Asymptotic expansion of exponential integral $\int_ 0^1 \frac {e^{-t} dt} {1+x^2t^3}$

Question

How to find the full asymptotic behavior of $$\int_ 0^1 \frac {e^{-t} dt} {1+x^2t^3}$$ as $x$ tends to $0^+$ ?

Answer 1

Note that $$\frac{1}{1+x^2t^3}=\sum_{n=0}^\infty (-1)^nx^{2n}t^{3n}$$

Hence, we have

$$\begin{align} \int_0^1 \frac{e^{-t}}{1+x^2t^3}\,dt&=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1t^{3n}e^{-t}\,dt\\\\ &=\sum_{n=0}^\infty (-1)^nI_nx^{2n} \end{align}$$

where $I_n=\int_0^1 t^{3n}e^{-t}\,dt$ and satisfies the recurrence relationship

$$I_n=-e^{-1}\left(1+3n+(3n)(3n-1)\right)+(3n)(3n-1)(3n-2)I_{n-1}$$

The first few terms of the expansion are

$$\int_0^1\frac{e^{-t}}{1+x^2t^3}\,dt=\left(\frac{e-1}{e}\right)-2\left(\frac{3e-8}{e}\right)x^2+\left(\frac{720e-1957}{e}\right)x^4+O(x^6)$$