How to get details about how Mathematica did a definite Integral

Question

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

Answer 1

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:

In our case:

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

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
...

Answer 2

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.