Integral $\int_{0}^{\infty} u^{2}\left(e^{-4x(u^{-12}-u^{-6})}-1\right) \,\mathrm du$

Question

I need to find the integration of

$$I(x) = \int_{0}^{\infty} u^{2}\left(e^{-4x(u^{-12}-u^{-6})}-1\right) \,\mathrm du$$

I am given

$$e^{-4x(u^{-12}-u^{-6})}-1 \approx \begin{cases} -1 & \text{for}& u \leq 1 \\ -4x(u^{-12}-u^{-6}) & \text{for} & u > 1 \end{cases}$$

Thank you for your help.

Answer 1

Using your approximation: \begin{align*} I(x) &= \int_0^\infty u^2\left( \exp \left(-4x(u^{-12}-u^{-6})\right) - 1\right) \,\mathrm du \\ &\approx \int_0^1- u ^2 \,\mathrm du + \int_1^\infty -4xu^2(u^{-12}-u^{-6}) \,\mathrm du\\ &= -\frac 13 -4x\int_1^\infty u^{-10}-u^{-4}\,\mathrm du \\ &=-\frac 13 -4x\left( \bigg[ \frac{u^{-9}}{-9}\bigg]^\infty_1 - \bigg[ \frac{u^{-3}}{-3}\bigg]^\infty_1\right) \\ &= \frac 89x-\frac 13 \end{align*}