maplelearner

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These are replies submitted by maplelearner

@ecterrab

 

Hi,

Here are the evaluations in Mathematica:

N[MeijerG[{{0, 1/2}, {}}, {{0, 1}, {-1, -1}}, 10, 1/2]] = 0.320375

N[MeijerG[{{0, 1/2}, {}}, {{0, 1}, {-1, -1}}, 10^(1/2)]] = 1.13144

Hence MeijerG[{{0, 1/2}, {}}, {{0, 1}, {-1, -1}}, 10, 1/2] ≠ MeijerG[{{0, 1/2}, {}}, {{0, 1}, {-1, -1}}, 10^(1/2)]

@Thomas Richard 

 

Hi Thomas,

Thanks alot for your response.

 

I issued following commands in maple. However it is not evaluating the function as expected. Here are the outputs:

with(MmaTranslator):

Input1:    convert("MeijerG[{{0, 1/2}, {}}, {{0, 1}, {-1, -1}}, a, 1/2]", FromMma); 
Output:  MeijerG([[0, 1/2], []], [[0, 1], [-1, -1]], a, 1/2)

 

Attached is the file for your convenience.MeijerG.mw

 

Thanks

 

HEre is the file:

Heaviside_Int_undefi.mw

@one_man Thanks alot..

@Preben Alsholm 

 

Hi Alsholm,  Thanks for your tip..

Hi Carlove,

Thanks for your reply!

You are correct.  What if I change the lower limits as 10, i.e. the HankelH1 is moving valid in the domain such that variable 10< y <= Infinity. Hence, I would like to find the integration.

 

 

 

@Carl Love 

 

Carl love,

Thanks for your attention.

Let me look at the following integral

 

 

Maple: 

evalf(int(r BesselJ(1,r)* BesselJ(0,r), r = 0..infinity)) 

Float(undefined)

Mathematica: 

NumberForm[ NIntegrate[BesselJ[0, x]*BesselJ[1, x]*x, {x, 0, Infinity},   AccuracyGoal -> 20], 15]

1.7262346598671985

The plot of the integrand shows that the integration is possible. However I am not able to get the integration get done though Mathematica evaluated the integral with out any warning.

 

 

@Carl Love 

 

Hi Carl,

 

It is not working without quotes.

 

 

Thanks. Sorted out as you told. I am using Maple 16

Thanks. Sorted out as you told. I am using Maple 16

Hi Carl,

Thanks for the reply.

 

Now I am able to think of it. Since the indefinite integral leads to one contact of integration. Which is going to take care of this additional contstant.

 

Thanks...

Hi Carl,

Thanks for the reply.

 

Now I am able to think of it. Since the indefinite integral leads to one contact of integration. Which is going to take care of this additional contstant.

 

Thanks...

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