Fourier transform in terms of special function?

Question

I have a Fourier integral

$$\int\limits_{-\infty}^{\infty}\mathrm{d}t\,\frac{1}{t^2}\exp\left({\mathrm{i}\frac{t^3}{3}+\mathrm{i}Yt+\frac{\mathrm{i}\lambda^2}{4t}}\right),$$

where $Y$ and $\lambda$ are arbitrary real parameters.

Is it possible to express this integral in terms of some special functions, say hypergeometric functions or confluent hypergeometric functions?

Answer 1

$$I(Y,\lambda)=\int\limits_{-\infty}^{\infty}\mathrm{d}t\,\frac{1}{t^2}\exp\left(\frac{\mathrm{i}t^3}{3}+\mathrm{i}Yt+\frac{\mathrm{i}\lambda^2}{4t}\right)$$ $$=2\int\limits_{0}^{\infty}\mathrm{d}u\,\cos\left(\frac{1}{3u^3}+\frac{Y}{u}+\frac{u\lambda^2}{4}\right)$$

I had initially hoped for a simple answer, but was mistaken; since I started an answer, let me just record the result for $Y=0$, according to Mathematica, in terms of a hypergeometric function:

$$I(0,\lambda)= \tfrac{1}{6635520}\pi\, \mathbf{\lambda^{10} }\, _0F_7\left[;\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{5}{3},\frac{11}{6},2;2^{-20}3^{-8}\lambda^{12}\right]$$ $$-\,\tfrac{1}{165888}\ 3^{5/6} \, \Gamma \left(-\tfrac{5}{3}\right)\mathbf{\lambda^8 }\, _0F_7\left[;\frac{5}{6},\frac{7}{6},\frac{4}{3},\frac{4}{3},\frac{3}{2},\frac{5}{3},\frac{11}{6};2^{-20}3^{-8}\lambda^{12}\right]$$ $$-\,\tfrac{1}{3456} \,3^{1/6} \,\Gamma \left(-\tfrac{4}{3}\right) \mathbf{\lambda^6} \, _0F_7\left[;\frac{2}{3},\frac{5}{6},\frac{7}{6},\frac{7}{6},\frac{4}{3},\frac{3}{2},\frac{5}{3};2^{-20}3^{-8}\lambda^{12}\right]$$ $$+\,\tfrac{1}{72}\, \pi\, \mathbf{ \lambda^4} \, _0F_7\left[;\frac{1}{2},\frac{2}{3},\frac{5}{6},1,\frac{7}{6},\frac{4}{3},\frac{3}{2};2^{-20}3^{-8}\lambda^{12}\right]$$ $$-\,\tfrac{5}{108}\ 3^{5/6} \, \Gamma \left(-\tfrac{5}{3}\right) \,\mathbf{\lambda^2}\, _0F_7\left[;\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6},\frac{5}{6},\frac{7}{6},\frac{4}{3};2^{-20}3^{-8}\lambda^{12}\right]$$ $$-\,\tfrac{4}{9} \,3^{1/6} \,\Gamma \left(-\tfrac{4}{3}\right) \, _0F_7\left[;\frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{2}{3},\frac{5}{6},\frac{7}{6};2^{-20}3^{-8}\lambda^{12}\right]$$

This does answer the OP in the affirmative, at least for $Y=0$, although it's unlikely to be of much use.