# I Wave packet integral question

#### knockout_artist

Hi,

what kind if integration used on equation 1 so it turned into equation 2? this does not look like integration by parts. and where (x-x0) appeared from instead of (k-k0) ?

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#### DrClaude

Mentor
No integration was done. For instance, for the imaginary term, you multiply by the right-hand side by $\exp[i k_0(x-x_0)] \exp[-i k_0(x-x_0)]$, and move the second exponential inside the integral sign.

#### andrewkirk

Homework Helper
Gold Member
No integration has been done yet. It's the old mathematical trick of multiplying and dividing by the same thing.
Outside the integral is a factor of
$$\exp\left[-\frac{(x-x_0)^2}2(\Delta k)^2\right]$$
and inside the integral is a factor of the reciprocal
$$\exp\left[\frac{(x-x_0)^2}2(\Delta k)^2\right]$$
These cancel each other out.

They can be freely moved through the integral sign because they do not involve the integration variable $k$.

EDIT: Darn, Dr Claude jinxed me. Now I'm not allowed to talk for the rest of the day.

#### dextercioby

Homework Helper
I bet you are reading from a Russian book translated into English decades ago.

#### knockout_artist

I bet you are reading from a Russian book translated into English decades ago.
:D I have many of those books too.
But this image is from David Bohm's Quantum Theory.

"Wave packet integral question"

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