Laurent expansion of Meijer's G function

Question

I am considering the following equation (a Generalized hyper-geometric equation): $$\left(D-\beta_1\right)\left(D-\beta_2\right)f(x)-x\left(D+1-\alpha_1\right)\left(D+1-\alpha_2\right)f(x)=0$$ where in this case I specifically take $\alpha_1=\frac{1}{4}$, $\alpha_2=\frac{5}{4}$, $\beta_1=\frac{3}{4}$, $\beta_2=-\frac{1}{4}$. In fact this equation has the following two solutions: $$f_1(x)=x^{3/4}\,{}_2F_1\left(\frac{3}{2},\frac{1}{2};2;x\right)$$ The second solution is quite unfamiliar for me, which is denoted by MeijerG[{{}, {1/4, 5/4}}, {{-(1/4), 3/4}, {}}, x] in Mathematica. It has the following integral representation: $$f_2(x)=\frac{1}{2\pi i}\int_L\frac{\Gamma\left(s-\frac{1}{4}\right)\Gamma\left(\frac{3}{4}+s\right)}{\Gamma\left(s+\frac{1}{4}\right)\Gamma\left(\frac{5}{4}+s\right)}x^{-s}ds$$ with $L$ a proper contour. It can be seen that $\Gamma\left(s-\frac{1}{4}\right)\Gamma\left(\frac{3}{4}+s\right)$ has a simpole pole at $s=\frac{1}{4}$, and has double poles at $s=-\frac{3}{4}$, $s=-\frac{7}{4}$, $s=-\frac{11}{4}$,...I sum up the residue to get a series expansion: $$f_2(x)=\frac{2}{\pi}x^{-1/4}+\frac{x^{3/4}}{\pi^2}\sum_{n=0}^{\infty}\frac{\Gamma\left(\frac{3}{2}+n\right)\Gamma\left(\frac{1}{2}+n\right)}{n!\Gamma\left(n+2\right)}\left(-\psi\left(-n-\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)-\ln x\right)x^{n}$$ where I use $\psi(z)=\frac{\Gamma^{\prime}(z)}{\Gamma(z)}$ The problem is, does this series expansion converge on the whole complex plane? Can I use this expansion to determine the monodromy around both $0$ and $\infty$?

Thanks to the hint from @Mariusz Iwaniuk, I know that by using the following command in Mathematica: MeijerG[{{}, {1/4, 5/4}}, {{-(1/4), 3/4}, {}}, x] // FunctionExpand, the output may indicate that $f_2(x)$ vanishes outside the unit disk $0<|x|<1$. However, another question comes out: since the equation is invariant under $x\to \frac{1}{x}$, I can construct the third solution as $f_3(x)=f_2(1/x)$, which is nonzero only near the $\infty$. Then we just have three independent solutions at hand. What's the problem with my reasoning?

Answer 1

(recasted)

Let's rewrite your ode $$ \eqalign{ & 0 = \left( {D - \beta _1 } \right)\left( {D - \beta _2 } \right)f(x) - x\left( {D + 1 - \alpha _1 } \right)\left( {D + 1 - \alpha _2 } \right)f(x) = \cr & = \left( {D^2 - \left( {\beta _1 + \beta _2 } \right)D + \beta _1 \beta _2 } \right)f(x) + x\left( {D^2 + \left( {2 - \alpha _1 - \alpha _2 } \right)D + \left( {\alpha _1 \alpha _2 - \alpha _1 - \alpha _2 + 1} \right)} \right)f(x) = \cr & = \left( {\left( {1 + x} \right)D^2 + \left( {\left( {2 - \alpha _1 - \alpha _2 } \right)x - \left( {\beta _1 + \beta _2 } \right)} \right)D + \left( {\alpha _1 \alpha _2 - \alpha _1 - \alpha _2 + 1} \right)x} \right)f(x) = \cr & = \left( {\left( {1 + x} \right)D^2 + \left( {ax + b} \right)D + cx} \right)f(x) = \cr & = \left( {\left( {1 + x} \right)D^2 + \left( {a\left( {1 + x} \right) + b - a} \right)D + c\left( {1 + x} \right) - c} \right)f(x) = \cr & 1 + x \to \xi \cr & = \left( {\xi D^2 + \left( {a\xi + b - a} \right)D + c\xi - c} \right)f(\xi - 1) = \cr & = \left( {\xi D^2 + \left( {a\xi + b - a} \right)D + c\xi - c} \right)g(\xi ) = \cr & = \left( {D^2 + {{a\xi + b - a} \over \xi }D + {{c\xi - c} \over \xi }} \right)g(\xi ) \cr & \xi = 1/t \cr & = \left( {t^4 D^2 + 2t^3 D - {{a\xi + b - a} \over \xi }t^2 D + {{c\xi - c} \over \xi }} \right)h(t) = \cr & = \left( {D^2 + \left( {{{\left( {2 - b + a} \right)t^3 - at^2 } \over {t^4 }}} \right)D + {{c - ct} \over {t^4 }}} \right)h(t) \cr} $$

so that we get it into the canonical linear form ($D^2+p_1 (x) D+ p_0 (x) = 0$) $$ \bbox[lightyellow] { \left( {D^2 + {{ax + b - a} \over x}D + {{cx - c} \over x}} \right)g(x) = 0 } \tag{1}$$

Now, such an ode is not actually a hypergeometric equation. It is a 2nd degree linear ode with:

• a regular singular point at $x=0$,
• an irregular singular point at $x= \infty$.

According to the general theory (2), in the complex field:

• all the (two) solutions of (1) are analytic at a regular singular point and cannot have singular points other than those mentioned above ,
• they can be developed in a Taylor series around an ordinary point, and the series will have a convergence radius at least equal to the distance to nearest singularity,

At a regular singular point (our $z=x=0$), according to the same theory, eq. (1) might have an analytic solution.
If not, one solution is then for sure of the type $$ \left( {z - z_0 } \right)^\mu A(z) $$ and another (linearly independent from the above) of the form $$ \eqalign{ & \left( {z - z_0 } \right)^\nu B(z),\quad or \cr & \left( {z - z_0 } \right)^\nu B(z) + \left( {z - z_0 } \right)^\mu A(z)\ln \left( {z - z_0 } \right) \cr} $$ with $A(z), B(z), C(z)$ analytic till the nearest singularity.

That premised, instead (or at least before) resorting to a CAS, refer for example to where you have the advantage of having a table which states the solution in terms of various combinations of conditions on the parameters.

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(2)"Advanced Mathematical Methods for Scientists and Engineers" - C. M. Bender, S.A. Orszag