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I'd like to show: $$ I \equiv \frac{1}{(2 \pi {\rm i})^2} \int_{a - {\rm i} \infty}^{a + {\rm i} \infty} \int_{b - {\rm i} \infty}^{b + {\rm i} \infty} \frac{1}{{\rm Ai}(u) {\rm Ai}(v) (u-v)} \,dv \,du = \frac{1}{2} \;, $$ where each integral is over a vertical contour in the complex plane (like for an inverse Laplace transform) and the only restriction on $a$ and $b$ is: $$ 0 < b < a \;. $$ The Airy function of the first kind, $$ {\rm Ai}(z) \equiv \frac{1}{\pi} \int_0^\infty \cos \left( \frac{t^3}{3} + zt \right) \,dt \;, $$ satisfies ${\rm Ai}''(z) = z {\rm Ai}(z)$ and is an entire function with zeros on the negative real axis. Numerical integration gives me $I = \frac{1}{2}$, but how can I prove it?

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up vote 5 down vote accepted

The only zeros of the Airy function Ai are on the negative real axis. Furthermore, there is only a pole at $u=v$ which immediately shows that the integral does not depend on $a$ and $b$ chosen as long as $0\leq b<a$.

Because of the independence of $b$ and $a$, we choose $b=0$ and $a=\eta$ and let $v=i x$ and $u=\eta + i y$ with $x,y \in \mathbb{R}$. We can introduce center of mass and relative coordinates $c= (x+y)/2$ and $r= x-y$. Thus the integral reads $$I= \frac{1}{(2\pi)^2} \int_{-\infty}^\infty dc \int_{-\infty}^\infty dr\, \frac{i}{(r+i \eta) \mathop{\rm Ai}\left[i \left(c+\tfrac{r}2\right)\right] \mathop{\rm Ai}\left[i \left(c-\tfrac{r}2\right)\right]} \qquad (1)$$ where we already used the fact that we want to let $\eta\downarrow 0$ to eliminate $\eta$ in the argument of Ai.

In a next step, we use Sokhotsky's formula to simplify $$I= \frac{1}{(2\pi)^2}\int_{-\infty}^\infty dc \int_{-\infty}^\infty dr\,\left[ \pi \delta(r) + \mathcal{P} \frac{i}{r} \right] \frac{1}{\mathop{\rm Ai}\left[i \left(c+\tfrac{r}2\right)\right] \mathop{\rm Ai}\left[i \left(c-\tfrac{r}2\right)\right]}.$$

As the last factor is even in $r$ the principal value integral vanishes so we are left with $$I=\frac{1}{2(2\pi)}\int_{-\infty}^\infty dc \frac1{\mathop{\rm Ai}^2(ic)}.$$

This last integral could be another nice question. It turns out that WA knows the antiderivative. So $$\int_{-\infty}^\infty dc \frac1{\mathop{\rm Ai}^2(ic)} = - i \pi \frac{\mathop{\rm Bi}(ic)}{\mathop{\rm Ai}(ic)}\Biggr|_{-\infty}^\infty = 2\pi$$ from which the result follows.

Edit:

If you do not know Sokhotsky's formula you can obtain the result by adding to (1) the same integral with the substitution $r\mapsto -r$. Then $$2I = \frac{1}{(2\pi)^2} \int_{-\infty}^\infty dc \int_{-\infty}^\infty dr\, \underbrace{\left[\frac{i}{r+i \eta} +\frac{i}{-r+i \eta} \right]}_{\frac{2\eta}{r^2 + \eta^2}} \frac{1}{\mathop{\rm Ai}\left[i \left(c+\tfrac{r}2\right)\right] \mathop{\rm Ai}\left[i \left(c-\tfrac{r}2\right)\right]}.$$ As $\eta \downarrow 0$, the integral is dominated for $r$ close to 0 such that one can replace the last factor by its value at $r=0$.

Note that the last factor is exponentially decaying in both $c$ and $r$ as can be seen from the asymptotic expansion of Ai. We know that $|\mathop{\rm Ai}(ic)| \sim e^{\sqrt{2} |c|^{3/2}/3}/{2\sqrt{\pi} |c|^{1/4}}$ such that $$\left| \frac{1}{\mathop{\rm Ai}\left[i \left(c+\tfrac{r}2\right)\right] \mathop{\rm Ai}\left[i \left(c-\tfrac{r}2\right)\right]} \right| \sim 4 \pi \left| c+ \tfrac{r}2\right|^{1/4}\left| c- \tfrac{r}2\right|^{1/4} \exp \left[-\sqrt{2}\left(\left| c+ \tfrac{r}2\right|^{3/2} + \left| c- \tfrac{r}2\right|^{3/2} \right)/2 \right].$$ The exponential decay in both $r$ and $c$ is more than enough for our purposes.

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That's a very nice derivation! –  djws Jan 2 '13 at 2:50
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