Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.