# Closed form for $\int_0^{\infty}\sin(x^n)\mathbb{d}x$

I was wondering if anyone knows a closed form for

$$\mathrm{I} = \int_0^{\infty}\sin(x^n)\mathbb{d}x$$

Preliminary evaluations on Wolfram Alpha seem to yield something like this:

$$\mathrm{I} = k\sin\left(\frac{\pi}{2n}\right)\Gamma\left(\frac an \right)$$

where $k$ is a proportionality constant, normally $1$, and $a$ is a constant, normally $1$ as well.

• Mathematica gives $$\int_0^\infty \sin x^n dx = \Gamma\left(1 + \frac{1}{n}\right) \sin \frac{\pi}{2n}$$ – Henricus V. Nov 30 '15 at 4:39
• I wonder if it could be done as $Im\int_0^{\infty} \exp(i x^n) dx$. – marty cohen Nov 30 '15 at 4:45
• This question reminds me that I keep meaning to look up the details of the history of this integral. This is just my own speculation here, but I think the convergence of this integral is such a striking and counter-intuitive result that if a closed form value had not existed, they would have assigned it a Greek letter and called that a "closed form". :) – David H Nov 30 '15 at 5:52
• @DavidH: the convergence of this integral is such a striking and counter-intuitive result - See Riemann-Lebesgue lemma. – Lucian Nov 30 '15 at 6:44
• Hint: Use Euler's formula in conjunction with the integral expression for the $\Gamma$ function. – Lucian Nov 30 '15 at 6:47

It has a closed form

$$\int_{0}^{\infty} \exp(i x^{1/\alpha}) \, \Bbb{d}x = \Gamma(1+\alpha)e^{i\pi\alpha/2}, \quad \alpha \in (0, 1).$$

Applying the substitution $x \mapsto x^{\alpha}$, we can write

$$I(\alpha) := \int_{0}^{\infty} \exp(ix^{1/\alpha}) \, \Bbb{d}x = \int_{0}^{\infty} \alpha x^{\alpha-1} e^{ix} \, \Bbb{d}x, \tag{1}$$

which is much easier to work with.

Complex-analytic method. Rotate the contour of $\text{(1)}$ by $90^{\circ}$ degrees to have

$$I(\alpha) = \alpha \int_{0}^{i\infty} z^{\alpha-1} e^{iz} \, dz.$$

This is a standard technique, and we can justify this by applying the Cauchy integration formula to a quadrant contour and then utilizing the Jordan's lemma.

Then by the substitution $z \mapsto ix$, we have

$$I(\alpha) = \alpha i^{\alpha} \int_{0}^{\infty} x^{\alpha-1}e^{-x} \, dx = \Gamma(1+\alpha)i^{\alpha}.$$

Real-analytic method. Since the right-hand side of $\text{(1)}$ exists as improper-integral sense, it follows from the integral version of the Abel's theorem that

$$I(\alpha) = \int_{0}^{\infty} \alpha x^{\alpha-1} e^{ix} \, \Bbb{d}x = \lim_{\epsilon \downarrow 0} \int_{0}^{\infty} \alpha x^{\alpha-1} e^{ix}e^{-\epsilon x} \, \Bbb{d}x. \tag{2}$$

Using the formula

$$\frac{\Gamma(1-\alpha)}{x^{1-\alpha}} = \int_{0}^{\infty} u^{-\alpha}e^{-xu} \, \Bbb{d}u$$

and the Fubini's theorem, we can write

\begin{align*} \int_{0}^{\infty} \alpha x^{\alpha-1} e^{ix}e^{-\epsilon x} \, \Bbb{d}x &= \frac{\alpha}{\Gamma(1-\alpha)} \int_{0}^{\infty} \left(\int_{0}^{\infty} u^{-\alpha}e^{-xu} \, \Bbb{d}u\right) e^{-(\epsilon-i) x} \, \Bbb{d}x \\ &= \frac{\alpha}{\Gamma(1-\alpha)} \int_{0}^{\infty} \left(\int_{0}^{\infty} e^{-(u+\epsilon-i) x} \, \Bbb{d}x\right) u^{-\alpha} \, \Bbb{d}u \\ &= \frac{\alpha}{\Gamma(1-\alpha)} \int_{0}^{\infty} \frac{u^{-\alpha}}{u+\epsilon-i} \, \Bbb{d}u. \end{align*}

(Notice that we cannot apply the Fubini's theorem directly when $\epsilon = 0$ due to integrability issue. This explains why we are considering a regularized version $\text{(2)}$ of the original integral.)

Taking $\epsilon \downarrow 0$, we have

$$I(\alpha) = \frac{\alpha}{\Gamma(1-\alpha)} \int_{0}^{\infty} \frac{u^{1-\alpha} + iu^{-\alpha}}{u^2+1} \, \Bbb{d}u. \tag{3}$$

Using the beta function identity, the last integral in $\text{(3)}$ can be computed explicitly as

$$I(\alpha) = \frac{\alpha}{\Gamma(1-\alpha)} \cdot \frac{\pi}{2}\left( \frac{1}{\sin(\pi\alpha/2)} + \frac{i}{\cos(\pi\alpha/2)} \right).$$

Finally, from the Euler's reflection formula and the sine double-angle formula we have

$$I(\alpha) = \Gamma(1+\alpha) e^{i\pi\alpha/2}.$$

• Excellent Solutions! +1 – Mark Viola Nov 30 '15 at 5:25