0
$\begingroup$

In the course of my research I came across the following integral:

$$\int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b- \dfrac{ar}{\gamma}\right) \right) * \left(\operatorname{erf}\left(cx-d\right)-\operatorname{erf}\left(\frac{c}{\gamma}x-d-\frac{cr}{\gamma}\right)\right) \, dx$$
, where $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$ is the error function

Does anyone know if the integration given above can be calculated into closed-form?

Thank you in advance.

$\endgroup$
  • $\begingroup$ I doubt it. Have you tried Wolfram|Alpha? By the way, you can get proper formatting for functions like $\operatorname{erf}$ using the syntax \operatorname{erf}; you can get displayed equations (that have nicer integrals and fractions) by enclosing them in double instead of single dollar signs; and you can get proper parentheses that adapt to their content by preceding them with \left and \right. More generally, here's a basic tutorial and quick reference on how to typeset math on this site: math.meta.stackexchange.com/questions/5020. $\endgroup$ – joriki Jul 5 '18 at 8:36
  • 1
    $\begingroup$ @joriki Thank you for your advice. I tried this on Wolframa Alpha and also on Mathematica, but it failed $\endgroup$ – Yunho Choi Jul 5 '18 at 12:14
1
$\begingroup$

Let us take $a \ge 0$, $b \ge 0$, $c \ge 0$, $d \ge 0$, $\gamma \ge 0$ and $r \ge 0$ and let us define:

\begin{eqnarray} {\mathfrak I}^{(a,b)}_{c,d}(\gamma,r):= \int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b- \dfrac{ar}{\gamma}\right) \right) * \left(\operatorname{erf}\left(cx-d\right)-\operatorname{erf}\left(\frac{c}{\gamma}x-d-\frac{cr}{\gamma}\right)\right) \, dx \end{eqnarray} Then by differentiating with respect to the parameter $b$ and using the identity: \begin{equation} \int\limits_{-\infty}^\infty e^{-u^2} \text{erf}(A u + B) du=\sqrt{\pi} \text{erf} \left( \frac{B}{\sqrt{1+A^2}}\right) \end{equation} we get: \begin{eqnarray} &&\frac{\partial }{\partial b} {\mathfrak I}^{(a,b)}_{c,d}(\gamma,r) = \frac{2}{a} \left( \right. \\ && -\text{erf}\left(\frac{\frac{c (a r-b)}{a g}+d}{\sqrt{\frac{c^2}{a^2 g^2}+1}}\right)+\text{erf}\left(\frac{d-\frac{b c}{a}}{\sqrt{\frac{c^2}{a^2}+1}}\right)+g \left(\text{erf}\left(\frac{d-\frac{b c}{a}}{\sqrt{\frac{c^2}{a^2}+1}}\right)-\text{erf}\left(\frac{d-\frac{c (a r+b g)}{a}}{\sqrt{\frac{c^2 g^2}{a^2}+1}}\right)\right) \\ &&\left.\right) \end{eqnarray}

Now all we need to do is to integrate with respect to $b$ from minus infinity to $b$. Since $\int \text{erf}(x) dx = \exp(-x^2)/\sqrt{\pi} + x \text{erf}(x)$ and since in the expression for the partial derivative the parameter $b$ enters the error function in a linear manner only it is clear that it is possible to construct the anti-derivative. Having done this we just take the values at $b$ and at minus infinity . It turns out that the later value is equal to zero and therefore we are left with the value at $b$ only which we simplified by hand. Now the final result reads: \begin{eqnarray} &&{\mathfrak I}^{(a,b)}_{c,d}(\gamma,r) = \frac{2}{a c} \left( \right. \\ && -\frac{(g+1) \sqrt{a^2+c^2} e^{-\frac{(b c-a d)^2}{a^2+c^2}}}{\sqrt{\pi } } + \frac{\sqrt{a^2 g^2+c^2} e^{-\frac{(a c r+a d g-b c)^2}{a^2 g^2+c^2}}+\sqrt{a^2+c^2 g^2} e^{-\frac{(a c r-a d+b c g)^2}{a^2+c^2 g^2}}}{\sqrt{\pi } }+\\ && (g+1) (a d-b c) \text{erf}\left(\frac{b c-a d}{\sqrt{a^2+c^2}}\right)+\\ && (a c r-a d+b c g) \text{erf}\left(\frac{a c r-a d+b c g}{\sqrt{a^2+c^2 g^2}}\right)+(a c r+a d g-b c) \text{erf}\left(\frac{a c r+a d g-b c}{\sqrt{a^2 g^2+c^2}}\right)\\ \left. \right) \end{eqnarray}

For[count = 1, count <= 100, count++,
  {a, b, c, d, g, r} = RandomReal[{0, 1}, 6, WorkingPrecision -> 50];
  I1 = NIntegrate[(Erf[(a x - b)] - 
       Erf[(a/g x - b - a r/g)]) (Erf[c x - d] - 
       Erf[c/g x - d - c r/g]), {x, -Infinity, Infinity}, 
    WorkingPrecision -> 15];
  2/Abs[a] NIntegrate[ (-Erf[(xi c - d a)/Sqrt[a^2 + c^2]] + 
      Erf[(-d a g + c (xi - a r))/Sqrt[a^2 g^2 + c^2]] + 
      Abs[g] (-Erf[(xi c - d a)/Sqrt[a^2 + c^2]] + 
         Erf[(-d a + c (xi g + a r))/Sqrt[
          a^2 + c^2 g^2]])), {xi, -Infinity, b}];
  I2 = 2/(
    Abs[a] c) (-E^(-((b c - a d)^2/(a^2 + c^2))) Sqrt[a^2 + c^2] /
       Sqrt[\[Pi]] (1 + g) + 
      1/ Sqrt[\[Pi]] (E^(-((-b c + a d g + a c r)^2/(c^2 + a^2 g^2)))
           Sqrt[c^2 + a^2 g^2] + 
         E^(-((-a d + b c g + a c r)^2/(a^2 + c^2 g^2))) Sqrt[
          a^2 + c^2 g^2]) + (a d - b c) (1 + g) Erf[(b c - a d)/Sqrt[
        a^2 + c^2]] + (-a d + b c g + a c r) Erf[(-a d + b c g + 
         a c r)/Sqrt[
        a^2 + c^2 g^2]] + (-b c + a d g + a c r) Erf[(-b c + a d g + 
         a c r)/Sqrt[c^2 + a^2 g^2]]);
  If[Abs[I2/I1 - 1] > 10^(-3), 
   Print["results do not match", {a, b, c, d, g, r, {I1, I2}}]; 
   Break[]];
  If[Mod[count, 10] == 0, PrintTemporary[count]];
  ];
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.