ca.analysis and odes - Approximation to divergent integral

First, cut off the tail towards infinity:

$$f(x) = int_{x}^1 frac{Phi(t)}{t^5} dt + int_1^{infty} frac{Phi(t)}{t^5} dt.$$

The second term is a constant, so you can compute it numerically once and for all.

Write
$$e^{i pi u^2/2} = 1 + frac{i pi}{2} u^2 +R(u)$$
and
$$int_{0}^t e^{i pi u^2/2} du = t + frac{i pi}{6} t^3 + int_{0}^t R(u) du.$$

So
$$frac{Phi(t)}{t^5} = left( t^{-4} + frac{i pi}{6} t^{-2} + t^{-5} int_{0}^t R(u) du right) left( 1 + frac{i pi}{2} t^2 + R(t) right)=$$
$$t^{-4} + frac{2 pi i}{3} t^{-2} + left( t^{-4} R(t) - frac{pi^2}{12} + int_{0}^t R(u) du right).$$

So
$$int_{x}^1 frac{Phi(t)}{t^5} dt = frac{1}{3}left( x^{-3} - 1 right) + frac{2 pi i}{3} left( x^{-1} -1 right) + int_{x}^1 left( t^{-4} R(t) - frac{pi^2}{12} + int_{0}^t R(u) du right) du.$$

The integrands in the last term are bounded functions, and they are being integrated over bounded domains, so there is no problem approximating them numerically.

If you want an asymptotic formula, instead of a numerical approximation, you should be able to keep taking more terms out to get a formula like
$$f(x) = frac{1}{3} x^{-3} + frac{2 pi i}{3} x^{-1} + C + a_1 x + a_2 x^2 + cdots + a_n x^n + O(x^{n+1}) quad mathrm{as} x to 0.$$
You probably won't be able to get the constant $C$ in closed form, because it involves all those convergent integrals. The other $a_i$ will be gettable in closed form, although they will get worse and worse as you compute more of them.