So, I was recently working on the Sine Fresnal integral and was curious whether we could generalise for any Real Number, i.e.

$$I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx$$

I have formed a solution that I'm uncomfortable with and was hoping for qualified eyes to have a look over.

So, the approach I took was to employ Complex Numbers (I forget the name(s) of the theorem that allows this).


$$\sin\left(x^n\right) = \Im\left[-e^{-ix^n}\right]$$

And so, n

$$ I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx = \Im\left[\int_{0}^{\infty} -e^{-ix^n}\:dx \right]= -\Im\left[\int_{0}^{\infty} e^{-\left(i^{\frac{1}{n}}x\right)^{n}}\:dx \right]$$

Applying a change of variable $u = i^{\frac{1}{n}}x$ we arrive at:

\begin{align} I &= -\Im\left[i^{-\frac{1}{n}}\int_{0}^{\infty} e^{-u^{n}}\:du \right] \\ &= -\Im\left[i^{-\frac{1}{n}}\frac{\Gamma\left(\frac{1}{n}\right)}{n} \right]\\ &= \sin\left(\frac{\pi}{2n}\right)\frac{\Gamma\left(\frac{1}{n}\right)}{n} \end{align}

My area of concern is in the substitution. As $i^{-\frac{1}{n}} \in \mathbb{C}$, I believe the limits of the integral should have been from $0$ to $i^{-\frac{1}{n}}\infty$. Is that correct or not?

I'm also struggling with bounds on $n$ for convergence. Is this expression valid for all $n\in\mathbb{R}$

Any guidance would be greatly appreciated

  • 3
    $\begingroup$ Where does the negative sign in the imaginary part come from $\endgroup$ – Darkrai Nov 9 '18 at 10:23
  • $\begingroup$ @Manthanein - I wanted the integral to take the form of the Gamma Function and so to get the negative part in the exponential, I used the property that $e^{-ix} = \cos(x) -sin(x)i$ or $-e^{ix} =-\cos(x) + \sin(x)$ $\endgroup$ – user150203 Nov 9 '18 at 10:28
  • $\begingroup$ @Manthanein - I just realised I hadn't used that form in the question. Have now corrected. Sorry for any confusion caused. $\endgroup$ – user150203 Nov 9 '18 at 10:30
  • 2
    $\begingroup$ You have indeed moved the contour, so you need to show that the integral from $\infty$ to $i^{-1/n} \infty$ is zero, by an appropriate limiting argument. $\endgroup$ – Richard Martin Nov 9 '18 at 10:32
  • $\begingroup$ @RichardMartin - Is my result then a consequence of bad maths giving the right answer or is there a link that I'm unaware of? $\endgroup$ – user150203 Nov 9 '18 at 10:36

Another approach substitutes $y=x^n$ and writes $y^{1/n-1}$ in terms of a Gamma integral, viz. $$I=\Im\int_0^\infty\frac{1}{n}y^{1/n-1}\exp iy dy=\Im\int_0^\infty\int_0^\infty\frac{1}{n\Gamma(1/n)}z^{-1/n}\exp -y(z-i)dydz.$$By Fubini's theorem, and using $\Im\frac{1}{z-i}=\frac{1}{1+z^2}$,$$I=\int_0^\infty\frac{1}{n\Gamma(1/n)}\frac{z^{-1/n}}{1+z^2}dz.$$Then the substitution $z=\tan u$ obtains a Beta integral, which can be rewritten in terms of Gamma functions, and the result you've claimed is proven true, by the reflection formula of the Gamma function.


Some Hints:

$$I=\int_0^{\infty} \sin (x^n)dx $$ On substitution $x^n=t$ and using the series of $\sin$ we get $$I=\frac 1n \int_0^{\infty} t^{\frac 1n} \left(\sum_{k=0}^{\infty} (-1)^k \frac {t^{2k}k!}{(2k+1)!k!} \right) dt$$

On substituting $t^2=u$ we get $$ I= \frac {1}{2n} \int_0^{\infty} u^{\frac {1-n}{2n}}\left(\sum_{k=0}^{\infty} \frac {\frac {k!}{(2k+1)!}}{k!} (-u)^k \right) du$$

Now by Ramanujan's Master Theorem

$$I=\frac {1}{2n} \Gamma(s)\phi(-s)$$ where $\phi(k)=\frac {k!}{(2k+1)!}$ and $s=\frac {n+1}{2n}$

Hence along with properties of Gamma function, Mellin Transform and the Euler's reflection formula we get $$I=\frac {\pi}{2n\cos \left(\frac {\pi}{2n}\right)\Gamma \left(1-\frac 1n\right)}=\sin \left(\frac {\pi}{2n}\right)\frac {\Gamma\left(\frac 1n\right)}{n}$$

With a special case of $n=2$ we get the value of special integral popularly known as Fresnel integral with limit as $x$ tends to infinity


Start out with a couple of integration by parts: $$ \begin{align} \int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x &=-\frac1y\int_0^\infty\sin(x)\,\mathrm{d}e^{-xy}\tag1\\ &=\frac1y\int_0^\infty\cos(x)\,e^{-xy}\,\mathrm{d}x\tag2\\ &=-\frac1{y^2}\int_0^\infty\cos(x)\,\mathrm{d}e^{-xy}\tag3\\ &=\frac1{y^2}-\frac1{y^2}\int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x\tag4\\ &=\frac1{y^2+1}\tag5 \end{align} $$ Explanation:
$(1)$: prepare to integrate by parts
$(2)$: integrate by parts
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: add $\frac{y^2}{y^2+1}$ times $(4)$ to $\frac1{y^2+1}$ times the LHS of $(1)$

Now write $$ \begin{align} \int_0^\infty\sin\left(x^n\right)\,\mathrm{d}x &=\frac1n\int_0^\infty\sin(x)\,x^{\frac1n-1}\,\mathrm{d}x\tag6\\[3pt] &=\frac1{n\,\Gamma\!\left(1-\frac1n\right)}\int_0^\infty\sin(x)\int_0^\infty y^{-\frac1n}e^{-xy}\,\mathrm{d}y\,\mathrm{d}x\tag7\\ &=\frac1{n\,\Gamma\!\left(1-\frac1n\right)}\int_0^\infty y^{-\frac1n}\int_0^\infty\sin(x)\,e^{-xy}\,\mathrm{d}x\,\mathrm{d}y\tag8\\ &=\frac1{n\,\color{#C00}{\Gamma\!\left(1-\frac1n\right)}}\color{#090}{\int_0^\infty\frac{y^{-\frac1n}}{y^2+1}\,\mathrm{d}y}\tag9\\ &=\color{#C00}{\frac{\Gamma\!\left(\frac1n\right)\sin(\frac\pi{n})}{\color{#000}{n}\pi}}\color{#090}{\frac\pi2\sec\left(\frac\pi{2n}\right)}\tag{10}\\[9pt] &=\Gamma\!\left(1+\frac1n\right)\sin\left(\frac\pi{2n}\right)\tag{11} \end{align} $$ Explanation:
$\phantom{1}(6)$: substitute $x\mapsto x^{1/n}$
$\phantom{1}(7)$: $\int_0^\infty y^{-\frac1n}e^{-xy}\,\mathrm{d}y=x^{\frac1n-1}\Gamma\!\left(1-\frac1n\right)$
$\phantom{1}(8)$: Fubini
$\phantom{1}(9)$: apply $(5)$
$(10)$: $(4)$ from this answer for the green, and $(2)$ from the same answer for the red
$(11)$: simplify


Here is an alternative approach that avoids complex numbers and series altogether. To get round these two obstacles I will use a property of the Laplace transform.

Let $$I = \int_0^\infty \sin (x^n) \, dx, \qquad n > 1.$$ We begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives $$I = \frac{1}{n} \int_0^\infty \frac{\sin x}{x^{1 - 1/n}} \, dx.$$

The following useful property (does this result have a name? It would be so much nicer if it did!) for the Laplace transform will be used: $$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$ Noting that $$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$ and $$\mathcal{L}^{-1} \left \{\frac{1}{x^{1-1/n}} \right \} (t)= \frac{1}{\Gamma (1 - \frac{1}{n})} \mathcal{L}^{-1} \left \{\frac{\Gamma (1 - \frac{1}{n})}{x^{1-1/n}} \right \} (t) = \frac{t^{-1/n}}{\Gamma (1 - \frac{1}{n})},$$ then \begin{align} I &= \frac{1}{n} \int_0^\infty \sin x \cdot \frac{1}{x^{1 - \frac{1}{n}}} \, dx\\ &= \frac{1}{n} \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^{1 - \frac{1}{n}}} \right \} (t) \, dt\\ &= \frac{1}{n\Gamma (1 - \frac{1}{n})} \int_0^\infty \frac{t^{-1/n}}{1 + t^2} \, dt. \end{align} Enforcing a substitution of $t \mapsto \sqrt{t}$ yields \begin{align} I &= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \int_0^\infty \frac{t^{-\frac{1}{2} - \frac{1}{2n}}}{t + 1} \, dt\\ &= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \operatorname{B} \left (\frac{1}{2} - \frac{1}{2n}, \frac{1}{2} + \frac{1}{2n} \right )\\ &= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ). \tag1 \end{align} Applying Euler's reflexion formula we have $$\Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{2n} + \frac{\pi}{2} \right )} = \frac{\pi}{\cos \left (\frac{\pi}{2n} \right )},$$ and $$\Gamma \left (1 - \frac{1}{n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{n} \right ) \Gamma \left (\frac{1}{n} \right )}.$$ So (1) becomes $$I = \frac{\sin (\frac{\pi}{n} ) \Gamma (\frac{1}{n})}{2n \cos (\frac{\pi}{2n} )},$$ or $$I = \sin \left (\frac{\pi}{2n} \right ) \frac{\Gamma \left (\frac{1}{n} \right )}{n}, \qquad n > 1$$ where in the last line the double angle formula for sine has been used.


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