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How do I solve this integral?

Posted 10 years ago

Hi kind people of the Wolfram Community,

I'm having trouble solving this integral: Image of the Intregral .

Firstly I do not know how to do this in one calculation, due to the difference in boundary conditions. Secondly there is a more important problem: If I perform the integration on the first part:

  Integrate[(y^(2 a + b - 1)/Sqrt[y^(2 (a + b)) - 1] - y^(a - 1)), {y,   1, Infinity}]

I get the output:

 ConditionalExpression[ 1/a - (Sqrt[\[Pi]]  Gamma[-(a/(2 (a + b)))])/((2 a + b) Gamma[-1 + b/(2 (a + b))]),   Re[2 a + b] > 0 && Re[a] < 0]

So I assume this diverges if not "Re[2 a + b] > 0 && Re[a] < 0". However if I insert a=b=1 then Re[a]>0 and still the integral coverges:

Integrate[(y^(2 + 1 - 1)/Sqrt[y^(2 (2)) - 1] - y^(1 - 1)), {y, 1,  Infinity}]


1 - EllipticE[-1] + (2 - I) EllipticK[-1] - EllipticK[2]

Or if I NIntegrate I get the value 0.40093.

How do I get an accurate idea of the regions for which this expression converges? And how do I find the solutions in those regions where it does.

I would be very grateful if someone could help me out.

Thanks, M.

2 Replies

The series expansion at y=1 indicates all is fine there provided a!=-b. I suspect it will take careful analysis of behavior at infinity in order to figure out a full set of conditions for which you get convergence. From some manipulations I did by hand, I think what's required is that re(a+2b)>0.

POSTED BY: Daniel Lichtblau

It seems that Integrate is not getting the full region of integration as the solution as you notice. The ConditionalExpression does not mean that the integral is necessarily divergent or undefined outside of the region that it specifies--just that the solution that is given is valid inside of that region.

But perhaps someone with integration expertise at WRI can comment on why the result doesn't cover all permissible values of a and b.

Of course you can turn off the generation of the ConditionalExpression by using,

 y^(2*a + b - 1)/Sqrt[y^(2*(a + b)) - 1] - y^(a - 1), {y, 1, Infinity}, GenerateConditions -> False]

but this doesn't solve your question.

POSTED BY: David Reiss
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