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 am trying to evaluate the following integral:

$$\int_{a}^{\infty} \tfrac{(t-a)^2}{C^2} \cdot \exp ( - \tfrac{(t-a)^2}{C^2} ) \cdot \exp(iwt) \ dt$$

I thought about using the substitution $\beta \doteq \tfrac{(t-a)}{C}$ so the integral would become:

$$\int_{0}^{\infty} \beta^2 \cdot \exp (- \beta^2 ) \cdot \exp(i w (\beta C - a)) \ C \ d\beta$$

However, I don't know how to evaluate this either.

Any idea on how to do this?

EDIT: As Caran-d'Ache pointed out, I did in fact lose a minus! I'm still just learning Mathematica, how did you enter the integral?

share|improve this question
    
Are you sure you have got the right integrand? –  Mhenni Benghorbal May 19 '13 at 13:25
    
Unfortunately, yes, I'm sure! –  fabian789 May 19 '13 at 13:28
    
@fabian789 Mathematica says that the integral does not converge. So I guess you lost minus sign in the first exponent. –  Caran-d'Ache May 19 '13 at 16:46
    
@Caran-d'Ache I actually did lose one. Can you tell me the Mathematica command you used to evaluate the integral? –  fabian789 May 20 '13 at 9:33
add comment

3 Answers

up vote 1 down vote accepted

So one has the following integral:$$\mathrm{I}=\int_{0}^{\infty} \beta^2e^{- \beta^2} e^{i w (\beta C - a)} \ C \ d\beta$$ $$\mathrm{I}=Ce^{-i w a}\int_{0}^{\infty} \beta^2 e^{- \beta^2} e^{i w \beta C } d\beta$$ Let's for simplicity set $Ce^{-i w a}=\alpha$, $w C=s$, so: $$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- \beta^2} e^{i \beta s } d\beta$$ Joining the exponent terms and completing the square in the power of the resulting exponent: $$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- (\beta-\frac{i s}{2})^2} e^{-\frac{s^2}{4}} d\beta$$ Again $\tilde{\alpha}=\alpha \ e^{-\frac{s^2}{4}}$ and changing the variable (and the limits of integration) from $\beta$ to $t=\beta-\frac{i s}{2}$ one can obtain: $$\mathrm{I}=\tilde{\alpha}\int_{-\frac{i s}{2}}^{\infty-\frac{i s}{2}}\left(t+\frac{i s}{2}\right)^2 e^{- t^2} dt$$ Expanding the terms in the brackets: $$\mathrm{I}=\tilde{\alpha}\int_{-\frac{i s}{2}}^{\infty-\frac{i s}{2}}\left(t^2+i t s-\frac{s^2}{4}\right) e^{- t^2}=\frac{\tilde{\alpha}}{8} i \left(2 s-\sqrt{\pi } e^{-\frac{s^2}{4}} \left(s^2-2\right) \left(\text{erfi}\left(\frac{s}{2}\right)-i\right)\right)$$ where $ \text{erfi}$ - is the imaginary error function. After that one should assume all the substitutions what were made.

share|improve this answer
    
Thanks a bunch, this seems to be perfect! –  fabian789 May 20 '13 at 18:07
add comment

Or there is another way. One can look at $$\mathrm{I}=\alpha\int_{0}^{\infty} \beta^2 e^{- \beta^2} e^{i \beta s } d\beta$$ as the Laplace transform $\mathcal{L}_{\beta }\left[\beta ^2 e^{-\beta ^2 } \right](p)$ with $p=-is$. One can find (from the textbooks or using Mathematica) the answer. And after some simplification get the same result: $$\mathrm{I}=\frac{\tilde{\alpha}}{8} i \left(2 s-\sqrt{\pi } e^{-\frac{s^2}{4}} \left(s^2-2\right) \left(\text{erfi}\left(\frac{s}{2}\right)-i\right)\right)$$

share|improve this answer
    
Yes, the exercise I'm trying to solve actually asks me to Laplance transform this. I did however want to manually evaluate the integral. –  fabian789 May 20 '13 at 18:35
add comment

I think you may be able to integrate by parts twice if we have $e^{-(...)}$ instead of $e^{(...)}$, the integral in this current case does not converge

share|improve this answer
1  
I think this would be better as a comment. If you had tried it and it worked, then it would certainly be a hint type answer. Just some friendly advice. Regards –  Amzoti May 19 '13 at 16:45
    
thanks for advice. –  RETAS May 19 '13 at 16:59
add comment

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.