Can somebody show me how to calculate this integral?

$$ \int\limits_{-\infty}^\infty \exp \left[-\frac{(x-x_o)^2}{2 \sigma_x^2}-i (p - p_0) \frac{x}{\hbar}\right] \, dx $$

$x_0$, $p_0$, $\hbar$ are constants and $\sigma_x$ is a standard deviation of the Gaussian which we are integrating here. Somebody told me that i should complete the square.

EDIT: Thank you @Michael Hardy for a superb explaination. I did continue your calculation and got this:

$$ \begin{split} &\phantom{=}\int\limits_{-\infty}^\infty e^{-w^2} \cdot \underbrace{\exp \, -\left\{2x_o\dfrac{\sigma_x^2 i (p-p_0)}{\hbar} - \left( \frac{\sigma_x^2 i (p-p_0)}{\hbar} \right)^2 \right\}}_{constant}\, \mathrm{d} w =\\ &= \sqrt{\pi} \exp \, \left\{- 2x_o\dfrac{\sigma_x^2 i (p-p_0)}{\hbar} + \left( \frac{\sigma_x^2 \cdot i (p-p_0)}{\hbar} \right)^2 \right\} \end{split} $$

To get this result i used Gaussian integral. What i expected to get was the result in the below picture but my result is somewhat different. Why would that be? Was my integration wrong? In the picture there are some constants before the integral which do not play any signifficant role here.

Could someone explain, how author of the integral in the picture gets the result he does?

enter image description here


Complete the square, as follows:

\begin{align} \frac{(x-x_o)^2}{2 \sigma_x^2}+i (p - p_0) \frac{x}{\hbar} & = \frac{\hbar (x-x_o)^2 + 2\sigma_x^2 i (p-p_0)x}{2\sigma_x^2 \hbar} \\[8pt] & = \frac{(x^2 - 2x_ox + x_o^2) + \dfrac{2\sigma_x^2 i (p-p_0)}{\hbar}x}{2\sigma_x^2} \end{align}

Next, work on the numerator: \begin{align} & \phantom{{}=} (x^2 - 2x_ox + x_o^2) + \dfrac{2\sigma_x^2 i (p-p_0)}{\hbar}x \\[8pt] & = x^2 + \left(- 2x_o + \dfrac{2\sigma_x^2 i (p-p_0)}{\hbar}\right)x + x_o \\[8pt] & = \left[x^2 + \left(- 2x_o + \dfrac{2\sigma_x^2 i (p-p_0)}{\hbar}\right)x + \left(-x_o + \dfrac{\sigma_x^2 i (p-p_0)}{\hbar}\right)^2\right] + x_o^2 - \left(-x_o + \frac{\sigma_x^2 i (p-p_0)}{\hbar}\right)^2 \\[8pt] & = \left[x + \left(-x_o + \dfrac{\sigma_x^2 i (p-p_0)}{\hbar}\right)\right]^2 + \underbrace{2x_o\dfrac{\sigma_x^2 i (p-p_0)}{\hbar} - \left( \frac{\sigma_x^2 i (p-p_0)}{\hbar} \right)^2}_\text{constant} \\ & = w^2 + \text{constant}. \end{align} The word "constant" in this context means not depending on $x$.

The variable $w$ is the expression in square brackets.

Finally you need this: $dw=dx$.

Look at the integral again after this substitution.

  • $\begingroup$ Please give me some time to test this. When i suceed i will accept the anwser. And i will report back if i get any further questions regarding this. $\endgroup$ – 71GA Mar 3 '13 at 16:04
  • $\begingroup$ Thank you for your anwser. I thought that your explaination will help me to get my desired result, but my result is somewhat different from the one i should get. I edited my question accordingly. $\endgroup$ – 71GA Mar 4 '13 at 14:40

Complete the squares and use the formula $\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt \frac{\pi}{\alpha}$ for $\alpha>0$. Here http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ExpIntegrals.htm you can find a nice derivation of this formula.


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