13
$\begingroup$

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

$\int\nolimits_{-\infty}^{\infty}\operatorname{erf}(a+x)\operatorname{erf}(a-x)dx$

where $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$ is the familiar error function.

Does anyone know if this is solvable? If so, a suggestion on how to do this would be very much appreciated. I can tolerate a solution in terms of $\operatorname{erf}$ functions of $a$. I have tried my usual sources in these matters (G&R, the tables on the Wolfram website, Abramowitz's Handbook of Mathematical Formulas, the web) but couldn't find anything of use.

Thank you!

  • 0
    @Bullmoose: Sure. Double-check my calculations to be sure I got all the constants right (but I'm pretty sure I did). A proof of the entropy-power inequality based on a strengthened version of Young's inequality was [provided by A. Dembo](http://statistics.stanford.edu/~ckirby/techreports/NSF/COV%20NSF%2075.pdf). The Cover & Thomas text reproduces this in their chapter on information-theory inequalities.2011-09-12

2 Answers 2

13

$\operatorname{erf}(x)$ is an odd function, therefore, $ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x &=\lim_{L\to\infty}\;\int_{-L}^L(\operatorname{erf}(x+a)-\operatorname{erf}(x-a))\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{-L+a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{L-a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{L-a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{-L+a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=4a\tag{1} \end{align} $ since $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1$ and $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1$.

Furthermore, $ \begin{align} \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x &=\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x\\ &-\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x\tag{2} \end{align} $ To evaluate $ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x \end{align} $ note that $s\le a+x$ and $t\le a-x$; i.e. $s-a\le x\le a-t$ and $s+t\le2a$. Thus, $ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x &=\frac{4}{\pi}\int\int_{s+t\le2a}\int_{s-a}^{a-t}e^{-s^2-t^2}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\\ &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t \end{align} $ Change variables: $u=(s+t)/\sqrt{2}$ and $v=(s-t)/\sqrt{2}$ so that $s=(u+v)/\sqrt{2}$ and $t=(u-v)/\sqrt{2}$: $ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-\sqrt{2}u)_+\;e^{-u^2-v^2}\;\mathrm{d}u\;\mathrm{d}v\\ &=\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}(2a-\sqrt{2}u)\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\sqrt{2}u\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{2\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\;e^{-u^2}\;\mathrm{d}u^2\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2} \end{align} $ Therefore, $ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)+1\right)\left(\operatorname{erf}(a-x)+1\right)\;\mathrm{d}x =4a\left(\operatorname{erf}(\sqrt{2}a)+1\right)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{3} $ Thus, the convolution of $\operatorname{erf}(x)+1$ with itself is $2x(\operatorname{erf}(x/\sqrt{2})+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}e^{-x^2/2}$.

Subtract $4a$ from $(3)$ using $(1)$ and $(2)$ to get $ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x =4a\operatorname{erf}(\sqrt{2}a)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{4} $ My guess is you want either $(3)$ or $(4)$.

  • 0
    In fact, one has $\int_{-\infty}^\infty (\text{erf}(x+a)^2-1)\,dx=\int_{-\infty}^\infty (\text{erf}(x-a)^2-1)\,dx=\int_{-\infty}^\infty (\text{erf}(x)^2-1)\,dx,$ and apparently that last integral equals $-2\sqrt{2/\pi}$. That eliminates the outer terms, and the inner term corresponds to your identity (4). So all of those together give my identity---neat.2017-10-16
-2

$\int_{-\infty}^\infty 4/\pi \int_0^{a+x} e^{-t^{2}} dx $ . $ \int_0^{a-x} e^{-t^{2}} $ dx $= \int_{-\infty}^\infty 4/\pi [-(1/2t) e^{-t^{2}}]_0^{a+x} [-(1/2t) e^{-t^{2}}]_0^{a-x}$ = $ \int_{-\infty}^\infty (4/\pi) ((1/2a) e^{-a^{2}})^{2} - ((1/2x) e^{-x^{2}})^{2} $ = $[-((1/2x) e^{-x^{2}})^{2}]_{-\infty}^\infty$ = $\int_{-\infty}^\infty -((1/2x) e^{-x^{2}})^{2}$= $\int_{-\infty}^\infty -((1/4x^{2}) e^{-2x^{2}})] $ = $ [-(1/4x) e^{-2x^{2}}]_{-\infty}^\infty - \int_{-\infty}^\infty e^{-2x^{2}}$ = $[-(1/4x) e^{-2x^{2}}$+$(1/4x) e^{-2x^{2}}]_{-\infty}^\infty$ =0

  • 4
    @Angela, at the very least, it would be good of you to rewrite what you've $p$osted in a (more) readable format. At present, it reminds me of [this](http://www.campbellso$u$p.com/Images/products/2288.png).2011-09-10