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

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?

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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

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.

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Thanks a bunch, this seems to be perfect! – fabian789 May 20 '13 at 18:07

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: $$\mathrm{I}=\frac{\tilde{\alpha}}{8} i \left(2 s-\sqrt{\pi } e^{-\frac{s^2}{4}} \left(s^2-2\right) \left(\text{erfi}\left(\frac{s}{2}\right)-i\right)\right)$$

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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

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

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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