How is this integral computed??

78 Views Asked by At

I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes: 1)Consider the equation of a wave traveling in one dimension: $ u = ae^{-i(\omega t - kz)} $ where $\omega = 2 \pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${\partial z}/{\partial t} = {\omega}/{k}$ 2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = \sum a_n e^{-i(\omega_n t - k_n z)}$ 3)Consider the continuous case : $$ u(z) = \int_{-\infty}^\infty a(k)e^{-i(\omega t - kz)}dk $$ 4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $\Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = \int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{-i(\omega t - kz)}dk $$ 5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = \int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{i kz}dk $$

When the integration is carried out, the result is $$ u(z) = \Delta k e^{ik_0 z} \frac {\sin 1/2(\Delta kz)}{1/2(\Delta k z)}$$. My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.

1

There are 1 best solutions below

1
On BEST ANSWER

The antiderivative of $e^{ikz}$ with respect to $k$ is given by $\frac{e^{ikz}}{iz}$ for $z \neq 0$, so if we impose this condition on $z$, we have \begin{align}\int_{k_0 - \Delta k /2}^{k_0 + \Delta k /2} e^{i kz}dk &= \frac{1}{iz} \left[e^{i k z} \right]_{k_0 - \frac{\Delta k}{2}}^{k_0 + \frac{\Delta k}{2}} \\ &= \frac{1}{iz}e^{i k_0 z}\left(e^{i \frac{\Delta k}{2}} - e^{- i \frac{\Delta k}{2}} \right). \end{align}

Can you continue from here, perhaps using the fact that for any $\theta \in \mathbb{R}$, $e^{i \theta} = \cos{\theta} + i \sin{\theta}$?