I'm trying to integrate this:
$ U_{1}(Z_{1} + Z_{2},t) = C \int\nolimits_{0}^{\infty} (1-i\Phi) \exp[-(1+iV)g] \mathrm{d}g, $
with:
$ \Phi = \dfrac{\theta}{t_{c}} \int_{0}^{t} \dfrac{1}{1+2t'/t_{c}} \left( 1 - \exp \left[- \dfrac{2mg}{1+2t'/t_{c}} \right] \right) \mathrm{d}t'. $
Then, after doing $ \left|U_{1}(Z_{1} + Z_{2},t)\right|^{2} $, the answer is: $ I(t) = \left|\dfrac{C}{(1+iV)}\right|^{2} \left[\left[ 1 - \dfrac{\theta}{2} \tan^{-1} \left( \dfrac{2mV}{[(1+2m)^{2} + V^{2}](t_{c} / 2t) + 1 + 2m + V^{2}} \right) \right]^{2} + \right.$ $ + \left.\left[ \dfrac{\theta}{4} \ln \left( \dfrac{[1+2m/(1+2t/t_{c})]^{2} + V^{2}}{(1+2m)^{2} + V^{2}} \right) \right]^{2} \right] $
but my answer is:
$ I(t) = \left|\dfrac{C}{(1+iV)}\right|^{2} \left[ 1 -\dfrac{i \theta}{2} \left[ \text{ln} ( 1+2t/t_{c} ) + \text{Ei}\left(\dfrac{-2mg}{1+2t/t_{c}} \right) + \text{Ei}(-2mg) \right] \right]^{2}. $
Where did I go wrong?