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?


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: $$ I(p)=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_{0}^{+\infty}\frac{s^{-1/p}}{1+s^2}\,ds = \color{red}{\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\,\sec\left(\frac{\pi}{2p}\right)}.\tag{2}$$

  • $\begingroup$ Any relation between $\left(1\right)$ and the Mellin Transform? $\endgroup$ – jm324354 Sep 26 '15 at 22:46
  • $\begingroup$ @jm324354: well, sure, I applied the direct and inverse Laplace transform, but that can be seen as an application of the Mellin transform, too. $\endgroup$ – Jack D'Aurizio Sep 26 '15 at 23:08

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.

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.

  • $\begingroup$ May you provide some details using complex analysis? $\endgroup$ – Tolaso Sep 26 '15 at 20:33

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.


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