## 100 Reputation

11 years, 186 days

## Evaluated in Mathematica...

@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)]

## Thanks...

Hi Thomas,

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

## Attached is the file for your easy refer...

HEre is the file:

Heaviside_Int_undefi.mw

## Solved...

@one_man Thanks alot..

## @Preben Alsholm    Hi Alsholm,...

Hi Alsholm,  Thanks for your tip..

## Thanks I realized your point and able to...

Hi Carlove,

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.

## Lets consider this integral...

Carl love,

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,   ...

Hi Carl,

It is not working without quotes.

## Thanks. Sorted out as you told. I am usi...

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

## Thanks. Sorted out as you told. I am usi...

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

## Hi Carl, Thanks for the reply.   No...

Hi Carl,

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,