# "Sparse" Overflow error in NIntegrate

I want to evalutate an integral $$I = \int_{-0.01}^0 \int_{-0.01}^0 \left(\frac{1}{\sinh|t_2+t_1+T| \sinh|t_2+t_1|}\right)^{2/m} \left(\frac{\sinh|2t_1+t_2+T|}{\sinh|t_1+T|}\right)^{1/m} \left(\frac{\sinh|t_2|}{\sinh|t_1|}\right)^{1/m},$$ where $$m=3$$ is fixed and I want to vary $$T$$. This integral is mathematically shown to be convergent. For various values of $$T$$ within range $$10^{-6}, I sometimes get an overflow error, such as:

NIntegrate::inumri: The integrand ((Csch[Abs[t1]] Sinh[Abs[t2]])^(1/3) (Csch[Abs[5.01187*10^-6+t1]] Sinh[Abs[<<22>>+2 t1+t2]])^(1/3))/(Sinh[Abs[t1+t2]] Sinh[Abs[5.01187*10^-6+t1+t2]])^(2/3) has evaluated to Overflow, Indeterminate, or Infinity for all sampling points in the region with boundaries {{0,1},{0.,2.48466*10^-32}}.


I don't know a particular pattern for what values of $$T$$ does this error occurs. (However, the error does not occur for $$|T|>0.1$$.) Below is my relevant code:

exp1 = (Csch[Abs[t1]] Sinh[Abs[t2]])^(1/
m) (Sinh[Abs[t1 + t2]] Sinh[Abs[T + t1 + t2]])^(-2/
m) (Csch[Abs[T + t1]] Sinh[Abs[T + 2 t1 + t2]])^(1/m)
tab1 = Table[{T,
NIntegrate[exp1 /. {m -> 3}, {t1, -0.01, 0}, {t2, -0.01, 0}]}, {T,
Table[10^(-n), {n, 4, 6, 0.1}]}]


The result is:

{{0.0001, 0.225753 - 2.69464*10^-8 I}, {0.0000794328,
0.225296 - 2.4616*10^-8 I}, {0.0000630957,
0.224902 - 2.37137*10^-8 I}, {0.0000501187,
0.224563}, {0.0000398107, 0.22427}, {0.0000316228,
0.224019}, {0.0000251189, 0.223802}, {0.0000199526,
0.223616 - 3.04706*10^-8 I}, {0.0000158489,
0.223455}, {0.0000125893, 0.223317}, {0.00001,
0.223199}, {7.94328*10^-6, 0.223097}, {6.30957*10^-6,
0.223009}, {5.01187*10^-6,
NIntegrate[
exp1 /. {m -> 3}, {t1, -0.01, 0}, {t2, -0.01, 0}]}, {3.98107*10^-6,
NIntegrate[
exp1 /. {m -> 3}, {t1, -0.01, 0}, {t2, -0.01, 0}]}, {3.16228*10^-6,
0.222814}, {2.51189*10^-6, 0.222766}, {1.99526*10^-6,
0.222725}, {1.58489*10^-6, 0.22269}, {1.25893*10^-6,
0.22266}, {1.*10^-6, 0.222634}}


Note that sometimes the integral is left unevaluated.

Why does this error happen? Note that for $$T<0$$, the only singularity of the integral is at $$t_1=0$$, which is the boundary of the integration range.

• Independant of the value T there is a singularity at t1=t2=0! Oct 25 at 13:56
• @UlrichNeumann But that is again the boundary (corner) of the integration range! Oct 25 at 13:58

tab1 = Table[{T,

• Exact numbers have infinite precision. Oct 25 at 15:45