You may find the spherical coordinatefunction written as

$\mathit{\delta }\left(r-r_0\right)\mathbf{=}\mathit{\delta }\left(r-r_0\right)\mathit{\delta }\left(cos\theta -cos{\theta }_0\right)\mathit{\delta }\left(\varphi -{\varphi }_0\right)\mathbf{/}{\mathit{r}}^{\mathbf{2}}$

Show that this equation is equivalent to (11.22).

The given expressions is$\delta \left(r-r_0\right)=\frac{\left.\Delta \left(r-r_0\right)(\cos \theta -\cos {\theta }_1\right)\delta (\varphi -\varphi )}{r^2}$

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

Consider the below function.

$\delta \left(f\right(x\left)\right)=\sum _i\frac{\delta \left(x-x_i\right)}{\left|f^{\text{'}}\left(x_i\right)\right|}$

Where, this formula can be used only if $f\left(x_i\right)=0$and $f^{\text{'}}\left(x_i\right)\ne 0$, to clarify this formula this simply states, that if the given delta function is variable in a function $f\left(x\right),$such that this function have i-th roots "ith number of values of x, which would make the function $f\left(x\right)=0$, then we have to find those roots "for example, the roots are $x_1,x_2,\dots ,x_i$", finding the roots of the function $f\left(x\right),$we now differentiate the function $f\left(x\right),$to find $f(x{)}^{\text{'}}$, and then the delta function would be given by the following series

$f(x{)}^{\text{'}}\delta \left(f\right(x\left)\right)=\frac{\delta \left(x-x_1\right)}{\left|f^{\text{'}}\left(x_1\right)\right|}+\frac{\delta \left(x-x_2\right)}{\left|f^{\text{'}}\left(x_2\right)\right|}+\dots$

note: This divide the integral into ith-integrals, where not all of them would be non-zero, only those who have x_i which lies inside the limits integral would be evaluated.

Assume, $x_i={\theta }_0$

$$\begin{gathered} f\left(\theta \right)=\cos \theta -\cos {\theta }_0 \\ f\left(x_i\right)=0 \\ {f}^{\text{'}}\left(\theta \right)=-\sin \theta \\ {f}^{\text{'}}\left({\theta }_0\right)\ne 0 \end{gathered}$$

So, for a function of $\theta ,{\theta }_0$we can write,

$\delta \left(\cos \theta -\cos {\theta }_0\right)=\frac{\beta \left({\theta }_0\right)}{\sin {\theta }_{\text{k}}}$

Thus,

$\delta \left(\cos \theta -\cos {\theta }_0\right)=\frac{\beta \left({\theta }_0\right)}{\sin {\theta }_{\text{k}}}$