I'm reading through Clifford Truesdell's "An essay toward a unified theory of special functions", Princeton Univ. Press, 1948. All his exposition is based on the functional equation
$\frac{\partial}{\partial z}\mathrm F(z,\alpha)=\mathrm F(z,\alpha+1)$ He starts with
We are going to study functions $f (y, \alpha)$ satisfying a functional equation of the type
$\frac{\partial}{\partial y} f (y, \alpha) = \mathrm A(y, \alpha) f (y, \alpha) + \mathrm B(y, a) f (y, \alpha+1 )$
Then, we define
$g\left( {y,\alpha } \right) = f\left( {y,\alpha } \right)\exp \left\{ { - \int\limits_{{y_0}}^y {\mathrm A\left( {v,\alpha } \right)dv} } \right\}$
We verify that $g$ satisfies
$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = g\left( {y,\alpha + 1} \right)B\left( {y,\alpha } \right)\exp \left\{ { - \int\limits_{{y_0}}^y {\left[ {A\left( {v,\alpha + 1} \right) - A\left( {v,\alpha } \right)} \right]dv} } \right\}$
Thus we reduce the equation to
$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = C\left( {y,\alpha } \right)g\left( {y,\alpha + 1} \right)\tag {1}$
Now he states
In the case of nearly every special function that I know to satisfy an equation of type $(1)$, the coefficient $C(y, \alpha)$ is factorable, $C(y, \alpha)=Y(y)A( \alpha)$, so we asume
$\frac{\partial }{{\partial y}}g\left( {y,\alpha } \right) = Y(y)A( \alpha)g\left( {y,\alpha + 1} \right)$
Now he defines:
$z:= \int_{y_1}^y Y(v) dv$
and
$F(z,\alpha ): = g\left( {y,\alpha } \right)\exp \left\{ {\mathop {\mathrm S}\limits_{{\alpha _0}}^\alpha \log {\text{A}}\left( v \right)\Delta v} \right\}$
Now this is the operator that is troubling me
$\mathop {\mathrm S}\limits_{{\alpha _0}}^\alpha h\left( v \right)\Delta v = \mathop {\lim }\limits_{k \to {0^ + }} \left\{ {\int\limits_{{a_0}}^\infty {h\left( v \right){e^{ - kc\left( v \right)}}dv} - \sum\limits_{m = 0}^\infty {h\left( {a + m} \right){e^{ - kc\left( {a + m} \right)}}} } \right\}$
I can't find any reference to what $c(v)$ is. Is this known operator? What is $c$?
Anyways, I have a simple case I need to transform:
Let $\mathrm F\left( {x,\alpha } \right) = \int\limits_0^x {{{\left( {\frac{t}{{t + 1}}} \right)}^\alpha }} \frac{{dt}}{t}$
Then we have the functional equation
$\frac{\alpha }{x} \mathrm F\left( {x,\alpha } \right) - \frac{\alpha }{x} \mathrm F\left( {x,\alpha + 1} \right) = \frac{\partial }{{\partial x}} \mathrm F\left( {x,\alpha } \right)$
Following Truesdell's method, I define
$\mathrm G\left( {x,\alpha } \right) = \frac{{\mathrm F\left( {x,\alpha } \right)}}{{{x^\alpha }}}$
Then I have the functional equation
$\frac{\partial }{{\partial x}} \mathrm G\left( {x,\alpha } \right) = - \alpha \mathrm G\left( {x,\alpha + 1} \right)$
How can I transform it to the $\mathrm F$ equation using Truesdell's method?
The importance of the original $\mathrm F$ I define is that it can be used to show that
$\log (1+x)=\sum_{n=1}^\infty \frac{1}{n}\left(\frac x {x+1} \right)^n\text{ ; for } x > -\frac 1 2$
and maybe some other results can be derived. I still have a lot of exposition to read.