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$$\sum _{j=0}^{\infty } \frac{s^{2 j} }{j!2^{j}}\frac{\partial ^{2 j}\left(f(x) x^j\right)}{\partial x^{2 j}}$$

$$fs = \sum _{j=0}^{\infty } \frac{s^{2 j} }{j!2^{j}}\frac{\partial ^{2 j}\left(f(x) x^j\right)}{\partial x^{2 j}}$$

We see that the agreement is what is to expected in the sense of an asymptotic expansion.

Conclusion: the formula for $fs$ seems to be valid for a broader class of function.

Let me add the general expansion in the form of Till Hoffmann

It is very interesting that this expression seems to hold not only for the powers of y considered up to now but for more general function. I have successfully tested $f(x)=\frac{1}{y+1}$, and I shall continue this investigation.

Let me add the general expansion in the form of Till Hoffmann, which I call s-expansion in the following

It is very interesting that this expression seems to hold not only for the powers of y considered up to now but for more general functions f(y).

In order to verify this hypothesis we shall study examples for other functions than simple powers of the s-expansion fs of our integral.

The s-expansion with a function f is given by

fs[x_, s_, f_, n_] := Sum[s^(2 j) 1/(j! 2^j) D[x^(j) f, {x, 2 j}], {j, 0, n}] 

Example 1: Simple pole on the negative real y-axis

The integral is numerically

gg1[x_, s_] := NIntegrate[ Exp[-(x - y)^2/(2 y s^2)] 1/(1 + y) 1/Sqrt[2 \[Pi] y s^2], {y,0, \[Infinity]}] 

The first 3 terms of the s-expansion are

fs1[x_, s_] = Table[fs[x, s, 1/(1 + x), k], {k, 0, 2}] // Simplify (* {1/(1 + x), (-s^2 + (1 + x)^2)/(1 + x)^3, ( 3 s^4 - s^2 (1 + x)^2 + (1 + x)^4)/(1 + x)^5} *) 

Graphs up to order s^4 for two values of s are

With[{s = 1}, Plot[{gg1[x, s], {1/(1 + x), (-s^2 + (1 + x)^2)/(1 + x)^3, ( 3 s^4 - s^2 (1 + x)^2 + (1 + x)^4)/(1 + x)^5}}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400, PlotLabel -> Style["s-expansion of integral\nf(y) = 1/(1+y), s = " <> ToString[s] <> "\n", 14], AxesLabel -> {"x", "gg1[x]"}, Epilog -> {Text[ Style["Legend of curves:\nblue = integral, numeric (gg1)\nred = f(x)\n\ brown = O(\!\(\*SuperscriptBox[\(s\), \(2\)]\))\ngreen = \ O(\!\(\*SuperscriptBox[\(s\), \(4\)]\))", Medium], {1, 1.2}]}]] (* 150228_s-expansion_f1 _s1.jpg *) 

With[{s = 0.3}, Plot[{gg1[x, s], {1/(1 + x), (-s^2 + (1 + x)^2)/(1 + x)^3, ( 3 s^4 - s^2 (1 + x)^2 + (1 + x)^4)/(1 + x)^5}}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400, PlotLabel -> Style["s-expansion of integral\nf(y) = 1/(1+y), s = " <> ToString[s] <> "\n", 14], AxesLabel -> {"x", "gg1[x]"}, Epilog -> {Text[ Style["Legend of curves:\nblue = integral, numeric (gg1)\nred = \ f(x)\nbrown = O(\!\(\*SuperscriptBox[\(s\), \(2\)]\))\ngreen = O(\!\(\ \*SuperscriptBox[\(s\), \(4\)]\))", Medium], {1, 1.2}]}]] (* 150228_s-expansion_f1 _s0-3.jpg *) 

Example 2: Simple conjugate poles on the imaginary y-axis

The integral is numerically

gg2[x_, s_] := NIntegrate[ Exp[-(x - y)^2/(2 y s^2)] 1/(1 + y^2) 1/Sqrt[2 \[Pi] y s^2], {y, 0, \[Infinity]}] 

The first 3 terms of the s-expansion are

fs2[x, s]

(* {1/(1 + x^2), (s^2 x (-3 + x^2) + (1 + x^2)^2)/(1 + x^2)^3, ( s^2 x (-3 + x^2) (1 + x^2)^2 + (1 + x^2)^4 - 3 s^4 (1 - 10 x^2 + 5 x^4))/(1 + x^2)^5} *) 

Graphs up to order s^4 for two values of s are

With[{s = 1}, Plot[{gg2[x, s], {1/(1 + x^2), ( s^2 x (-3 + x^2) + (1 + x^2)^2)/(1 + x^2)^3, ( s^2 x (-3 + x^2) (1 + x^2)^2 + (1 + x^2)^4 - 3 s^4 (1 - 10 x^2 + 5 x^4))/(1 + x^2)^5}}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400, PlotLabel -> Style["s-expansion of integral\nf(y) = 1/(1+\!\(\*SuperscriptBox[\(y\), \ \(2\)]\)), s = " <> ToString[s] <> "\n", 14], AxesLabel -> {"x", "gg2[x]"}, Epilog -> {Text[ Style["Legend of curves:\nblue = integral, numeric (gg1)\nred = f(x)\n\ brown = O(\!\(\*SuperscriptBox[\(s\), \(2\)]\))\ngreen = \ O(\!\(\*SuperscriptBox[\(s\), \(4\)]\))", Medium], {1, 1.2}]}]] (* 150228_s-expansion_f2 _s0-9.jpg *) 

With[{s = 0.4}, Plot[{gg2[x, s], {1/(1 + x^2), ( s^2 x (-3 + x^2) + (1 + x^2)^2)/(1 + x^2)^3, ( s^2 x (-3 + x^2) (1 + x^2)^2 + (1 + x^2)^4 - 3 s^4 (1 - 10 x^2 + 5 x^4))/(1 + x^2)^5}}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400, PlotLabel -> Style["s-expansion of integral\nf(y) = \ 1/(1+\!\(\*SuperscriptBox[\(y\), \(2\)]\)), s = " <> ToString[s] <> "\n", 14], AxesLabel -> {"x", "gg2[x]"}, Epilog -> {Text[ Style["Legend of curves:\nblue = integral, numeric (gg1)\nred = \ f(x)\nbrown = O(\!\(\*SuperscriptBox[\(s\), \(2\)]\))\ngreen = O(\!\(\ \*SuperscriptBox[\(s\), \(4\)]\))", Medium], {1, 1.2}]}]] (* 150228_s-expansion_f2 _s0-4.jpg *) 

EDIT (Dr. Wolfgang Hintze, 28.02.15)

Let me add the general expansion in the form of Till Hoffmann

$$\sum _{j=0}^{\infty } \frac{s^{2 j} }{j!2^{j}}\frac{\partial ^{2 j}\left(f(x) x^j\right)}{\partial x^{2 j}}$$

It is very interesting that this expression seems to hold not only for the powers of y considered up to now but for more general function. I have successfully tested $f(x)=\frac{1}{y+1}$, and I shall continue this investigation.