# Problem with an integral

I'm trying to integrate this:

$U_{1}(Z_{1} + Z_{2},t) = C \int_{0}^{\infty} (1-i\Phi) \exp[-(1+iV)g] \mathrm{d}g,$

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: $$I(t) = \left|\dfrac{C}{(1+iV)}\right|^{2} \left[\left[ 1 - \dfrac{\theta}{2} \tan^{-1} \left( \dfrac{2mV}{[(1+2m)^{2} + V^{2}](t_{c} / 2t) + 1 + 2m + V^{2}} \right) \right]^{2} + \right.$$ $$+ \left.\left[ \dfrac{\theta}{4} \ln \left( \dfrac{[1+2m/(1+2t/t_{c})]^{2} + V^{2}}{(1+2m)^{2} + V^{2}} \right) \right]^{2} \right]$$

$$I(t) = \left|\dfrac{C}{(1+iV)}\right|^{2} \left[ 1 -\dfrac{i \theta}{2} \left[ \text{ln} ( 1+2t/t_{c} ) + \text{Ei}\left(\dfrac{-2mg}{1+2t/t_{c}} \right) + \text{Ei}(-2mg) \right] \right]^{2}.$$

Where did I go wrong?

-
Have you tried taking the absolute value of the answer you got? –  Please Delete Account Mar 21 '11 at 14:56
@Approximist Yes, but I don't know any trigonometric relation that give me $\arctan$ by manipulation of $\text{Ei}$. Is there any? –  Rodrigo Thomas Mar 21 '11 at 16:07
it looks rather suspicious that you have $g$ in your final result. Did you integrate over $g$? –  Fabian Mar 24 '11 at 9:03

Your result cannot be correct, as you have $g$ in the expression of $I(t)$ which you should have integrated over in the process of getting $U_1$.
To obtain the correct result, you split up the integral for $U_1$ into two parts $U_1= a + b$ with $$a= C \int_{0}^{\infty} dg e^{-(1+iV)g} = \frac{C}{1+i V} ,$$ $$b= -\frac{i \theta C}{t_c} \int_{0}^{\infty}dg \int_{0}^{t}dt' \frac{1}{1+2t'/t_{c}} \left( 1 - e^{- 2mg/(1+2t'/t_{c})} \right) e^{-(1+iV)g}.$$ Performing first the integral over $g$ in the expression for $b$ and then integrating over $t'$ you obtain the requested result easily...