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Assuming[{Element[S, Reals],S>0},Integrate[Exp[-I*S*w]/(w^2 + 1)^(3/2)
   ,{w, 0, Infinity}]]

gets

 1/2 S(I π BesselI[1, S] + 2 BesselK[1, S] - I π StruveL[-1, S])

However Mathematica cannot do the indefinte integral with a variable lower or upper limit so differentiation cannot be used to see how it did this integral. Is there any way to find out or person who can figure it out, please. Change of variable to Sw would give leading S as a divisor not a factor. Not clear where the 3 parts come from.

Thanks JM 

          LaplaceTransform[1/(w^2 + 1)^(3/2), w, s] 
         -(1/2) π s (BesselY[1, s] + StruveH[-1, s])

is down to 2 terms. But still unclear where they come from

Greg Hurst
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simon
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2 Answers2

5

I'm not sure how Integrate found the answer, but here's a way it could have conceivably used.

If we can express your integrand as the product of two MeijerG functions, there's a rich list of integral identities we can use. In particular we can treat your integral as a Mellin transform at s == 1 and use the general identity here:

enter image description here

In our case:

mei = MeijerGReduce[Exp[-I*S*w]/(w^2 + 1)^(3/2), w]

enter image description here

int = MellinTransform[mei, w, 1]

MeijerG[{{1/2}, {}}, {{0, 1/2, 1}, {}}, -S^2/4]/π

FullSimplify[FunctionExpand[int], S > 0]

1/2 I S (-2 + π BesselI[1, S] - 2 I BesselK[1, S] - π StruveL[1, S])

We can see int (defined above) appears in the TracePrint of the Integrate call and so maybe this is the method it chose:

TracePrint[
  Integrate[Exp[-I*S*w]/(w^2 + 1)^(3/2), {w, 0, Infinity}],
  _MeijerG,
  TraceInternal -> True
]

MeijerG[{{1/2}, {}}, {{0, 1/2, 1}, {}}, 16/S^4, -2]
...
...
long spew omitted
...
Greg Hurst
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    Indeed, you can recast the integral as a Mellin convolution: Assuming[S > 0, MellinConvolve[Exp[-I S w]/w^2, (1 + w^2)^(-3/2), w, 1]] – J. M.'s missing motivation Mar 27 '19 at 16:40
  • Neither MeijerGReduce or MellinConvolve copied and pasted seem to work in v7, but maybe has them internally to do the integral . i tried TracePrint on the original integral fortunately in a new notebook but it has spewed for 10 minutes at least. Had to abort with TaskManager! – simon Mar 28 '19 at 02:51
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    @simon, the Mellin stuff are new in version 11, so you can't use them in version 7 (and you made no mention that you were using an old version). – J. M.'s missing motivation Mar 28 '19 at 04:56
  • Chip could you perhaps please put the meijerG result in non-png form so I can copy and paste. and see what the two parts are in old v7? I suspect a difference as the Imaginary part is a small differnce of two exponential increases. What Is the symbol similiar to theta abd where defined? – simon Mar 29 '19 at 14:28
  • The Fourier Transform gives real Part SK1[S} which just declines exponentially. Using Sign[w] in the FT gives the imaginary part which is the small difference of two exponential increasing functions which has a computing limit but doesnot FullSimplify to analytically subtract. Non-ideal feature must originate in the MeijerG decomposition but sorry about v7 can't regenerate that or paste in Chip Hurst's answer as .png – simon Mar 29 '19 at 14:31
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    The output is (2*Inactive[MeijerG][{{-1/2}, {}}, {{0}, {}}, w^2]*Inactive[MeijerG][{{}, {}}, {{0, 1/2}, {}}, I*S*w/2, 1/2])/Pi. The reason I posted it as an image is the presence of Inactive, which doesn't exist in V7. You can just pattern match it away, but be aware that these particular MeijerG expressions auto evaluate. – Greg Hurst Mar 29 '19 at 14:43
  • Thanks Chip, Ok all my V7 can do is show the Meijer 'factoring' was of the complex Exponential and the inverse 3/2 power. So it is no allowing me to get at my (revised) question which is now where do the two near cancelling imaginary terms1/2 S(I π BesselI[1, S] - I π StruveL[-1, S]) originate? with a view to analytically subtract them into one net function. I have been trying reverse Fourier sin transforms of these two but it can't do any such inverses – simon Mar 30 '19 at 13:41
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    That terms has a series expansion 2 π I Sum[(-S/2)^n/(n Gamma[n/2]^2), {n, 1, ∞}]. Maybe it's possible to get a more concise expression from that. – Greg Hurst Mar 30 '19 at 14:16
  • Thamks Chip I was thinking more of the asymptotic series showing the cancellation which I intermittently get, as this -I s - 3 I s^3 - 45 I s^5 ,where s=1/S.... How do I get the general infinite inverse series? – simon Mar 31 '19 at 00:15
0

Thanks to everyone for their input. The real and imaginary parts of this integral are in fact given in Abramovitz and Stegun, eqns. 9.6.25 and 12.2.3 resp. sections on $K_1$ and $L_1$, though that still doesn't explain how they were worked out originally.

$$I_{-\nu}(x)-\mathbf{L}_\nu(x)=\frac{2(x/2)^\nu}{\sqrt{\pi}\,\Gamma(\nu+\tfrac12)}\int_0^\infty\sin(tx)(1+t^2)^{\nu-\tfrac12}\mathrm dt\quad \Re\nu<\frac12,\,x>0$$

There is a as much info there on the imaginary part difference function BesselI[1, S] - StruveL[-1, S]) as StruveL individually including the asymptotic series so their small difference has been noted before. But the Taylor series that Chip found is not there, so thanks.

Unfortunately Mathematica does not recoginise this difference function and does not use the differnce asymptotic series so its computation broke down at S=35 for me.

J. M.'s missing motivation
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