Evaluate $\mathbf{D}$and $\mathbf{X}$(Equations and ). Check your answers against Equations $7.47$and $7.48$.

An integral is a number that is assigned to a function in order to explain displacement, area, volume, and other notions that occur when infinitesimal data is combined.

Integration is the term used to describe the process of locating integrals.

Calculate two integrals:

$\begin{array}{l}D=a⟨{\psi }_0\left(r_1\right)\left|\frac{1}{r_2}\right|{\psi }_0\left(r_1\right)⟩\\ X=a⟨{\psi }_0\left(r_1\right)\left|\frac{1}{r_1}\right|{\psi }_0\left(r_2\right)⟩\end{array}$

where${\psi }_0\left(r\right)=\frac{e^{-r/a}}{\sqrt{\pi a^3}}$, and$r_1=r\text{\hspace{0.17em}and\hspace{0.17em}}{r}_2=\sqrt{r^2+R^2-2rR\cos \theta }$

The first integral is,

$\begin{array}{c}D=\frac{1}{\pi a^2}\int e^{-r/a}\frac{1}{r_2}{e}^{-r/a}{r}^{2}dr\sin \theta d\theta d\phi \\ =\frac{2}{a^2}{\int }_0^{\infty }{r}^2{e}^{-2r/a}dr{\int }_0^{\pi }\frac{\sin \theta d\theta }{\sqrt{r^2+R^2-2rR\cos \theta }}\end{array}$

As,

$\begin{array}{l}u=r^2+R^2-2rR\cos \theta \\ du=2rR\sin \theta \end{array}$

The limits will be

$0\to {(r-R)}^2\pi \to {(r+R)}^2$

$\begin{array}{c}D=\frac{1}{R{a}^2}{\int }_0^{\infty }r{e}^{-2r/a}dr{\int }_{{(r-R)}^2}^{{(r+R)}^2}\frac{du}{\sqrt{u}}\\ =\frac{2}{R{a}^2}{\int }_0^{\infty }r{e}^{-2r/a}[r+R-|r-R|]dr\\ =\frac{2}{R{a}^2}[{\int }_0^{R}r{e}^{-2r/a}(r+R-R+r)dr+{\int }_R^{\infty }r{e}^{-2r/a}(r+R-r+R)dr]\\ =\frac{2}{R{a}^2}[2{\int }_0^R{r}^2{e}^{-2r/a}dr+2R{\int }_R^{\infty }r{e}^{-2r/a}dr]\end{array}$

Further, evaluate and get,

$\begin{array}{c}D=\frac{4}{R{a}^2}\{\frac{a^3}{8}[e^{-2r/a}(-\frac{4R^2}{a^2}-\frac{4R}{a}-2)+2]+R\frac{a^2}{4}{e}^{-2R/a}(\frac{2R}{a}+1)\}\\ =\frac{4}{R{a}^2}\cdot \frac{a^2}{4}\{\frac{a}{2}{e}^{-2R/a}(-\frac{4R^2}{a^2}-\frac{4R}{a}-2)+a+R{e}^{-2R/a}(\frac{2R}{a}+1)\}\\ =\frac{1}{R}\{-2R{e}^{-2R/a}-a{e}^{-2R/a}+a+R{e}^{-2R/a}\}\\ =\frac{a}{R}-\frac{a}{R}{e}^{-2R/a}-e^{-2R/a}\\ D=\frac{a}{R}-(1+\frac{a}{R})e^{-2R/a}\end{array}$

Now, the second integral,

$\begin{array}{c}X=\frac{1}{\pi a^2}\int e^{-r/a}\frac{1}{r}{e}^{-r_2/a}{r}^{2}dr\sin \theta d\theta d\phi \\ =\frac{2}{a^2}{\int }_0^{\infty }r{e}^{-r/a}dr{\int }_0^{\pi }\exp (-\frac{\sqrt{r^2+R^2-2rR\cos \theta }}{a})\sin \theta d\theta \end{array}$

As,

$\begin{array}{l}u=r^2+R^2-2rR\cos \theta \\ du=2rR\sin \theta \end{array}$

Thus, The limits will be $0\to {(r-R)}^2\pi \to {(r+R)}^2$

Use this in the integral and get,

$\begin{array}{c}X=\frac{1}{R{a}^2}{\int }_0^{\infty }{e}^{-r/a}dr{\int }_{{(r-R)}^2}^{{(r+R)}^2}{e}^{-\sqrt{u}/a}du\\ =\frac{1}{R{a}^2}{\int }_0^{\infty }{e}^{-r/a}2a^2[e^{-|r-R|/a}(\frac{|r-R|}{a}+1)-e^{-R/a}{e}^{-r/a}(\frac{r}{a}+\frac{R}{a}+1)]dr\end{array}$

Now separate the integral, which contains the absolute value, into two parts:

$\begin{array}{c}X=\frac{2}{R}\{{\int }_0^{\infty }{e}^{-r/a}{e}^{-|r-R|/a}(\frac{|r-R|}{a}+1)-e^{-R/a}{\int }_0^{\infty }{e}^{-2r/a}(\frac{r}{a}+\frac{R}{a}+1)dr\}\\ =\frac{2}{R}\{{\int }_0^R{e}^{-r/a}{e}^{-R/a}{e}^{r/a}(\frac{R}{a}-\frac{r}{a}+1)dr+{\int }_R^{\infty }{e}^{-r/a}{e}^{R/a}{e}^{-r/a}(-\frac{R}{a}+\frac{r}{a}+1)dr\\ -e^{-R/a}{\int }_0^{\infty }[\frac{r}{a}{e}^{-2r/a}+(\frac{R}{a}+1)e^{-2r/a}]dr\}\\ =\frac{2e^{-R/a}}{R}\{\frac{R^2}{a}-\frac{R^2}{2a}+R+e^{2R/a}{\int }_R^{\infty }{e}^{-2r/a}(-\frac{R}{a}+\frac{r}{a}+1)dr-\frac{1}{a}\frac{a^2}{4}-(\frac{R}{a}+1)\frac{a}{2}\}\\ =\frac{2e^{-R/a}}{R}\{\frac{R^2}{2a}+R+e^{2R/a}\left[\frac{1}{a}\frac{a}{4}{e}^{-2R/a}(a+2R)+(1-\frac{R}{a})\frac{a}{2}{e}^{-2R/a}\right]-\frac{a}{4}-\frac{a}{2}(\frac{R}{a}+1)\}\end{array}$

Further evaluate as:

$\begin{array}{c}=\frac{2e^{-R/a}}{R}\left\{\frac{R^2}{2a}+R+\frac{1}{4}(a+2R)+\frac{a}{2}(1-\frac{R}{a})-\frac{a}{4}-\frac{a}{2}(\frac{R}{a}+1)\right\}\\ =\frac{2e^{-R/a}}{R}(\frac{R^2}{2a}+\frac{R}{2})\\ X=e^{-R/a}(1+\frac{R}{a})\end{array}$

Thus, after calculating the two integrals, it is found that both yield the same outcome as equations $7.45$and $7.46$.