I have the following integral to compute

$$ \int_{-\infty}^{A} \frac{1}{(4 \pi)} \Big[ \text{erf}\Big(\frac{k M - x + 2 \pi}{\sqrt{2} k S}\Big) - \text{erf}\Big(\frac{k M - x}{\sqrt{2} k S}\Big) \Big] dx $$

Where $\text{erf}$ is the error function. Mathematica does compute the indefinite integral, but not the definite one like this. If I see the indefinite integral, it looks like that at $-\infty$, the function is not converging. However, looking at the function, I feel at $-\infty$, the integrand of this must be $0$.

The indefinite integral from Mathematica:

$$\frac{1}{4\pi} \Bigg[ \left(k M-x\right) \text{erf}\left(\frac{k \mu_{\omega}-x}{\sqrt{2} k S}\right) \\ +\left(-k M+x-2 \pi \right) \text{erf}\left(\frac{k M-x+2 \pi }{\sqrt{2} k S}\right)+\sqrt{\frac{2}{\pi }} k S \left(e^{-\frac{\left(x-k M\right)^2}{2 k^2 S^2}}-e^{-\frac{\left(k M-x+2 \pi \right)^2}{2 k^2 S^2}}\right) \Bigg] $$

================== EDIT based on – user293787's comment:

I am trying the following input:

L[x_]  := 1/(4 Pi) ( 
   Erf[(k M - x + 2 Pi)/(Sqrt[2] k S)] -  Erf[(k M - x )/(Sqrt[2] k S)] )

Integrate[L[x], {x, -Infinity, A}]
  • 2
    $\begingroup$ The definite integral Integrate[Erf[2*Pi-x]-Erf[-x],{x,-Infinity,A}] is evaluated immediately. Please provide the code that does not return a result, you may just have created confusion with all your constants and simply need the appropriate Assumptions. $\endgroup$
    – user293787
    Commented Aug 11, 2022 at 7:57
  • $\begingroup$ I have edited the question with the code! $\endgroup$
    – CfourPiO
    Commented Aug 11, 2022 at 10:01

1 Answer 1


OP probably has information about k and S that should be conveyed to Mathematica via Assumptions. This will make it easier for Integrate to decide, for example, if an integral converges or not.

Two examples:

  • If k and S are both positive, one can use:
  • If k and S are real and nonzero, one can use:

Both these Integrate are evaluated explicitly and give a result in terms of Erf.

  • 1
    $\begingroup$ Thank you for the answer. I was looking for this. This solved my issue. For me, S was non negative real number and k an integer greater than 0. I used these assumptions. $\endgroup$
    – CfourPiO
    Commented Aug 11, 2022 at 10:27

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