# The function is real, while its integral is complex [duplicate]

Suppose the function

SingleCrossSection[Ss_, s2_, m_, m2_, mp_] =
-(1/(288 π^3 s2^3 Ss^4 (mp^2 - Ss)^2))
((m^2 - s2) (1/((m + m2)^2 - (mp - Sqrt[
Ss])^2) (s2 - (mp - Sqrt[
Ss])^2) Sqrt[-2 mp^2 ((m + m2)^2 + Ss) + ((m + m2)^2 -
Ss)^2 +
mp^4] (m^4 (-6 mp^8 + 3 mp^6 (s2 + 24 Ss) +
mp^4 (3 s2^2 - 27 s2 Ss - 172 Ss^2) +
mp^2 Ss (-30 s2^2 - 19 s2 Ss + 152 Ss^2) +
Ss^2 (35 s2^2 + 35 s2 Ss - 46 Ss^2)) +
m^2 s2 (3 mp^8 - 6 mp^6 (s2 + 12 Ss) +
mp^4 (3 s2^2 + 18 s2 Ss + 110 Ss^2) -
2 mp^2 Ss (15 s2^2 + 17 s2 Ss + 8 Ss^2) +
Ss^2 (11 s2^2 + 38 s2 Ss - 25 Ss^2)) -
s2^2 (-3 mp^8 - 3 mp^6 (s2 - 24 Ss) +
mp^4 (6 s2^2 + 63 s2 Ss - 134 Ss^2) +
mp^2 Ss (-60 s2^2 + 19 s2 Ss + 64 Ss^2) +
Ss^2 (46 s2^2 - 71 s2 Ss + Ss^2))) +
12 Ss^2 (m^4 (2 mp^6 - 3 mp^4 (s2 + 2 Ss) + 6 mp^2 Ss^2 +
s2^3 + 3 s2 Ss^2 - 2 Ss^3) +
m^2 s2 (2 mp^6 + 3 mp^4 (s2 - 2 Ss) -
6 mp^2 (s2^2 + s2 Ss - Ss^2) + s2^3 + 3 s2 Ss^2 -
2 Ss^3) +
s2^2 (-mp^6 + 3 mp^4 (s2 + Ss) - 3 mp^2 Ss^2 - 2 s2^3 +
6 s2^2 Ss - 3 s2 Ss^2 + Ss^3)) Log[-(
s2/(-2 mp^2 + 4 mp Sqrt[Ss] + s2 - 2 Ss))]) -
12 Ss^2 Log[
m2^2/(-m^2 - 2 m m2 +
s2)] (-3 s2^3 (mp^2 - Ss)^2 (m^2 + mp^2 - Ss) Log[-(
s2/(-2 mp^2 + 4 mp Sqrt[Ss] + s2 - 2 Ss))] -
1/((m + m2)^2 - (mp - Sqrt[Ss])^2) (m^2 -
s2) (s2 - (mp - Sqrt[
Ss])^2) Sqrt[-2 mp^2 ((m + m2)^2 + Ss) + ((m + m2)^2 -
Ss)^2 +
mp^4] (m^4 (-2 mp^4 + mp^2 (s2 + 4 Ss) + s2^2 + s2 Ss -
2 Ss^2) +
m^2 s2 (-2 mp^4 + mp^2 (4 Ss - 5 s2) + s2^2 + s2 Ss -
2 Ss^2) +
s2^2 (mp^4 - 2 mp^2 (s2 + Ss) - 2 s2^2 + 4 s2 Ss +
Ss^2))));


defined on

{s2, (m+m2)^2, (Sqrt[Ss]-mp)^2}, Ss > (mp+m2+m)^2, m > 0, m2>0, mp>0


It is real function of its arguments on the domain of the definition.

I'm trying to evaluate its integral over s2:

CrossSectionIntegral[Ss_, s2_, m_, m2_, mp_] =
Simplify[Integrate[SingleCrossSection[Ss, s2, m, m2, mp],
s2, Assumptions -> Ss > 0 && Sqrt[Ss] > mp + m2 + m]];
CrossSection[Ss_, m_, m2_, mp_] =
Collect[CrossSectionIntegral[Ss, (Sqrt[Ss]-mp)^2, m, m2, mp] -
CrossSectionIntegral[Ss, (m+m2)^2, m, m2, mp], m] ;


The huge output for the integration contains, as far I can see, the combinations of polylogarithms with positive arguments larger than 1 and logarithms with negative arguments, and thus seems to be comples. However, because of an identity relating a polylogarithm with the argument larger than one with a polylogarithm with the argument less than one, all the imaginary terms becomes zero. Mathematica seems to know these identities, or at least evaluates the imaginary part to zero.

Now, I'm going to check this graphically for particular values of Ss, mp,m2. I write

Plot[{CrossSection[6.5, m, 0.1, 1]}, {m, 0, Sqrt[6.5] - 1}]


The output is not smooth, there are some discontinuites on the plot (the yellow curve, please ignore the blue curve here and below): From the other side, if I plot the real part,

Plot[{Re[CrossSection[6.5, m, 0.1, 1]]}, {m, 0, Sqrt[6.5] - 1}]


the graphic is the same, but discontinuites are lost (the yellow curve): Please ignore any errors during obtaining the output (such as "Infinite expression 1/0 encountered"), since the reason for them is just slightly incorrect upper limit for the plot (it must be smaller).

What can be the reason of such problem? I expect that the integral of the function over the domain where it is real and continuous will be real quantity, so it seems that the problem may be in misinterpretation of the output by Mathematica.

## marked as duplicate by Mr.Wizard♦ functions StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 7 '17 at 18:32

• Does it help if you increase the points used for plotting using PlotPoints ? Have I understood the question; are the discontinuities the problem of the appearance or another curve when you take the real part? – Hugh Aug 7 '17 at 17:13
• @Hugh : sorry, another curve is given from another plot. The only problem worrying me are discontinuities. As for PlotPoints, i'm not sure. Actually the function from my qiestion was obtained by the approximation of another function (whose integral corresponds to the blue curve), and its plot (obtained by numerical integration) is smooth. – John Taylor Aug 7 '17 at 17:16
• Try using PlotPoints-> 500. This will give 500 starting points and this may find all the intervals. – Hugh Aug 7 '17 at 17:19
• @Hugh : But why then the plot of real part is smooth? – John Taylor Aug 7 '17 at 17:24
• Try plotting the imaginary part. Are there small artifacts? This could spoil the algorithm for plotting points. Mathematica leaves a gap if the function is complex. – Hugh Aug 7 '17 at 17:27

If we just plot the function, the real part and the imaginary part around the discontinuity we find the following

  Plot[{CrossSection[6.5, m, 0.1, 1]}, {m, 0.7, 0.82}, PlotRange -> All]
Plot[{Re[CrossSection[6.5, m, 0.1, 1]]}, {m, 0.7, 0.82},
PlotRange -> All]
Plot[{Im[CrossSection[6.5, m, 0.1, 1]]}, {m, 0.7, 0.82},
PlotRange -> All]   So my guess is that the small imaginary part is too big for the plotting algorithm and it does not plot because it sees an imaginary part.

Note that you are hoping for imaginary parts to subtract to zero. This is not quite managed because of numerical precision. As Mr.Wizard mentions using Chop would be an option. It gets rid of small values.

This may be a duplicate but I can't find it. So the work around is to plot the real part or use Chop.

• You can also Chop away the imaginary noise: Plot[Chop @ CrossSection[6.5, m, 0.1, 1], {m, 0.7, 0.82}, PlotRange -> All] – Mr.Wizard Aug 7 '17 at 17:52
• @Mr.Wizard : is this imaginary noise just the artifact of Mathematica? – John Taylor Aug 7 '17 at 18:14
• See my additional comment in the answer you are hoping that imaginary parts will subtract to zero. – Hugh Aug 7 '17 at 18:23