# Fractional differential equation of ln

When $$0<\alpha\leq1$$ , what is the "Riemann-Liouville" fractional derivative of :$$D^\alpha\left(\alpha\ln\left(\frac{c1}{\alpha}t+c2\right)\right)=?$$ The Riemann-liouville fractional derivative is defined as follows:$$D^\alpha f(t)=D^nJ^{n-\alpha}f(t)$$where $$n=[\alpha]+1$$ and$$J^{n-\alpha}f(t)=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-x)^{n-\alpha-1}f(x)dx,$$then$$D^{\alpha}f(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^n}{dt^n}\int_0^t(t-x)^{n-\alpha-1}f(x)dx.$$

• What have you tried? What are the definitions you are working with? Please provide more context. – Ramanujan May 29 '19 at 22:42

$$\begin{cases} \frac{d^{\alpha}}{dt^{\alpha}}f(t)=\frac{1}{\Gamma(-\alpha)}\int_0^t(t-x)^{-\alpha-1}f(x)dx \\ f(t)=\alpha\ln\left(\frac{c_1}{\alpha}t+c_2\right) \end{cases}$$ $$\frac{d^{\alpha}}{dt^{\alpha}}f(t)=\frac{1}{\Gamma(-\alpha)}\int_0^t(t-x)^{-\alpha-1}\alpha\ln\left(\frac{c_1}{\alpha}x+c_2\right)dx$$ $$\int(t-x)^{-\alpha-1}\ln\left(\frac{c_1}{\alpha}x+c_2\right)dx= \frac{ 1 }{(\alpha)^2}\left( (\alpha)(t-x)^{n-\alpha}\ln\left(\frac{c_1t}{a}+c_2 \right)+ \left(\frac{c_1}{c_2\alpha+c_1x} \right)^{\alpha}\,_2 F_1\left(\alpha\:,\:\alpha-n\:;\:\alpha-n+1\:;\:(\alpha)\ln\left(\frac{c_1x}{\alpha}+c_2 \right) \right) \right)$$
$$\boxed{ \frac{d^{\alpha}}{dt^{\alpha}}f(t)= -\frac{1}{\alpha\Gamma(-\alpha)} \left( \left(\frac{c_1}{c_2\alpha+c_1t} \right)^{\alpha}\,_2 F_1\left(\alpha\:,\:\alpha\:;\:\alpha+1\:;\:\alpha\ln\left(\frac{c_1t}{\alpha}+c_2 \right) \right) - \alpha t^{-\alpha}\ln\left(\frac{c_1t}{a}+c_2 \right)+ \left(\frac{c_1}{c_2\alpha} \right)^{\alpha}\,_2 F_1\left(\alpha\:,\:\alpha\:;\:\alpha+1\:;\:(\alpha)\ln(c_2) \right)\right)}$$ $$_2F_1$$ is the Gauss hypergeometric function.
• In order to clarify the result is given only for $0<\alpha<1$. – JJacquelin Jun 2 '19 at 9:47