# Short integration involving $e$ and $i$

Durring our lectures our professor calculated an integral like this:

$$\Psi = \Psi_0 \int\limits^{k+\delta k}_{k-\delta k} \! e^{ikx}\, \textrm{d}k = \Psi_0 \int\limits^{k+\delta k}_{k-\delta k} \! \frac{ix \, e^{ikx}}{ix}\, \textrm{d}k = \frac{\Psi_0}{ix} e^{ikx} \Big|^{k+\delta k}_{k-\delta k}$$

I dont understand how the last part is done. Could someone show me how it is possible that:

$$\Psi_0 \int\limits^{k+\delta k}_{k-\delta k} \! \frac{ix \, e^{ikx}}{ix}\, \textrm{d}k = \frac{\Psi_0}{ix} e^{ikx} \Big|^{k+\delta k}_{k-\delta k}$$

$\Psi_0$ is a constant.

-
this is trivial just differentiate e^(i k x) to get i k e^x, so dividing both by the constant gives that integral of 1/(ik) e^(ikx) is e^(ikx). – user58512 Feb 19 '13 at 20:40

$$\Psi_0 \int\limits^{k+\delta k}_{k-\delta k} \! \frac{ix \, e^{ikx}}{ix}\, \textrm{d}k = \frac{\Psi_0}{ix} \int\limits^{k+\delta k}_{k-\delta k} \! ix \, e^{ikx}\, \textrm{d}k =\frac{\Psi_0}{ix} e^{ikx} \Big|^{k+\delta k}_{k-\delta k}$$

Because $$\frac{\textrm{d}e^{ikx}}{\textrm{d}k} = ix e^{ikx}$$

What I don't really get is why you have $k$ in both the integral limits and you are integrating with respect to it, should this be another term?

-
This is some physics problem when we integrate the wave function $\Psi$ over a wave vector $k$. This means that with integral we get a superposition of multiple wave functions which is quite localized and enables us to determine position of a particle more acurately. This is a part of Quantum Mechanics. – 71GA Feb 19 '13 at 20:52
I wanted to say that it isn't good style to use the same variable both for the limits of the integral and for integrating, because when integrating the boundaries of the integral are fixed. If you have more complicated integrals, you could lose track of what you are doing. Just saying that it would be better style to use $t$ in the integral. – Stefan Feb 19 '13 at 20:55
Thank you for your advice! It will come usefull i am for sure. One more thing. Do we treat $x$ as a constant if we integrate over $k$? – 71GA Feb 19 '13 at 21:04
As $x$ doesn't depend on $k$, yes we treat it as a constant (so we can pull $\frac{1}{ix}$ to the outside of the integral). – Stefan Feb 19 '13 at 21:07
Thanks this is perfectly clear to me now. – 71GA Feb 19 '13 at 21:54

We have $$\frac{ixe^{ikx}}{ix} = e^{ikx}$$ and $i$ and $x$ should be considered as constants because you are integrating with respect to $k$. So you just use the rule for integration of an exponential: $$\int e^{ak}\mathrm{d}k = \frac{1}{a}e^{ak} + C$$

-