# Integral of involving Airy function without using its antiderivative

Inspired by my answer to this question, I am interested in evaluating the following definite integral $$\frac{1}{2\pi i} \int_{c-i \infty}^{c+i\infty} \frac{dz}{\mathop{\rm Ai}^2(z)} =1$$ without using its antiderivate. At first, it seems we need to have $c\geq 0$ but as I will point out below, any $c\in\mathbb{R}$ such that $\mathop{\rm Ai}(c) \neq 0$ will give the same result.

The integral can be solved using the antiderivative $\frac{\pi\mathop{\rm Bi}(z)}{\mathop{\rm Ai}(z)}$ together with the well-known asymptotic expansion for Ai and Bi as I pointed out in the answer to the question.

All the zeros of the Airy function lie on the negative real axis. The curious fact why the above mentioned result holds for almost all $c$ is that all the residues at the corresponding poles of $\mathop{\rm Ai}^{-2}(z)$ vanish. The reason is that due to Airy's differential equation $\mathop{\rm Ai}''(z) - z \mathop{\rm Ai}(z) =0$ every zero of Ai comes together with a vanishing second derivative. Thus for any pole $z^*$ of $\mathop{\rm Ai}^{-2}(z)$ we have $$\frac{1}{\mathop{\rm Ai}^{2}(z)} =\frac{1}{[\alpha (z- z^*) + \mathcal{O}(z-z^*)^3 ]^2} = \frac{1}{\alpha^2 (z- z^*)^2} + \mathcal{O}(1)$$ and the residue vanishes.

My question is if/how the definite integral can be solved without resorting to the antiderivative?

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## 1 Answer

A solution is given in the appendix of the following paper: Knessl, C. Exact and Asymptotic Solutions to a PDE That Arises in Time-Dependent Queues. Adv. Appl. Prob., 32(1) 256--283, 2000.

Let me describe this solution and highlight some subtle points.

${\rm Ai}(z)$ has different asymptotic approximations for large $|z|$ valid in different regions of the complex plane. We are going to use $${\rm Ai}(z) \sim \frac{1}{2 \pi^{\frac{1}{2}} z^{\frac{1}{4}}} \left( e^{-\frac{2}{3} z^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} z^{\frac{3}{2}}} \right) \;,$$ which is valid as $|z| \to \infty$ assuming $\frac{\pi}{3} < {\rm arg}(z) < \frac{5 \pi}{3}$, to replace $$f(z) = \frac{1}{{\rm Ai}^2(z)} \;,$$ with $$g(z) = \frac{4 \pi z^{\frac{1}{2}}} {\left( e^{-\frac{2}{3} z^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} z^{\frac{3}{2}}} \right)^2} \;.$$

We need to consider the following three contours (with which we will take $R \to \infty$): \begin{eqnarray} C_1 &=& \{ {\rm i} s \}_{s = -R}^{s = R} \;, \\ C_2 &=& \{ R e^{{\rm i} \theta} \}_{\theta = \frac{\pi}{2}}^{\theta = \frac{3 \pi}{2}} \;, \\ C_3 &=& \{ e^{\frac{4 {\rm i} \pi}{3}} s \}_{s = R}^{s = 0} \cup \{ e^{\frac{2 {\rm i} \pi}{3}} s \}_{s = 0}^{s = R} \;. \end{eqnarray}

The given problem is to evaluate $$I \equiv \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1} f(z) \,dz \;.$$ By Cauchy's residue theorem, since the residues of the singularities of $f(z)$ are all zero, we have $$\lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1 \cup C_2} f(z) \,dz = 0 \;,$$ thus $$I = - \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_2} f(z) \,dz \;.$$ In view of the asymptotic approximation, it follows that $$I = - \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_2} g(z) \,dz \;.$$ Again by Cauchy's residue theorem, since the residues of the singularities of $g(z)$ are all zero, we have $$\lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1 \cup C_2} g(z) \,dz = 0 \;,$$ thus $$I = \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1} g(z) \,dz \;.$$ Finally we may deform $C_1$ to $C_3$, i.e., $$I = \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_3} g(z) \,dz \;.$$

We can evaluate this integral by explicitly parametrizing $C_3$. For the first part of $C_3$, $z = e^{\frac{4 \pi {\rm i}}{3}} s$, where $s$ ranges from $\infty$ to $0$, and note that, in particular, $z^{3/2} = s^{3/2}$. For the second part of $C_3$, $z = e^{\frac{2 \pi {\rm i}}{3}} s$, where $s$ ranges from $0$ to $\infty$, and here $z^{3/2} = -s^{3/2}$. We have

$$I = \frac{2}{{\rm i}} \int_\infty^0 \frac{s^{\frac{1}{2}}} {\left( e^{-\frac{2}{3} s^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} s^{\frac{3}{2}}} \right)^2} \,ds + \frac{2}{{\rm i}} \int_0^\infty \frac{-s^{\frac{1}{2}}} {\left( e^{\frac{2}{3} s^{\frac{3}{2}}} + {\rm i} e^{-\frac{2}{3} s^{\frac{3}{2}}} \right)^2} \,ds \;.$$

By multiplying and dividing each integral by the square of the complex conjugate of the term appearing in the denominator (the usual trick to make the denominator real-valued) and combining the integrals and simplifying we arrive at \begin{eqnarray} I &=& 8 \int_0^\infty \frac{s^{\frac{1}{2}}} {\left( e^{\frac{4}{3} s^{\frac{3}{2}}} + e^{-\frac{4}{3} s^{\frac{3}{2}}} \right)^2} \,ds \\ &=& \int_0^\infty \frac{1}{{\rm cosh}^2(u)} \,du \\ &=& {\rm tanh}(u) \big|_0^\infty \\ &=& 1 \;, \end{eqnarray} as required.

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It is indeed very subtle. What confuses me a bit is that on first sight the integral of $f$ (or of $g$ for that matter) along the contour $C_2$ vanishes because $g\to 0$ sufficiently fast. What is wrong with my argument? –  Fabian Jan 1 '13 at 6:43
That confused me too. But recall ${\rm Ai}(z) \to 0$ as $z \to -\infty$. Thus $f(z) \to \infty$ as $z \to -\infty$. Also $g(-t) = \frac{2^{3/2} \pi t^{1/2}}{1 + {\rm sin}(4/3 t^{3/2})} \to \infty$ as $t \to \infty$. –  djws Jan 1 '13 at 7:56
but still it is a single point and away from the real axis the function approaches 0 exponentially! So I guess one can think about the $g$ at infinity to be something like a $\delta$ function. –  Fabian Jan 1 '13 at 9:27
By the way: how did you find this paper? –  Fabian Jan 1 '13 at 9:48
I already knew about the paper as the main problem there of Brownian motion with simple time-dependent drift arose in my research. –  djws Jan 1 '13 at 18:30