I'm working through a homework problem and came to the point where I needed to evaluate this integral.
$$\int_{-(z^2 - 2z)}^{z^2 - 2z} \dfrac{-1}{\alpha z} \dfrac{1}{\sqrt{ \dfrac{2}{m} \bigg (E - z^2 - 2z - \dfrac{l^2 \alpha^2}{2m(\ln(z))^2} \bigg )} } dz$$
I'll give some background on the problem quickly. I'm trying to find the equation of motion of a particle of mass $m$ in the 1D Morse potential, $$V(x) = D \big ( e^{-2\alpha x} - 2e^{-\alpha x} \big ).$$ In consulting with my professor, he gave me the hint of setting $z = e^{-\alpha x}$ and solving for $x$. This gives $x = -\frac{1}{\alpha}\ln(z)$. He then suggested I use an equation from Goldstein's Classical Mechanics, which was $$ t = \int_{r_0}^r \dfrac{dr}{\sqrt{\dfrac{2}{m}\big ( E - V - \dfrac{l^2}{2mr^2} \big )}}$$
I did as he suggesting using $r=x$. I chose my integration bounds to be the turning points of the potential, $E = \pm V(x) = \pm(z^2 -2z)$. Now I'm stuck with this seemingly horrendous integral that I don't know how to solve. Once I get this, it's a simple matter of inverting that equation from $t(x)$ to $x(t)$ to get my equation of motion.
Any help would be greatly appreciated.
Thank you!
There are some issues with your integral, like how $z$ is both present as the dummy variable and in the limit of integration. If I understand correctly, $z$ is real, so how can you integrate something with $\ln z$ over a negative domain? So instead, I'll address the modified integral from the comments. We need $$ \int_{-(z^2-2z)}^{(z^2-2z)}\frac{1}{\alpha t\sqrt{\frac2m(E-t^2+2t)}}dt. $$ Notice that the argument of the square root in the denominator can be expressed as $(\frac{2E}m+1)-(t-1)^2$, allowing an easy trigonometric substitution to complete the integral. Then, just integrate your resulting integral just like you typically would in an elementary calculus course. (I have a feeling that you got your signs wrong somewhere, since the limits of integration do not work out nicely at all, but since I'm not acquainted with the physics of the problem I can't say for certain. Either way, this is probably the best method.)