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Can anyone suggest a closed form for either of these integrals?
and
where denotes the Bessel function of the first kind, is a positive integer, is any real number, and is constant.There are various formulae out there for integrals of this kind: for instance, equation 5.52 in Gradshteyn-Ryzhik tells us that
and Wolfram's website (for instance, here) lists similar formulae, mostly involving generalised hypergeometric functions.
(I have been able to reduce the problems I have been posting in other threads and MathSE down to these integrals.)
Last edited by zetafunc (2017-01-21 09:10:03)
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How can I represent
?In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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May I see your M code and pardon my slow replies, I am having computer problems.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Integrate[r^(-1) BesselJ[1, r], {r, t, Infinity}]
returns something involving Struve functions, which is not a closed form solution. However, those Struve functions can be bounded.
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What are you trying to do?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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The first integral seems to be close to 1 for |k| small and close to 0 for |k| large.
How did you determine that?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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What did you plot?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Wolfram Alpha gives a plot of the integral here: http://www.wolframalpha.com/input/?i=in … %5B1,+x%5D
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Mathematica thinks
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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No;
Integrate[BesselJ[1, x]/x, {x, 0, \[Infinity]}]
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Oh, I see. The definite integral returns:
ConditionalExpression[
1/2 (2 + BesselJ[1, t] (2 - \[Pi] t StruveH[0, t]) +
t BesselJ[0, t] (-2 + \[Pi] StruveH[1, t])),
Re[t] > 0 && Im[t] == 0]
Here t = |k| so both conditional expressions are automatically satisfied. Unfortunately I can't see a nice way of bounding those Struve functions without getting a power of |k| that is too large.
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How did you get that, mine just spits out a 1 for
See you later, I need to go offline.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Also
Integrate[BesselJ[d/2, x]/x, {x, t, \[Infinity]}]
returns
ConditionalExpression[
2/d - 2^(-1 - d/2) t^(d/2)
Gamma[d/
4] HypergeometricPFQRegularized[{d/4}, {1 + d/4, 1 + d/2}, -(t^2/
4)], Re[t] > 0 && Im[t] == 0]
Last edited by zetafunc (2017-01-21 22:05:21)
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Hi;
We can clean that up a bit if we make the assumption that t is real and greater than 0.
Assuming[t > 0, Integrate[BesselJ[d/2, x]/x, {x, t, \[Infinity]}]]
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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What kind of answer are you expecting?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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