# About the Use of Generalized Forms of Derivatives in the Study of Electromagnetic Problems
## 2 Lamb-Bateman Equation, Operational Methods, and Fractional Calculus
: _In order to obtain a more complete picture of the formalism we have foreseen, we consider the solution of Equation (23) for the cases with \(\alpha=\frac{1}{3}\) and \(\alpha=\frac{2}{3}\). To this aim, we remind that [39, 40]_  
\[g_{\frac{1}{3}}(\eta)=\frac{1}{3\pi}\frac{K_{\frac{1}{3}}\left(\frac{2}{3 \sqrt{3\,\eta}}\right)}{\sqrt{\eta^{3}}},\quad\eta\geq 0,\qquad\qquad K_{ \alpha}(x)=\frac{1}{2}\left(\frac{x}{2}\right)^{\alpha}\int_{0}^{\infty}e^{-t- \frac{x^{2}}{4t}}\ t^{-p-1}dt \tag{26}\]  
_with \(K_{\alpha}(x)\) being the second kind modified Bessel function and_  
\[ \begin{split}& g_{\frac{3}{3}}(\eta)=\sqrt{\frac{3}{\pi}}\ \frac{e^{-\frac{\eta}{2}}}{\eta}\mathcal{W}^{\left(\frac{1}{2},\frac{1}{2} \right)}(y),\qquad\qquad\qquad y=\frac{4}{27\eta^{2}},\\ & W^{(\kappa,\mu)}(x)=e^{-\frac{x}{2}}x^{\mu+\frac{1}{2}}U\Big{(} \mu-\kappa+\frac{1}{2},2\mu+1;x\Big{)},\\ & U(a,c;x)=\frac{1}{\Gamma(a)}\int_{0}^{\infty}e^{-xt}t^{\mu-1}( 1+t)^{c-a-1}dt,\qquad\qquad Re(a)>0,\end{split}  \tag{27}\]  
_where \(W^{(\kappa,\mu)}(x)\) is the Whittaker function and \(U(a,c;x)\) the confluent hyper-geometric function of the second kind. The comparison between the solution of Equation (23) with \(\alpha=\frac{1}{3},\frac{2}{3}\) is eventually reported in Figure 3._  
Figure 2: Comparison between the solutions of Equations (17) and (23) for \(\
u=\frac{1}{2}\) and different values of \(t\). (**a**) \(t=0.5\); (**b**) \(t=1\). As a further example of application of the composite transform method, we consider the solution of the problem given below. **Example 2**. : _Let, \(\forall\
u,\beta\in\mathbb{R}:0<\
u<1\) and \(\beta>0\)_  
\[\hat{D}_{x}^{(\
u,\beta)}\ u(x)=f(x),\qquad\qquad\hat{D}_{x}^{(\
u,\beta)}=(1+ \partial_{x}^{\
u})^{\beta}, \tag{28}\]  
_where the unknown function \(u(x)\) is expressed in terms of \(f(x)\) as_  
\[ \begin{split} u(x):&=u_{(\
u,\beta)}(x)=\hat{D}_{x }^{(\
u,-\beta)}f(x)=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}e^{-s}s^{\beta-1} e^{-s\partial_{x}^{\
u}}ds\ f(x)=\\ &=\frac{1}{\Gamma(\beta)}\int_{0}^{\infty}e^{-s}s^{\beta-1}\bigg{(} \int_{0}^{\infty}g_{\
u}(\eta)f\Big{(}x-\eta\ s^{\
u-1}\Big{)}d\eta\bigg{)}ds. \end{split}  \tag{29}\]  
_In the following Figures 4-6 we provide the functions \(u(x)\) for different values of \(\beta\) and \(\
u\) and \(f(x)\) being Gaussian or super-Gaussian function._  
**Observation 2**. : _Before concluding this introductory section, we note that for \(0<\beta<1\), we can obtain the solution of the Example 2 in alternative way. We consider the fractional differential equation_  
\[\hat{D}_{x}^{(\
u,-\beta)}\ u(x)=f(x) \tag{30}\]  
_and recast it according to the following_  
\[ \begin{split}& u_{(\
u,-\beta)}(x)=\hat{D}_{x}^{(\
u,\beta)}\ f(x)=\hat{D}_{x}^{(\
u,-(1-\beta))}\hat{D}_{x}^{( \
u,1)}f(x)=\frac{1}{\Gamma(1-\beta)}\int_{0}^{\infty}e^{-s}s^{-\beta}e^{-s \partial_{x}^{\
u}}\kappa(x)ds,\\ &\kappa(x)=(1+\partial_{x}^{\
u})f(x)\end{split}  \tag{31}\]  
_where the action of the fractional exponential operator on the function \(\kappa(x)\) can be specified as_  
\[e^{-s\partial_{x}^{\
u}}\kappa(x)=\int_{0}^{\infty}g_{\
u}(\eta)\Big{[}f\Big{(} x-\eta\ s^{\
u^{-1}}\Big{)}+f^{(\
u)}\Big{(}x-\eta\ s^{\
u^{-1}}\Big{)}\Big{]}d\eta, \qquad\qquad u^{(\
u)}(x)=\partial_{x}^{\
u}u(x). \tag{32}\]  
_It is worth stressing that the use of Equation (12)_  
\[f^{(\
u)}\Big{(}x-\eta\ s^{\
u^{-1}}\Big{)}=\frac{1}{\Gamma(1-\
u)}\int_{0}^{ x-\eta\ s^{\
u^{-1}}}\Big{(}x-\xi-\eta\ s^{\
u^{-1}}\Big{)}^{-\
u}f^{\prime}(\xi)d\xi \tag{33}\]  
Figure 5: Equation (29) for different values of \(\beta\), \(\
u=\frac{1}{3}\) and \(f(x)\) Gaussian or super-Gaussian function. **(a)**\(f(x)=e^{-x^{2}}\); **(b)**\(f(x)=e^{-x^{4}}\). eventually yields_  
\[ \begin{split} e^{-s\partial_{x}^{v}}(\partial_{x}^{v}f(x))& =\frac{1}{\Gamma(1-v)}\int_{0}^{\infty}g_{v}(\eta)\int_{0}^{x-\eta\;s^{v-1}} \Big{(}x-\xi-\eta\;s^{v-1}\Big{)}^{-v}f^{\prime}(\xi)d\xi d\eta=\\ &=\int_{0}^{\infty}g_{v}(\eta)f^{(v)}\Big{(}x-\eta\;s^{v-1}\Big{)} d\eta.\end{split}