Question

Evaluate the integral in Equation 11.91, to confirm the expression on the right.

Short answer

Therefore, the integral equation was evaluated.

$f\left(\theta \right)=-\frac{2m\beta }{{\sout{h}}^2\left({\mu }^2+{\mathbb{k}}^2\right)}$

Step-by-step solution

Given.

The amplitude of the scattering in Born approximation is given by equation11.91 as:

$$\begin{gathered} f\left(\theta \right)\cong -\frac{2m\beta }{{\sout{h}}^2\mathbb{k}}{\int }_0^{\infty }{e}^{-\mu r}\sin \left(\mathbb{k}r\right)dr \\ f\left(\theta \right)=-\frac{2m\beta }{{\sout{h}}^2\left({\mu }^2+{\mathbb{k}}^2\right)} \end{gathered}$$

To evaluate the integral equation.

We need to evaluate the integral to get the RHS of the equation, we need to convert the sine function into the complex form, that is:

$\sin \left(\mathbb{k}r\right)=\frac{1}{2i}\left(e^{i\mathbb{k}r}-e^{-i\mathbb{k}r}\right)$

So the integral becomes,

$$\begin{gathered} {\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{\mu r}}\sin \left(\mathbb{k}\mathrm{r}\right)\mathrm{dr}=\frac{1}{2\mathrm{i}}{\int }_0^{\infty }\left[{\mathrm{e}}^{-(\mathrm{\mu }-\mathrm{i\mu })\mathrm{r}}-{\mathrm{e}}^{-(\mathrm{\mu }+\mathrm{i\mu })\mathrm{r}}\right]\mathrm{dr} \\ =\frac{1}{2\mathrm{i}}{\overline{)\left[\frac{{\mathrm{e}}^{-(\mathrm{\mu }-\mathrm{i\mu })\mathrm{r}}}{-(\mathrm{\mu }-\mathrm{i}\mathbb{k})}-\frac{{\mathrm{e}}^{-(\mathrm{\mu }+\mathrm{i\mu })\mathrm{r}}}{-(\mathrm{\mu }+\mathrm{i}\mathbb{k})}\right]}}_0^{\infty } \\ =\frac{1}{2\mathrm{i}}\left[\frac{1}{\mathrm{\mu }-\mathrm{i}\mathbb{k}}-\frac{1}{\mathrm{\mu }+\mathrm{i}\mathbb{k}}\right] \\ =\frac{\mathbb{k}}{{\mathrm{\mu }}^2+{\mathbb{k}}^2} \end{gathered}$$

Thus,

$$\begin{gathered} f\left(\theta \right)=-\frac{2m\beta }{{\sout{h}}^2\mathbb{k}}\frac{\mathbb{k}}{{\mu }^2+{\mathbb{k}}^2} \\ f\left(\theta \right)=-\frac{2m\beta }{{\sout{h}}^2({\mu }^2+{\mathbb{k}}^2)} \end{gathered}$$

Therefore, the integral equation was evaluated.