Convert Airy Integral to cosine integral?

186 Views Asked by At

I'm doing some research on the Airy function and its derivation and I am confused on how to arrive from the Airy function to Jeffrey's form, as is presented in his paper The Effect of Love Waves of Heterogeneity in the Lower Layer. He jumps from the Airy Integral to a function containing cosine:

\begin{align} Ai(x)&=\frac{1}{2\pi i}\int^{i\infty}_{-i\infty}e^{\kappa x+\frac{1}{3}\kappa^3}d\kappa \\ &= \frac{1}{\pi}\int^\infty_0\cos(kx-\frac{1}{3}k^3)dk. \end{align}

I'm a little (a lot) confused by this. Initially I tried writing $e^{f(\kappa)}$ as $\kappa=\alpha+i\beta$, so I'd have something in the form of $re^{i\theta}$ which is form of the complex number $\kappa$, but this is quite clearly wrong.

I would really like some clarification on how this can be evaluated, so any help would be much appreciated!

1

There are 1 best solutions below

2
On BEST ANSWER

The path is such that $\kappa$ is purely imaginary, let $\kappa=ik$.

Then

$$\frac{1}{2\pi i}\int^{i\infty}_{\kappa=-i\infty}e^{\kappa x+\frac{1}{3}\kappa^3}d\kappa =\frac{1}{2\pi}\int^{\infty}_{k=-\infty}e^{ikx-\frac{1}{3}ik^3}d\kappa.$$

As the complex exponential has an even real part and an odd imaginary part, only the first remains, giving

$$\frac{1}{\pi}\int^{\infty}_{0}\Re\left(e^{ikx-\frac{1}{3}ik^3}\right)d\kappa.$$