I need to find a model to fit my data, defined on [0, ∞[ or even on ]0, ∞[. It seems that some kind of (reversed) sigmoid function would work, but my data aren't symmetric, so traditional sigmoid functions don't really work.
I tried tweaking a Gomertz function, but as you can see below, it decreases too slowly at first, and then too fast. Any ideas?
And here, the raw data:
x y
0.3535533906 9.9979731306
0.5 9.9979441096
0.7071067812 9.997896157
0.7071067812 9.9978749407
1 9.997780566
1 9.9977588907
1.2247448714 9.9976603783
1.4142135624 9.9975407811
1.4142135624 9.9975199294
1.4142135624 9.9975177632
1.5811388301 9.9973989145
1.7320508076 9.9972495691
2 9.997075222
2 9.9970653382
2 9.9970071052
2.2360679775 9.9968735327
2.4494897428 9.9965729325
2.4494897428 9.9965735888
2.8284271247 9.9963045626
2.8284271247 9.9961509644
2.8284271247 9.996102247
2.8284271247 9.9962285165
3.1622776602 9.9956701877
3.1622776602 9.9956411184
3.4641016151 9.9950417481
3.4641016151 9.9952492462
3.7416573868 9.9946102874
4 9.9942744597
4 9.9943302815
4 9.9939916438
4 9.9943702025
4.472135955 9.9931768043
4.472135955 9.9932582478
4.8989794856 9.9924335386
4.8989794856 9.9919089218
4.8989794856 9.9927108557
5.2915026221 9.9914500289
5.6568542495 9.9913424021
5.6568542495 9.9900219326
5.6568542495 9.9907046142
5.6568542495 9.9898771885
5.6568542495 9.9904639461
6.3245553203 9.9879377579
6.3245553203 9.9879743157
6.3245553203 9.9897649623
6.9282032303 9.9851938132
6.9282032303 9.9868618753
6.9282032303 9.9877988638
7.4833147735 9.9849382687
7.4833147735 9.9859169334
8 9.9830432438
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8.94427191 9.9792529317
8.94427191 9.9802033408
8.94427191 9.9823864194
9.7979589711 9.9747179652
9.7979589711 9.9749611618
9.7979589711 9.9760116806
9.7979589711 9.9776693712
10.5830052443 9.9714823251
11.313708499 9.9686710997
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11.313708499 9.9661118054
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12.6491106407 9.9591613419
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12.6491106407 9.9626980052
13.8564064606 9.9521806585
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14.9666295471 9.9446790369
14.9666295471 9.9498590399
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16.9705627485 9.9346889776
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19.5959179423 9.9100270227
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19.5959179423 9.919521431
21.1660104885 9.8947578693
22.627416998 9.8918094719
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22.627416998 9.8798409365
22.627416998 9.8777822652
22.627416998 9.8715468945
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25.2982212813 9.8382348743
25.2982212813 9.841944607
25.2982212813 9.8685173357
27.7128129211 9.8196631997
27.7128129211 9.8234759383
27.7128129211 9.8308509026
28.2842712475 9.824205574
29.9332590942 9.7912290073
29.9332590942 9.7981835254
32 9.7689532343
32 9.7678748253
32 9.7545795668
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32 9.7655441336
33.941125497 9.7415241942
33.941125497 9.7222751527
35.77708764 9.7071517554
35.77708764 9.7131108475
35.77708764 9.7596412415
39.1918358845 9.636118575
39.1918358845 9.6217432379
39.1918358845 9.6325205631
39.1918358845 9.6792384173
39.5979797464 9.6866221408
40 9.6450847668
42.332020977 9.5850964021
45.2548339959 9.5794257293
45.2548339959 9.5220969239
45.2548339959 9.5387735428
45.2548339959 9.5203534668
45.2548339959 9.5485919631
45.2548339959 9.5483055814
48 9.499226795
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50.5964425627 9.4072535945
50.5964425627 9.4027464987
50.5964425627 9.4676509042
50.9116882454 9.4323128908
55.4256258422 9.3174138587
55.4256258422 9.3008138587
55.4256258422 9.3999171698
56 9.3143904143
56.5685424949 9.3465622319
56.5685424949 9.271269093
58.7877538268 9.2476393817
59.8665181884 9.2211094876
59.8665181884 9.2274917118
64 9.132537622
64 9.107278809
64 9.0319794364
64 9.1088696748
64 9.0975565848
67.8822509939 9.0844153404
67.8822509939 8.9932419602
67.8822509939 8.98866032
71.55417528 8.8683766313
71.55417528 9.0182329381
71.55417528 9.0916227983
72 8.9556913491
75.894663844 8.7364886005
78.3836717691 8.7381626778
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101.8233764909 8.0734844887
101.8233764909 8.0099313959
107.33126292 7.894393358
110.8512516844 7.8206052335
110.8512516844 7.7661708866
110.8512516844 7.9481519641
112 7.8145896471
112 7.5834543001
113.1370849898 7.7650979962
113.1370849898 7.6457065055
113.1370849898 7.747631427
117.5755076536 7.4402250135
117.5755076536 7.5244375513
119.7330363768 7.4775244535
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126.4911064067 7.2773311199
127.2792206136 7.3919916031
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128 7.3325210272
128 7.3311883257
135.7645019878 7.1130947152
135.7645019878 7.1199026643
135.7645019878 7.1155703141
137.1714255959 6.9092688699
138.5640646055 7.0195246551
143.10835056 6.8011022721
143.10835056 6.9297952756
143.10835056 7.365793704
144 6.8812037556
144 6.6660381104
151.7893276881 6.560975023
151.7893276881 6.6885403071
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160 6.5292012988
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169.3280839081 6.1050638233
176.3632614804 5.9712312801
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178.8854382 6.1642171555
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1915.7285820283 0.7945652559
1995.3225303194 0.7721619479
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2023.8577025078 0.719840126
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2194.7428095337 0.7006571374
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2821.8121836862 0.5396575193
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2896.3093757401 0.4923064599
3135.3468707625 0.4714507449
3219.9378875997 0.4931032241
3258.3480477076 0.4435256704
3527.2652296078 0.4241201511
3620.3867196751 0.3924015328
4072.9350596345 0.346517912
I would suggest plotting the data on a $\log$-$\log$ scale. I've done so, and that high end tail is definitely a power law (straight line on a $\log$-$\log$ scale). You can't use a straight Pareto distribution, though, because of the curvature at the low end.
If you don't particularly care about the theoretical content, you could always throw a smoothly broken power law at it. Given a positive function, $f(x)$, you can add a break to it around $x_n$ by multiplying it by: $$B(x, s, \beta, x_n) = \left(1 + \left[\frac{x}{x_n}\right]^{|\beta|s}\right)^{\operatorname{sgn}(\beta)/s}.$$ This will add a bend to the function around $x_n$ with a width controlled by $s$ (the sharpness - the larger $s$ the sharper the bend, with $s\rightarrow \infty$ leading to a corner). If $f$ is a power law and $\lim_{x\rightarrow 0} \frac{\operatorname{d} \ln [f(x)\, B(x)]}{\operatorname{d} \ln x} = m$, then $\lim_{x\rightarrow \infty} \frac{\operatorname{d} \ln [f(x)\, B(x)]}{\operatorname{d} \ln x} = m + \beta$.
For this sigmoid the low $x$ behavior is obviously $F(x) = \mathrm{const}$, but I don't know if you'll need one break or two to model it's behavior.
Note that unless there's something interesting about the second derivative of this function, this process of throwing breaks at it won't reveal anything particularly interesting. That's because if you look at $B$ in $\log$-$\log$ space then as $s\rightarrow \infty$, $\log B(x) \rightarrow \beta R(\log (x/x_n))$, with $R$ the ramp function. Because the ramp function is a Green's function for the second derivative operator, any twice differentiable map from $\mathbb{R}^+ \rightarrow \mathbb{R}^+$ can be modeled arbitrarily well if you use enough breaks with large enough sharpness.
Looking at the CDF ($=1 - \mathrm{SF}$) your data at the low end is approximately: $$F(x \ll 1) \approx 9.9979731306 - 2\times 10^{-4} x^2 + \ldots $$ At the high end, it's approximately: $$F(x \gg 1000) \approx \frac{1.5\times 10^3}{x}.$$
Looking at a graph where I tweaked the parameters using "$\chi$ by eye", a broken power law with $\beta = -1$, $s=2$, $x_n=150$, and normalization of $9.9979731306$ does an okay job. That is: $$F(x) \approx 9.9979731306 \left(1 + \left[\frac{x}{150}\right]^2\right)^{-1/2}.$$ You could, of course, achieve better by actually fitting the curve using an optimization package.