The expression in question is (we have replaced sigma by s)
g = Integrate[ Exp[-(x - y)^2/(2 y s^2)] f[y] 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] $\int_0^{\infty } \frac{e^{-\frac{(x-y)^2}{2 s^2 y}} f[y]}{\sqrt{2 \pi } \sqrt{s^2 y}} \, dy$
First of all we notice that for y>0
"Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0] = DiracDelta[x - y]" Formally
Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0, Assumptions -> {y > 0, x > 0, y != x}] (* Out[37]= 0 *) In[39]:= Limit[Exp[-(x - y)^2/(2 y s^2)] 1/Sqrt[2 \[Pi] y s^2], s -> 0, Assumptions -> {y > 0, x > 0, y == x}] (* Out[39]= \[Infinity] *) so that
Limit[g, s -> 0] = f[x] Assuming now that f[y] can be expanded into a power series about y = 0 and consider
g1 = Integrate[ Exp[-(x - y)^2/(2 y s^2)] y^k/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}, Assumptions -> {s > 0, x^2 > 0}] (* Out[16]= (E^(x/s^2) Sqrt[2/\[Pi]] Abs[x]^(1/2 + k) BesselK[1/2 + k, Abs[x]/s^2])/s *) Now the series expansion about s = 0 becomes
sg1 = Series[g1, {s, 0, 2}, Assumptions -> s > 0] // Normal (* Out[27]= E^(x/s^2 - Abs[x]/s^2) (1/2 k (1 + k) s^2 Abs[x]^(-1 + k) + Abs[x]^k) *) Let us distinguish the region x<0 and x>0:
Simplify[sg1, x < 0] (* Out[28]= 1/2 E^((2 x)/s^2) (k (1 + k) s^2 - 2 x) (-x)^(-1 + k) *) Obviously, the exponential factor cannot be expanded about s=0.
Simplify[sg1, x > 0] (* Out[31]= 1/2 x^(-1 + k) (k s^2 + k^2 s^2 + 2 x) *) In the limit
Limit[%, s -> 0] (* Out[32]= x^k *) i.e. we rediscover the effect of the delta function.