Convoluting inverse square root with Gaussian

Question

Bug introduced in 9.0 and persisting through 11.0.1 or later

I would like to convolute the inverse square root on the interval [0,inf] with a Gaussian function, like so:

f[x_] := 1/Sqrt[x]
g[x_, sigma_] := Exp[-x^2/(2 sigma^2)]
conv[x_, sigma_] := Integrate[f[y] g[x - y, sigma], {y, 0, \[Infinity]}, Assumptions -> {sigma > 0}]
conv[x,sigma]

The result from Mathematica (9) reads

Exp[-x^2/(4 sigma^2)] Sqrt[-x] BesselK[1/4, x^2/(4 sigma^2)]/Sqrt[2]

which is purely imaginary for x>0.

This is certainly not the answer to the question I had in mind, which should be a positive real-valued function for all real-valued x (correct?).

Please tell me what is going wrong and (if possible) what I need to change to get the result I am looking for.

P.S. Adding assumptions like x>0 also doesn't seem to help (it simply will replace Sqrt[-x] by I Sqrt[x]).

Actually, I asked this exact question last year! It's a bug, according to Daniel Lichtblau. See link above. — DumpsterDoofus, Aug 12 '14 at 16:03
@DumpsterDoofus It is indeed almost identical, except for the fact that here I am using the inverse square root. In the answer given to your question the assumption n>0 needs to be removed, then it seems to provide the correct result also here. — leopold.talirz, Aug 12 '14 at 16:45
It's a bug, which I will report. If you use x^e instead of 1/Sqrt[x], specify e>-1, and substitute e->-1/2 at the end, you might get a more reliable result (at least it worked for a simpler case that has the same problem). — Daniel Lichtblau, Aug 12 '14 at 19:31
@Daniel It seems to work even without any prior assumption on the exponent of x. If desired, I can create an answer with details (but it would be quite similar to the answer for DumpsterDoofus' question). — leopold.talirz, Aug 12 '14 at 22:54
Glad something worked. Up to you whether you think it's worthwhile to create an answer. — Daniel Lichtblau, Aug 12 '14 at 22:55
With Mathematica 8.0.0.0 I get ConditionalExpression[(Pi*Sqrt[x]*(BesselI[-1/4, x^2/(4*sigma^2)] + BesselI[1/4, x^2/(4*sigma^2)]))/(2*E^(x^2/(4*sigma^2))), x > 0] — celtschk, Aug 13 '14 at 12:15
The problem was introduced in version 9, hence the better result in version 8. Unfortunately the newer code is actually useful and it's going to take some work to understand how/why it goes astray in this case. — Daniel Lichtblau, Aug 13 '14 at 15:13
@DanielLichtblau interestingly, in 13.2.0 we have this result by following the OP or this after following the comments. Presumably, they should be equal(?) I am tagging you because I know you will see it :-) sorry.... — bmf, Jan 11 '23 at 15:06
@bmf What you show is a phenomenon I've seen often enough. Actually some of this is undergoing work for 13.3 so these might change (for the better, I hope). — Daniel Lichtblau, Jan 11 '23 at 17:13

Convoluting inverse square root with Gaussian

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