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What is this MeijerG Function

Qiang Lu

Qiang Lu, Shorter University

Posted 26 days ago

After following integral, I got a MeijerG result, Integrate[(1/(2 r (1 + r^2))Sin[r])^2r, {r, 0, \[Infinity]}] Out=1/16 Sqrt[\[Pi]] MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1] I don't know what is MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, z], I tried Activate command, it gives same result. So I checked online and found MeijerG function in Mathematica should be: MeijerG[{{a1,...,an},{an+1,...,ap}},{{b1,...,bm},{bm+1,...,bq}},z,] So I believe bm+1 should be greater than bm and so does bq is greater than bm too. However, the MeijerG function I got is MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1], bm=2 and bm+1=0 and bq=1/2, which means both bm+1 and bq is less than bm. I am confused. Anyone can help me? I don't think this is a mistake of Mathematica.

POSTED BY: Qiang Lu

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Michael Rogers

Michael Rogers, Emory University

Posted 26 days ago

Numerically the result checks out: Integrate[(1/(2 r (1 + r^2)) Sin[r])^2 r, {r, 0, Infinity}] % // N NIntegrate[(1/(2 r (1 + r^2)) Sin[r])^2 r, {r, 0, Infinity}] (* 1/16 Sqrt[\[Pi]] MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1] 0.074985 0.074985 ) If you apply `FunctionExpand[]` you get a large expression in terms of other special functions: 1/16 Sqrt[\[Pi]] MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1] // FunctionExpand ( 1/16 Sqrt[\[Pi]] (2/Sqrt[\[Pi]] - 1/( 75 E^2 Sqrt[\[Pi]])(150 E^2 - 150 EulerGamma - 150 E^2 EulerGamma - 150 I E^2 \[Pi] + 225 E^2 Cosh[2] + 150 E^2 EulerGamma Cosh[2] + 150 CoshIntegral[2] - 300 E^2 (-(1/(8 E^2)) - ExpIntegralEi[-2]/2) - 300 E^2 (1/(2 E^2) + ExpIntegralEi[-2]) - 300 E^2 ((3 E^2)/8 + 1/2 (-I \[Pi] - ExpIntegralEi[2])) - 450 E^2 Log[2] + 75 E^2 Log[16] + 300 E^2 (Cosh[2] - Sinh[2]/2) - 450 E^2 Sinh[2] - 150 E^2 EulerGamma Sinh[2] + 150 SinhIntegral[2] + 150 E^2 SinhIntegral[2] - 64 E^2 ((75 Cosh[2])/16 - (75 Sinh[2])/8 + (225 SinhIntegral[2])/ 32))) *) BTW, `an+1` means the parameter $a_{n+1}$ not $a_n+1$. You must have been reading a source that didn't or couldn't do the proper typesetting.

Numerically the result checks out:

Integrate[(1/(2 r (1 + r^2)) Sin[r])^2 r, {r, 0, Infinity}]
% // N
NIntegrate[(1/(2 r (1 + r^2)) Sin[r])^2 r, {r, 0, Infinity}]

(*
1/16 Sqrt[\[Pi]] MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1]
0.074985
0.074985
*)

If you apply FunctionExpand[] you get a large expression in terms of other special functions:

1/16 Sqrt[\[Pi]]
   MeijerG[{{1, 1}, {}}, {{1, 2}, {0, 1/2}}, 1] // FunctionExpand
(*
1/16 Sqrt[\[Pi]] (2/Sqrt[\[Pi]] - 1/(
   75 E^2 Sqrt[\[Pi]])(150 E^2 - 150 EulerGamma - 
     150 E^2 EulerGamma - 150 I E^2 \[Pi] + 225 E^2 Cosh[2] + 
     150 E^2 EulerGamma Cosh[2] + 150 CoshIntegral[2] - 
     300 E^2 (-(1/(8 E^2)) - ExpIntegralEi[-2]/2) - 
     300 E^2 (1/(2 E^2) + ExpIntegralEi[-2]) - 
     300 E^2 ((3 E^2)/8 + 1/2 (-I \[Pi] - ExpIntegralEi[2])) - 
     450 E^2 Log[2] + 75 E^2 Log[16] + 
     300 E^2 (Cosh[2] - Sinh[2]/2) - 450 E^2 Sinh[2] - 
     150 E^2 EulerGamma Sinh[2] + 150 SinhIntegral[2] + 
     150 E^2 SinhIntegral[2] - 
     64 E^2 ((75 Cosh[2])/16 - (75 Sinh[2])/8 + (225 SinhIntegral[2])/
        32)))
*)

BTW, an+1 means the parameter $a_{n+1}$ not $a_n+1$. You must have been reading a source that didn't or couldn't do the proper typesetting.

POSTED BY: Michael Rogers