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I know that: $$\int_{-\infty}^0 Ai(x)dx = \frac{2}{3}, \qquad \int_0^{+\infty} Ai(x) dx = \frac{1}{3}$$ But I don't understand why according to DLMF are valid:

$$\int_{-\infty}^x Ai\left(t\right) dt = \pi \left[ Ai(x) Hi'(x) - Hi(x)Ai'(x) \right]$$

And: $$\int_{-\infty}^x Bi\left(t\right) dt =\pi\left[\mathrm{Bi}\left(x\right)\mathrm{Hi}'\left(x\right)-\mathrm{Bi}'% \left(x\right)\mathrm{Hi}\left(x\right)\right]. $$

Where Hi e Gi are the Scorer Functions: $$Hi\left(x\right) = \frac{1}{\pi} \int_0^{+\infty} e^{\left( xt - \frac{t^3}{3}\right)}dt$$ $$Gi\left(x\right) = \frac{1}{\pi} \int_0^{+\infty} sin \left(xt + \frac{t^3}{3} \right) dt$$ solutions of $$\frac{d^2f}{dz^2} - zf\left(z\right) = \pm\frac{1}{\pi}$$

steph
  • 117

1 Answers1

1

It suffices to differentiate, e.g.

$$\begin{gathered}\left(\mathrm{Ai}(x)\mathrm{Hi}'(x)-\mathrm{Ai}'(x)\mathrm{Hi}(x)\right)'= \mathrm{Ai}(x)\mathrm{Hi}''(x)-\mathrm{Ai}''(x)\mathrm{Hi}(x)=\\= \mathrm{Ai}(x)\left(x\mathrm{Hi}(x)+\tfrac1\pi\right)-x\mathrm{Ai}(x)\mathrm{Hi}(x)=\tfrac1\pi\mathrm{Ai}(x)\end{gathered}.$$ We see that the derivatives of the left and right side of the first identity coincide, which implies that the corresponding functions can only differ by a constant. To show that this constant is zero, it suffices to check that the right side vanishes as $x\to\infty$.

P.S. The proof of the second identity is exactly the same. Note that you have copied incorrectly its right side from DLMF.

Start wearing purple
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