/2 - 1/\bracks{2p}} \over \Gamma\pars{1 - 1/p}} \\[5mm] = &\ {1 \over 2p}\,{\pi/\sin\pars{\pi\braces{1/\bracks{2p} + 1/2}} \over \Gamma\pars{1 - 1/p}} = \bbx{\pi\sec\pars{\pi/\bracks{2p}} \over 2p\,\Gamma\pars{1 - 1/p}} \\ &\ \end{align}