It has been a long time since I've needed to do integration... hope you can help
What is the result of the following where $\alpha$ is a constant;
$$\int_0^\infty \exp[-(\Delta^{-2/3}-\alpha)^2]\,\Delta^{-5/2}\,d\Delta$$
Does the following substitution help?
$$x=\Delta^{-2/3}-\alpha$$
which gives
$$-\frac{3}{2}\int^{\infty}_{-\alpha}(x+\alpha)^{5/4}\,e^{-x^2}\,dx$$
Any hints would be gratefully appreciated.
OK, here is Maple's result... $$ \int_0^\infty \frac{\exp\left(-(w^{-2/3}-a)^2\right)\;dw}{w^{5/2}} = U_{13}L^{(1/2)}_{3/8}(a^2)+ U_{33}L^{(3/2)}_{3/8}(a^2)+ U_{17}L^{(1/2)}_{7/8}(a^2)+ U_{37}L^{(3/2)}_{7/8}(a^2) $$
where $$ U_{13} = \frac{-3\pi^{3/2} a \sqrt{2}\sqrt{2-\sqrt{2}}(64a^4-40a^2-15)e^{-a^2}}{244\;\Gamma(7/8)} \\ U_{33} = \frac{3\pi^{3/2} a^3 \sqrt{2}\sqrt{2-\sqrt{2}}(2a-1)(2a+1)e^{-a^2}}{14\;\Gamma(7/8)} \\ U_{17} = \frac{-\sqrt{\pi}\;\Gamma(5/8)(128 a^6-256 a^4 -42 a^2-7)(1+\sqrt{2})e^{-a^2}}{88} \\ U_{37} = \frac{\sqrt{\pi}\;\Gamma(5/8)a^2(64 a^4-72 a^2-7)(1+\sqrt{2})e^{-a^2}}{44} $$ and $L_n^{(\alpha)}(x)$ is the generalized Laguerre function. I don't have its definition when $n$ and $\alpha$ are not integers, though. Presumably Maple does.
For example, when $a=1$, we get $2.936418$, both numerically from the integral, and from the Laguerre function version.