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I have the following function:

$$f(x) = \frac{2\arctan(\sqrt{\omega^2 + 1} - \omega)}{1 + \frac{1}{2}\ln(1 + \omega^2) - \ln\omega}$$

And something strange happens when I am going to plot it with different PlotRange choice. Here is the plot according to the code

Plot[qbec[ω], {ω, 0, 3}, PlotRange -> Automatic]

enter image description here

And here is what happens when I plot it with the choice

Plot[qbec[ω], {ω, 0, 3}, PlotRange -> {0,2}]

enter image description here

P.s. Yes, in my case I called the function above as "qbec" because of some reason.

The fact is this: this function has to be plotted with three other functions together, and there is no problem in the final plot except for this "gap" or whatever it may be.

I NEED to choose a PlotRange of {0,3} because the plot needs to be clear, but when I do it, this function behaves bad. I already tried with PlotRange: Full, All and Automatic. No way to solve it.

Is there a trick or a way to fix it?

What I need is this function to start from $0$, as it shall.

Thank you!!

EDIT - THE GAP

Here is a zoom for the second plot, in order to make you to see what I mean with "gap".

enter image description here

EDIT - qbec function

qbec[ω_] := ArcTan[1/ω]/(1 + 1/2*Log[(1 + ω^2)/ω^2])
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So, as noticed the problem is that your function converges very slowly with ω approaching 0.:

LogLinearPlot[qbec[1/t], {t, 1, 10^25}
  , Frame -> True
  , PlotRangePadding -> Scaled[.05]
  , PlotRange -> {0, Automatic}
  , GridLines -> {None, {0}}
  , FrameLabel -> {"1/ω", "qbec[1/ω]"}
]

enter image description here

and is Indeterminate at x=0. So it may be tricky get that right automatically.

You can do this dirty trick though, just add a point at {0,0}:

Show[
    Plot[qbec[ω], {ω, 0., 3}
      , PlotRange -> {0, 2}
      , PlotStyle -> Thickness@.01
    ] /. Line[{pts__}] :> Line[{{0, 0}, pts}]
  , Plot[Sin[x] + 1, {x, 0, 3}
      , PlotStyle -> Red
    ]
]

Show and Plot[Sin are there to mimic your background case where you want to add more plots.

enter image description here

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  • $\begingroup$ +1 More generally, qbec[0] = qbec[0.] = Limit[qbec[w], w -> 0] and Line[{pts__}] :> Line[{{0, qbec[0]}, pts}] $\endgroup$ – Bob Hanlon Jun 27 '17 at 13:00
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ParametricPlot[{Exp[-x], qbec[Exp[-x]]}, {x, -Log[3], 300},
 PlotRange -> {{-.5, 3}, {0, 2}}, AspectRatio -> 1/GoldenRatio, 
 PlotPoints -> 100, PlotStyle -> Red, AxesOrigin -> {-.5, 0}]

enter image description here

moved the axis just for illustration.

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