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I know that the official standard formula can be found here, but I am having a very hard time simplifying this special case:

MeijerG[{{}, {1, 1}}, {{0, 0, A}, {}}, x]

where $A$ is a positive integer and $x > 0$. What is the analytic form of this function?

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Wolfram|Alpha returns an expression involving the generalized hypergeometric function and the digamma function. –  joriki Dec 14 '11 at 21:53
1  
"What is the analytic form of this function?" - the Meijer function is already an analytic form. You meant to ask if there's a simpler form for your Meijer G expression... –  J. M. Dec 14 '11 at 23:57

3 Answers 3

up vote 5 down vote accepted

Let $$ f(a,x) = G_{2,3}^{3,0}\left(x\left| \begin{array}{c} 1,1 \\ 0,0,a \\ \end{array} \right.\right) $$ Then it is not hard to determine that, for positive $x$, $$ f(1,x) = - \mathrm{Ei}(-x) $$ Additionally, it follows from Mellin-Barnes representation of the Meijer G-function that $$ f(a+1, x) = a \cdot f(a,x) - x \partial_x f(a,x) = -x^{a+1} \partial_x \left( x^{-a} f(a, x) \right) $$ This means that $$ x^{-n} f(n, x) = (-1)^n \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}} \left( x^{-1} \mathrm{Ei}(-x) \right) $$ or

$$ f(n, x) = (-x)^n \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}} \left( x^{-1} \mathrm{Ei}(-x) \right) $$

Verification in Mathematica:

In[38]:= {MeijerG[{{}, {1, 1}}, {{0, 0, 7}, {}}, 
   x], (-x)^7 D[ExpIntegralEi[-x]/x, {x, 6}]} /. x -> 1`16

Out[38]= {1348.4143043955619, 1348.41430439556}
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1  
who said anything about a being an integer! (+1) –  Simon Dec 15 '11 at 2:20
    
If $a$ is not an integer, then we can interpret this as the differintegral of the exponential integral... :) –  J. M. Dec 15 '11 at 2:43
1  
The naive, analytic continuation approach to "fractional" calculus is the best! (I use it all the time in dimensional regularization) –  Simon Dec 15 '11 at 2:47
    
@Simon Quoting the OP, 'where A is a positive integer and x>0.` –  Sasha Dec 15 '11 at 6:00
    
@Sasha: So maybe reading isn't my strong suit... –  Simon Dec 15 '11 at 7:15

Using Maple, it appears that the function is $\frac{x^A}{A^2} {}_2F_2(A,A; A+1,A+1;-x)$ for integers $A \ge 2$.

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We've seen Mathematica and Maple, so for comparison, here's SymPy. Running the IPython console for SymPy version 0.7.1 gives the same answer as FunctionExpand[MeijerG[{{}, {1, 1}}, {{0, 0, A}, {}}, x]] in Mathematica or (Wolfram|Alpha)

In [1]: from sympy.abc import a

In [2]: hyperexpand(meijerg([],[1,1],[0,0,a],[], x), allow_hyper=True)
Out[2]: 
 a      2  ┌─  ⎛    a, a     │   ⎞                                     
x ⋅Γ(-a) ⋅ ├─  ⎜             │ -x⎟                                     
          2╵ 2 ⎝a + 1, a + 1 │   ⎠                                     
────────────────────────────────── - log(x)⋅Γ(a) + Γ(a)⋅polygamma(0, a)
                     2                                                 
            Γ(-a + 1)                                                  

In [3]: expand(simplify(_))
Out[3]: 
                                    a  ┌─  ⎛    a, a     │   ⎞
                                   x ⋅ ├─  ⎜             │ -x⎟
  log(x)⋅a!   a!⋅polygamma(0, a)      2╵ 2 ⎝a + 1, a + 1 │   ⎠
- ───────── + ────────────────── + ───────────────────────────
      a               a                          2            
                                                a            

Props to Tom Bachmann for implementing this code. If you want to understand how these reductions are done, it is worth reading his code, his blog and and the references to Kelly B. Roach's original algorithm.

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