Airy function value and derivative value at 0

I see that from https://en.wikipedia.org/wiki/Airy_function, $ Ai(0) = \frac{1}{3^{\frac{2}{3}} * \Gamma (\frac{2}{3})} $ and $Ai'(0) = - \frac{1}{3^{\frac{1}{3}} * \Gamma (\frac{1}{3})} $. But how can I see these from the contour integral representation of Airy function : $ Ai(z) = \frac{1}{2 \pi i} \int_C e^{(\frac{t^3}{3} - zt)} dt$, since $ Ai(0) = \frac{1}{2 \pi i} \int_C e^{(\frac{t^3}{3})} dt$, and $Ai'(0) = \frac{1}{2 \pi i} \int_C t^2 * e^{(\frac{t^3}{3})} dt$? and how can I connect them with the gamma function. Thank you for any help.

Solution 1:

Another definition of the Airy Ai function will have an imaginary part of zero because the imaginary part is even:

$$\text{Ai}(z)=\frac1\pi \int_\Bbb R e^{\frac i3t^3+i zt}dt\implies \text{Ai}(0)=\frac2\pi\int_0^\infty e^{\frac i3 t^3}dt,\text Ai’(0)= \frac{2i}\pi\int_0^\infty te^{\frac i3t^3}dt $$

Now let $-x=\frac i3 t^3$:

$$\text{Ai}(0)=\frac{2i(3i)^\frac13}{3\pi}\int_0^\infty e^{-x} x^{\frac13-1}dx= \frac{2i(3i)^\frac13}{3\pi} \Gamma\left(\frac13\right)$$

You can now do the same for $\text{Ai}’(0)$:

$$\text{Ai}’(0)=\frac{2i}\pi\frac{(3i)^\frac23}{3}\int_0^\infty e^{-x} x^{\frac23-1}dx=\frac{2i(3i)^\frac23}{3\pi} \Gamma\left(\frac23\right) $$ Please correct me and give me feedback!