i am new to Mathematica. Here is the situation:

I have a exponential function of n,t that I want to calculate integral over t for different values of n. I want to define this integral as a function of n. I do it with := and it works fine if I provide a number for n but if I put a variable lets say x it tries to evaluate the integral and get an expression of x. This expression leads to precision error and the result for replacing x with a number gives zero at the output which is not correct. This is a sample code that I use:

A[t_]=Exp[-2.005*10^8 t];
B[n_]:=Integrate[A[t] Exp[-4000 I n \[Pi] t], {t, 0, 1/2000}];


General:"Exp[-100250.] is too small to represent as a normalized machine \
number; precision may be lost"
0. + 0. I

4.9*10^-9 even if not precise is OK for my application.

I have another function as:

C[f_, g_, n_, m_, max_] := 
  ParallelSum[(f /. n :> it ) g /. n :> (m - it), {it, -max, max}];

I want to use B and another function for example F in C like


Now for every m I get zero instead of a very small value. The precision loss is ok for me but I would like to have some value instead of zero to work with in other parts of my application.


1 Answer 1


If you rationalize A[t] integration works

A[t_] = Exp[-2.005*10^8 t] // Rationalize[#, 0] &;
B[n_]:=Integrate[A[t] Exp[-4000 I n \[Pi] t], {t, 0, 1/2000}]

B[x] (*-((I - I E^(-100250 - 2 I x \[Pi]))/(4000 (-50125 I + x\[Pi])))*)
B[0] (*(-1 + E^100250)/(200500000 E^100250)*)
  • $\begingroup$ The integration by itself works the problem is using C even for a large F for example F[x_] = Exp[100000]; I get Exp[-50125.] is too small to represent as a normalized machine number; precision may be lost and 0+0I $\endgroup$ Commented Aug 23, 2023 at 15:16
  • 1
    $\begingroup$ @AminGholizad You might increase WorkingPrecision or use an appropriate time scale. $\endgroup$ Commented Aug 23, 2023 at 15:39
  • $\begingroup$ If I do C[B[x],F[x],x,m,100]; it gives zero with precision error but if I do ParallelSum[Integrate[A[t] Exp[-4000 I n \[Pi] t], {t, 0, 1/2000}],F[m-x],{x,-100,100}] it works as expected. $\endgroup$ Commented Aug 24, 2023 at 7:33

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