Wrong divergence with numerical value

Question

this is my first question in this forum.

I'm trying to evaluate some complicated function of, say, $x$ near $x=0$ (in order to integrate it later). The problem is that the numerical value of this function when $x \rightarrow 0$ goes to infinity while I know for sure that the limit of this function at the origin is $0$. This is happening because the function has the following structure (but much more complicated):

$f(x) \sim (A+\frac{B}{x}) x^2$

and for some reason mathematica is sending the second term to infinity faster than $x^2$ goes to zero.

Here I post a screenshot of my output, where it can be seen that the limit in the origin is indeed $0$:

Do you know any way of dealing with this problem?

Thank you.

Since I think that I wasn't accurate enough i will explain my problem better:

I'm trying to solve the following integral

$f(\theta) \propto \int_0^\infty dq \frac{q}{(\mu^2+q^2)^2} \int_0^{2 \pi} dx \frac{q^2 \cos(x)^2}{[q^2+c_1-2q c_2 cos(x)]^2[q^2+c_1'-2q c_2' cos(x+\theta)]^2[q^2+c_1''+c_3'' cos(x+\theta)-2q c_2'' cos(x)]^2} $

The code that I made solves the $x$ integral using the residue theorem. So, defining $z=e^{ix}$, $a_1=\frac{q^2+c_1}{2 q c_2}$, $a_2=\frac{q^2+c_1'}{2 q c_2'}$, $a_3=\frac{q^2+c_1''}{2 q c_2''}$, $b=\frac{c_3''}{2 q c_2''}$ and $c=\theta$ (where all the constants are real) the integral transforms to

$\int_0^\infty dq \frac{q}{(\mu^2+q^2)^2q^2} \int_0^{2 \pi} dx \frac{C}{q^4} \frac{q^2 \cos(x)^2}{[a_1 - cos(x)]^2[a_2 - cos(x+c)]^2[a_3+b cos(x+c)-cos(x)]^2} $,

where $C$ is some constant. The code that I have made to solve this integral is

z1 = (a3 + Sqrt[(a3)^2 - 1 - b^2 + 2 b Cos[c]])/((1 - b Exp[I c]));
z2 = (a3 - Sqrt[(a3)^2 - 1 - b^2 + 2 b Cos[c]])/((1 - b Exp[I c]));
z3 = a1 + Sqrt[a1^2 - 1];
z4 = a1 - Sqrt[a1^2 - 1];
z5 = (a2 + Sqrt[a2^2 - 1]) Exp[-I c];
z6 = (a2 - Sqrt[a2^2 - 1]) Exp[-I c];
f[z_] := ((16 E^(-4 I c)
   z^2 A6 (1 + E^(2 I c) z^2)^2))/((z - z3)^2 (z - z4)^2 (z - 
   z5)^2 (z - z6)^2 ((1 - b E^(I c) ) (z - z1) (z - z2))^2) z
peq1 = 2 Pi (Residue[f[z], {z, z2}] + Residue[f[z], {z, z4}] + 
 Residue[f[z], {z, z6}]);

Check if the integral is fine:

In[204]:= N[peq1 /. {A6 -> 6, a1 -> 2, a2 -> 2, a3 -> 3, b -> 1, c -> 2}]
NIntegrate[( 6 Cos[2 + x]^2)/((2 - Cos[x])^2 (2 - 
Cos[2 + x])^2 (3 + 1 Cos[x + 2] - Cos[x])^2), {x, 0, 2 Pi}]

Out[204]= 0.338923 + 8.99042*10^-13 I

Out[205]= 0.338923

In[207]:= prue = peq1 /. {A6 -> c3^2};

In[208]:= 
prue2[k1_, k2_, q_, \[Mu]_, \[Mu]p_, \[CapitalDelta]\[Phi]_] :=prue /. {c3 -> 1/(4 q k1), a1 -> (k1^2 + q^2 + \[Mu]p^2)/(2 k1 q),a2 -> (k2^2 + q^2 + \[Mu]p^2)/(2 k2 q), a3 -> (q^2 + k1^2 + k2^2 + \[Mu]^2 +2 k1 k2 Cos[\[CapitalDelta]\[Phi]])/(2 q k1), b -> -k2/k1,c -> -\[CapitalDelta]\[Phi]}

In[209]:= example[q_] := prue2[1, 1, q, 1/10, 2/10, 1/10]

In[210]:= N[example[1/10000]]

Out[210]= -2.82812*10^10 - 7.3361*10^11 I

In[211]:= Limit[example[q], q -> 0]

Out[211]= 0

I think that the problem with the divergence have to be in the definition of the variables $a_1$, $a_2$, $a_3$ and $a_4$ but I don't how to solve the $x$ integral in a way that there is no $1/q$ factors.

Thank you all.

Answer 1

If you make a plot you can see that Mathematica has trouble evaluating example[q] near q=0:

LogLinearPlot[Re[example[q]], {q, 0, 2}, PlotRange -> {-100000, 300000}]
LogLinearPlot[Im[example[q]], {q, 0, 2}, PlotRange -> All]

If the purpose is to establish a value to begin integration, it seems that you could pick a number around q = 0.1.