$$I(x)=\int^{\infty}_{x}\exp(-t^{4}) dt \text{ as }x \rightarrow \infty$$
I have in my notes that:
\begin{align}I(x)&=\dfrac{-1}{4}\int^{\infty}_{x}\dfrac{1}{t^3}\dfrac{d}{dt}(e^{-t^{4}}) dt\\\\ &=\dfrac{-1}{4} \left[\dfrac{e^{-t^4}}{t^3}\right]^{\infty}_{x}-\dfrac{3}{4}\int^{\infty}_{x}\dfrac{1}{t^4}e^{-t^4}dt\\\\ & = \dfrac{e^{-x^4}}{4x^3}-\dfrac{3}{4}\int^{\infty}_{x}\dfrac{1}{t^4}e^{-t^4}dt\end{align}
I understand that to obtain the asymptotic approximation of the function we need to integrate by parts but could someone explain the above steps.
The point is that we know how to integrate $t^3 e^{-t^4}$ by just substituting $u=t^4$. So you force in a factor of $t^3$:
$$\int_x^\infty e^{-t^4} dt = \int_x^\infty \frac{1}{t^3} (t^3 e^{-t^4}) dt$$
and then integrate by parts from there, with $dv=t^3 e^{-t^4} dt$. In the end you get a new integral. You still can't actually calculate it, but it decays faster as $x \to \infty$ than your original integral (because it has a factor of $\frac{1}{t^4}$ in the integrand).