# About the Use of Generalized Forms of Derivatives in the Study of Electromagnetic Problems
## 2 Lamb-Bateman Equation, Operational Methods, and Fractional Calculus  
We give an idea of the underlying technicalities, also exploited in this article, by discussing the solution of the Lamb-Bateman equation [26; 27; 28]. **Example 1**. : _The Lamb-Bateman equation, originally proposed for the solution of a solitary wave, is written as_  
\[\int_{0}^{\infty}u(x-y^{2})dy=f(x) \tag{1}\]_where \(u(x)\) is a function to be determined in terms of \(f(x)\). The relevant solution is readily achieved by the use of operational methods (29), allowing one to recast it in the form_  
\[\hat{L}\ u(x)=f(x),\qquad\qquad\qquad\hat{L}=\int_{0}^{\infty}e^{-y^{2}\partial_ {x}}dy\, \tag{2}\]  
_which is just a consequence of the shift operator identity (29)_  
\[e^{a\partial_{x}}f(x)=f(x+a). \tag{3}\]  
_If we take the freedom of treating the derivative operator, in the definition of \(\hat{L}\), as an ordinary algebraic quantity, we can work out the Gaussian integral and end up with the explicit definition of the operator \(\hat{L}\) as_  
\[\hat{L}=\frac{1}{2}\sqrt{\frac{\pi}{\partial_{x}}}, \tag{4}\]  
_thus obtaining_  
\[\hat{L}^{-1}=\frac{2}{\sqrt{\pi}}\ \partial_{x}^{\frac{1}{2}} \tag{5}\]  
_and eventually_  
\[u(x)=\frac{2}{\sqrt{\pi}}\ \partial_{x}^{\frac{1}{2}}f(x). \tag{6}\]  
_We accordingly expressed the solution of the Lamb-Bateman equation in terms of the derivative of order \(\frac{1}{2}\) of the function \(f(x)\)._  
The next step is that of stipulating the action of a fractional derivative on a given function. **Exercise 1**. : _We recall first the Laplace transform identity_  
\[a^{-\
u}=\frac{1}{\Gamma(\
u)}\int_{0}^{\infty}e^{-as}s^{\
u-1}ds,\qquad Re(a ),Re(\
u)>0\, \tag{7}\]  
_which, by using Equation (3), allows writing of the following_  
\[\partial_{x}^{-\frac{1}{2}}g(x)=\frac{1}{\Gamma\Big{(}\frac{1}{2}\Big{)}}\int _{0}^{\infty}e^{-s\partial_{x}}s^{-\frac{1}{2}}ds\ g(x)=\frac{1}{\Gamma\Big{(} \frac{1}{2}\Big{)}}\int_{0}^{\infty}s^{-\frac{1}{2}}g(x-s)ds. \tag{8}\]  
_Changing the integration variable and arranging the limits of integration, we eventually find_  
\[a\partial_{x}^{-\frac{1}{2}}g(x)=\frac{1}{\sqrt{\pi}}\int_{a}^{x}(x-\xi)^{- \frac{1}{2}}g(\xi)d\xi. \tag{9}\]  
_The lowest integration limit, following from (8), is \(-\infty\); we put a generic \(a\) to be consistent with the current definition of the differintegral operator. If \(a=0\) the sub-index on the left of the derivative operator will be neglected. The use of the semigroup property of the fractional differintegral operator (1; 2; 3; 4; 5; 6)_  
\[\partial_{x}^{\frac{1}{2}}=\partial_{x}^{-\frac{1}{2}}\partial_{x} \tag{10}\]  
_finally leads to_  
\[\hat{L}\ u(x)=f(x),\qquad\qquad\qquad\hat{L}=\int_{0}^{\infty}e^{-y^{2}\partial_ {x}}dy\, \tag{2}\]0  
_The apex in front of \(g(x)\) denotes first-order derivative._In the previous Exercise, we adopted 0 as the lower limit. The relevant value should however be chosen in agreement with the specific problem under study. Further comments will be provided in the final comments. More in general, any derivative of order \(\alpha\) can be written as  
\[\hat{L}\ u(x)=f(x),\qquad\qquad\qquad\hat{L}=\int_{0}^{\infty}e^{-y^{2}\partial_ {x}}dy\, \tag{2}\]1  
which is the definition of the fractional derivative according to [PERSON] [30; 31]. Equation (12) states, with the ordinary derivative, that the fractional derivative of a constant is vanishing. This feature is not contained in other forms of fractional derivatives (e.g., the Riemann-Liouville and Euler derivatives [1; 2; 3; 4; 5; 6]). In the following, we limit ourselves to the derivative order with \(0<\alpha<1\) and just touch on the cases with \(\alpha>1\) later in the article. The Caputo derivative (as well as the others) is a non-local integral operator, not dissimilar from those entering the treatment of pseudo-differential operators in relativistic quantum mechanic [9; 10; 11; 13; 32; 33].