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I found the following integral evaluation very interesting to me:

Integral of product of two error functions (erf)

and I hoped that I could use that result to evaluate the following integral: $$ \int_{-\infty}^{\infty}\exp\left(-t^{2}\right)\,\mathrm{erfc}\left(t-c\right)\,\mathrm{erfc}\left(d-t\right)\,\mathrm{d}t=\frac{4}{\pi}\int_{-\infty}^{\infty}\exp\left(-t^{2}\right)\int_{t-c}^{\infty}\int_{d-t}^{\infty}\exp\left(-u^{2}-v^{2}\right)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}t $$

So I note that $u\geq t-c$, $v\geq d-t$,
thus $t\leq u+c$ and $t\geq d-v$,
thus $d-v\leq t\leq u+c$ and $u+v\geq d-c$.

Hence $$ \frac{4}{\pi}\int_{-\infty}^{\infty}\exp\left(-t^{2}\right)\int_{d-t}^{\infty}\int_{t-c}^{\infty}\exp\left(-u^{2}-v^{2}\right)\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}t= $$ $$ \qquad\qquad =\frac{4}{\pi}\int\!\int_{u+v>d-c}\exp\left(-u^{2}-v^{2}\right)\,\int_{d-v}^{u+c}\exp\left(-t^{2}\right)\,\mathrm{d}t\,\mathrm{d}u\,\mathrm{d}v $$

I know that $$ \int_{d-v}^{u+c}\exp\left(-t^{2}\right)\,\mathrm{d}t=\frac{1}{2}\sqrt{\pi}\left(\mathrm{erf}\left(u+c\right)-\mathrm{erf}\left(d-v\right)\right) $$

but I don't quite understand how I should deal with $$\frac{4}{\pi}\int\!\int_{u+v>d-c}\exp\left(-u^{2}-v^{2}\right)\mathrm{d}u\,\mathrm{d}v\,. $$ What limits of integration I should use there? Thanks for any suggestions.

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3 Answers 3

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Let $c \in {\mathbb R}$ and $d \in {\mathbb R}$. Without loss of generality we can consider a slightly different function: \begin{equation} f(c,d):= \int\limits_{-\infty}^\infty \exp(-t^2) \mbox{erfc}(t-c) \mbox{erfc}(t-d) dt \end{equation} Now we differentiate with respect to $c$. We have: \begin{eqnarray} \partial_c f(c,d) &=& \lim\limits_{\epsilon \rightarrow 0} \frac{\sqrt{2}}{\sqrt{\pi}} \exp(-\frac{c^2}{2}) \left( \sqrt{\pi} + 2 \sqrt{\pi} T(\epsilon,\frac{1}{\sqrt{2}},\frac{2d-c}{\sqrt{2}}) - 2 \sqrt{\pi} T(\epsilon,\frac{1}{\sqrt{2}},\frac{-2d+c}{\sqrt{2}})\right)\\ &=&\sqrt{2} \exp(-\frac{c^2}{2}) \left( 1-erf(\frac{c-2 d}{\sqrt{6}})\right) \end{eqnarray} where $T(h,a,b)$ is the generalized Owen's T function Generalized Owen's T function .

Now, all we need to do is to integrate. Since $f(-\infty,d)=0$ we integrate over $c$ from minus infinity to $c$. We have: \begin{eqnarray} f(c,d) &=& \sqrt{2} \left( \sqrt{\frac{\pi}{2}}(1+erf(\frac{c}{\sqrt{2}}) + 2erf(\frac{d}{\sqrt{2}})) + \sqrt{2 \pi} 2 T(c,\frac{1}{\sqrt{3}},-\frac{2 d}{\sqrt{3}})\right) \\ &=& \frac{1}{3 \sqrt{\pi}} \left( 12 \pi T\left(c,\frac{c-2 d}{\sqrt{3} c}\right)+12 \pi T\left(d,\frac{d-2 c}{\sqrt{3} d}\right)-6 \arctan\left(\frac{c-2 d}{\sqrt{3} c}\right)-6 \arctan\left(\frac{d-2 c}{\sqrt{3} d}\right)+3 \pi \text{erf}\left(\frac{c}{\sqrt{2}}\right)+3 \pi \text{erf}\left(\frac{d}{\sqrt{2}}\right)+4 \pi\right) \end{eqnarray} where $T(h,a)$ is Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function .

In[1060]:= {c, d} = RandomReal[{-2, 2}, 2, WorkingPrecision -> 50];
NIntegrate[Exp[-t^2] Erfc[t - c] Erfc[t - d], {t, -Infinity, Infinity}]
(4 \[Pi] - 6 ArcTan[(c - 2 d)/(Sqrt[3] c)] - 
 6 ArcTan[(-2 c + d)/(Sqrt[3] d)] + 3 \[Pi] Erf[c/Sqrt[2]] + 
 3 \[Pi] Erf[d/Sqrt[2]] + 12 \[Pi] OwenT[c, (c - 2 d)/(Sqrt[3] c)] + 
 12 \[Pi] OwenT[d, (-2 c + d)/(Sqrt[3] d)])/(3 Sqrt[\[Pi]])

Out[1061]= 0.483318

Out[1062]= 0.48331807775609703646923225386370751256977344715

Update: Now let us take four real numbers $a_1 \in {\mathbb R}$, $a_2 \in {\mathbb R}$, $c\in {\mathbb R}$ and $d \in {\mathbb R}$ and consider a more general integral: \begin{equation} f^{(a_1,a_2)}(c,d):= \int\limits_{-\infty}^\infty \exp(-t^2) \mbox{erfc}(a_1 t-c) \mbox{erfc}(a_2 t-d) dt \end{equation} Then by doing the same calculations as above we easily arrive at the following formula: \begin{eqnarray} &&f^{(a_1,a_2)}(c,d)= \frac{1}{\sqrt{\pi}} \left(\right.\\ && 4 \pi T\left(\frac{\sqrt{2} c}{\sqrt{a_1^2+1}},\frac{a_1 a_2 c-\left(a_1^2+1\right) d}{c \sqrt{a_1^2+a_2^2+1}}\right)+4 \pi T\left(\frac{\sqrt{2} \sqrt{a_1^2+1} d}{\sqrt{a_1^2 a_2^2+a_1^2+a_2^2+1}},\frac{a_1 a_2 d-\left(a_2^2+1\right) c}{d \sqrt{a_1^2+a_2^2+1}}\right) +\\ &&-2 \arctan\left(\frac{a_1 a_2 c-\left(a_1^2+1\right) d}{c \sqrt{a_1^2+a_2^2+1}}\right)-2 \arctan\left(\frac{a_1 a_2 d-\left(a_2^2+1\right) c}{d \sqrt{a_1^2+a_2^2+1}}\right)+\\ &&\pi \text{erf}\left(\frac{\sqrt{a_1^2+1} d}{\sqrt{a_1^2 a_2^2+a_1^2+a_2^2+1}}\right)+\pi \text{erf}\left(\frac{c}{\sqrt{a_1^2+1}}\right)+\\ &&\pi +2 \arctan\left(\frac{a_1 a_2}{\sqrt{a_1^2+a_2^2+1}}\right)\\ &&\left.\right) \end{eqnarray}

For[count = 1, count <= 200, count++,
  {a1, a2, c, d} = RandomReal[{-5, 5}, 4, WorkingPrecision -> 50];
  I1 = NIntegrate[
    Exp[-t^2] Erfc[a1 t - c] Erfc[a2 t - d], {t, -Infinity, Infinity}];
  I2 = 1/Sqrt[\[Pi]] (\[Pi] + 
      2 ArcTan[(a1  a2)/Sqrt[(1 + a1^2 + a2^2)]] - 
      2 ArcTan[ (a1 a2 c - (1 + a1^2) d)/(
        Sqrt[(1 + a1^2 + a2^2)] c)] - 
      2 ArcTan[ (a1 a2 d - (1 + a2^2) c)/(
        Sqrt[(1 + a1^2 + a2^2)] d)] + \[Pi] Erf[c/Sqrt[
        1 + a1^2]] + \[Pi] Erf[(Sqrt[(1 + a1^2)] d)/ Sqrt[
        1 + a1^2 + a2^2 + a1^2 a2^2]] + 
      4 \[Pi] OwenT[(Sqrt[2] c)/Sqrt[
        1 + a1^2],  (a1 a2 c - (1 + a1^2) d)/(
        Sqrt[(1 + a1^2 + a2^2)] c)] + 
      4 \[Pi] OwenT[(Sqrt[2] Sqrt[(1 + a1^2)] d)/ Sqrt[
        1 + a1^2 + a2^2 + a1^2 a2^2],  (a1 a2 d - (1 + a2^2) c)/(
        Sqrt[(1 + a1^2 + a2^2)] d)]);
  If[Abs[I2/I1 - 1] > 10^(-3), 
   Print["results do not match", {a1, a2, c, d, {I1, I2}}]; Break[]];
  If[Mod[count, 10] == 0, PrintTemporary[count]];
  ];
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    $\begingroup$ Where were you when I needed this answer the most? ;-) This question is so old! Nevertheless I'll try to dust off my the research I needed the answer for, simply out of curiosity. I love that you added Mathematica code! Thank you! $\endgroup$
    – petru
    Commented Mar 23, 2019 at 18:01
  • $\begingroup$ @petru I was wondering in what context did you come across this problem? $\endgroup$
    – Przemo
    Commented Mar 25, 2019 at 10:48
  • $\begingroup$ It was theoretical analysis of a very simple evolutionary algorithm. Part of what I was trying to achieve was more or less to evaluate a triple integral of a product of three Gaussian functions, $\int_{-\infty}^{\infty}\int_{-\infty}^{x_1}\int_{-\infty}^{x_2}\ldots \text{d}x_3\text{d}x_2\text{d}x_1$, to give you a general idea. $\endgroup$
    – petru
    Commented Mar 28, 2019 at 18:18
  • $\begingroup$ @Przemo Mant thanks for your great answer!!! $\endgroup$
    – drandran12
    Commented Jul 27, 2021 at 8:49
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You can rotate coordinates. To do so, apply the following change of variables: $$\begin{cases} x = \frac{1}{\sqrt{2}} (u - v) \\ y = \frac{1}{\sqrt{2}} (u+v) \end{cases}$$ This leads to nicer integral: $$\int\int_{u+v > d-c} e^{-(u^2 + v^2)} \, du\,dv = \int_{\frac{1}{\sqrt{2}}(d-c)}^\infty \int_{-\infty}^{+\infty} e^{-x^2 - y^2} \, dx \,dy$$

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Ok, so after some manipulations and using the integral $$ \int_{-\infty}^{\infty}\exp\left(-b^{2}(x-c)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x=\frac{\sqrt{\pi}}{b}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right), \quad b>0 $$

I ended with

$$ \int_{-\infty}^{\infty}\exp\left(-t^{2}\right)\,\mathrm{erfc}\left(t-c\right)\,\mathrm{erfc}\left(d-t\right)\,\mathrm{d}t= $$

$$ =2\int_{\frac{1}{\sqrt{2}}\left(d-c\right)}^{\infty}\exp\left(-y^{2}\right)\left[\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y+\sqrt{2}c\right)\right)+\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y-\sqrt{2}d\right)\right)\right]\,\mathrm{d}y. $$

I checked it numerically and it's correct!

The last thing I need to do is to find an evaluation of a general integral $$ \int_{z}^{\infty}\exp\left(-y^{2}\right)\mathrm{erf}\left(m\left(y-n\right)\right)\,\mathrm{d}y. $$

but unfortunately I don't know how.

EDIT

And incidentally

$$ \int_{-\infty}^{\infty}\exp\left(-t^{2}\right)\,\mathrm{erfc}\left(t-c\right)\,\mathrm{erfc}\left(d-t\right)\,\mathrm{d}t= $$ $$ =2\int_{\frac{1}{\sqrt{2}}\left(d-c\right)}^{\infty}\exp\left(-y^{2}\right)\left[\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y+\sqrt{2}c\right)\right)+\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y-\sqrt{2}d\right)\right)\right]\,\mathrm{d}y = $$ $$ = -2\int_{-\infty}^{\frac{1}{\sqrt{2}}\left(c-d\right)}\exp\left(-y^{2}\right)\left[\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y-\sqrt{2}c\right)\right)+\mathrm{erf}\left(\frac{1}{\sqrt{3}}\left(y+\sqrt{2}d\right)\right)\right]\mathrm{d}y= $$ $$ {\small = 2\int_{0}^{\infty}\exp\!\left(-\left(y-\frac{1}{\sqrt{2}}\left(c-d\right)\right)^{2}\right)\left[\mathrm{erf}\!\left(\frac{1}{\sqrt{3}}\left(y+\frac{1}{\sqrt{2}}\left(c+d\right)\right)\right)+\mathrm{erf}\!\left(\frac{1}{\sqrt{3}}\left(y-\frac{1}{\sqrt{2}}\left(c+d\right)\right)\right)\right]\mathrm{d}y } $$

$$ {\small = -2\int_{-\infty}^{0}\exp\!\left(-\left(y+\frac{1}{\sqrt{2}}\left(c-d\right)\right)^{2}\right)\left[\mathrm{erf}\!\left(\frac{1}{\sqrt{3}}\left(y+\frac{1}{\sqrt{2}}\left(c+d\right)\right)\right)+\mathrm{erf}\!\left(\frac{1}{\sqrt{3}}\left(y-\frac{1}{\sqrt{2}}\left(c+d\right)\right)\right)\right]\mathrm{d}y. } $$

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    $\begingroup$ @ petru: The integral which, as you said, you do not know how to calculate is related to the Owen's T function as explained in here math.stackexchange.com/questions/3087504/… . $\endgroup$
    – Przemo
    Commented Mar 21, 2019 at 19:41

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