1
$\begingroup$

So I know that $$\displaystyle \int_{0}^{\infty} \text{erf}(x) dx$$ does not converge so I am assuming that $$\displaystyle \int_{0}^{\infty} \frac{\text{erf}(x)}{x} dx$$ does not converge? Is there anyway to estimate these integrals?

By the way I arrived at that equation from the following:

$$\displaystyle \int_{0}^{p} \frac{\text{exp}(-r^2)r^2}{p} dr \approx \frac{\text{erf}(p)}{p}$$

so maybe my evaluation of that integral is wrong? Thanks for the help guys!

$\endgroup$
6
  • 3
    $\begingroup$ $$\int_0^{\infty} dx \operatorname{erfc}{x} = \frac1{\sqrt{\pi}} $$ $\endgroup$
    – Ron Gordon
    Nov 30, 2015 at 2:40
  • $\begingroup$ erf is essentially the constant $1$ for large $x.$ So the integrand in your second integral is larger than $1/(2x)$ and the integral larger than $(1/2) \log N$ where the integral is from $1$ out to $N$ $\endgroup$
    – Will Jagy
    Nov 30, 2015 at 2:46
  • $\begingroup$ @RonGordon Wait i'm confused... wolfram alpha seems to think that does not converge wolframalpha.com/input/… $\endgroup$ Nov 30, 2015 at 3:13
  • 2
    $\begingroup$ @user2879934: It converges. Integrate by parts. Think for yourself, don't substitute common sense with Dr. Wolfram's robot. $\endgroup$
    – Ron Gordon
    Nov 30, 2015 at 3:37
  • 2
    $\begingroup$ @user2879934: yes, but the point is that the relationship between erf and erfc should give you insight into how the integral over erf diverges. $\endgroup$
    – Ron Gordon
    Nov 30, 2015 at 3:50

2 Answers 2

1
$\begingroup$

If $f$ is pdf of a standard normal ($f'=-xf$), integration by parts yields:

$$I(a,b)=\int_{a}^{b} r^2f(r)\,dr = \Phi(b)-\Phi(a)+af(a)-bf(b)$$

For $a=0$ and $b$ large we get asymptotics: $$I(0,p)\approx\frac 1 2-(b+\frac 1 b)f(b)\approx \frac 1 2-bf(b)$$ You can translate the above into your integral.

$\endgroup$
0
$\begingroup$

Since: $$ \text{erf}(x)=\int_0^x\frac{2}{\sqrt{\pi}}\exp\left(-y^2\right)\mathrm{d}y $$ Thus, since all the functions are non-negative, we can change the order of integration - take care of the change of the integration limits (due to Fubini): $$ \displaystyle \int_{0}^{\infty} \frac{\text{erf}(x)}{x}\mathrm{d}x=\int_0^\infty\int_{0}^x\frac{2}{\sqrt{\pi}x}e^{-y^2}\mathrm{d}y\mathrm{d}x=\int_{0}^\infty\int_{y}^\infty \frac{2}{\sqrt{\pi}x}e^{-y^2}\mathrm{d}x\mathrm{d}y $$ And the inner integral does not converge for any $y$.

$\endgroup$
1
  • $\begingroup$ Thanks Akos! Is there any way to approximate that reliably? Or if not is there any way to solve the last integral in my question without erf? $\endgroup$ Nov 30, 2015 at 3:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .