Error Function Integration

Calculus Level 4

Evaluate: ∫ 0 ∞ ( 1 − erf ( x ) ) d x \large \int_0^\infty\left(1-\text{erf}(x)\right)\ dx

Notation: erf ( x ) = 2 π ∫ 0 x e − t 2 d t \displaystyle\text{erf}(x)=\frac2{\sqrt\pi} \int_0^x e^{-t^2}dt denotes the error function .


The answer is 0.56418958.

Joseph Newton by Joseph Newton , 20, Australia

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2 solutions

Chew-Seong Cheong
Sep 27, 2018

I = ∫ 0 ∞ ( 1 − erf ( x ) ) d x Complimentary error function erfc ( z ) = 1 − erf ( z ) = ∫ 0 ∞ erfc ( x ) d x erfc ( z ) = 2 π ∫ z ∞ e − t 2 d t = x erf ( x ) ∣ 0 ∞ − ∫ 0 ∞ 2 x e − x 2 π d x By integration by parts = lim ⁡ x → ∞ erf ( x ) 1 x − 0 − e − x 2 π ∣ 0 ∞ A 0/0 case, L’H o ˆ pital’s rule applies. = lim ⁡ x → ∞ 2 e − x 2 π − 1 x 2 − 0 − 0 + 1 π Differentiate up and down w.r.t. x = 0 + 1 π ≈ 0.564 \begin{aligned} I & = \int_0^\infty (1-\text{erf}(x)) \ dx & \small \color{#3D99F6} \text{Complimentary error function erfc}(z) = 1 - \text{erf}(z) \\ & = \int_0^\infty \text{erfc}(x) \ dx & \small \color{#3D99F6} \text{erfc}(z) = \frac 2{\sqrt \pi} \int_z^\infty e^{-t^2} dt \\ & = x\text{ erf}(x) \ \bigg|_0^\infty - \int_0^\infty \frac {2xe^{-x^2}}{\sqrt \pi} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6}\lim_{x\to \infty} \frac {\text{erf}(x)}{\frac 1x}} - 0 - \frac {e^{-x^2}}{\sqrt \pi} \ \bigg|_0^\infty & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = {\color{#3D99F6}\lim_{x\to \infty} \frac {\frac {2e^{-x^2}}{\sqrt \pi}}{-\frac 1{x^2}}} - 0 - 0 + \frac 1{\sqrt \pi} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = {\color{#3D99F6}0} + \frac 1{\sqrt \pi} \approx \boxed{0.564} \end{aligned}

7 Helpful 0 Interesting 1 Brilliant 0 Confused
Brian Moehring
Sep 30, 2018

Note that 1 − erf ( x ) = 2 π ∫ 0 ∞ e − t 2 d t − 2 π ∫ 0 x e − t 2 d t = 2 π ∫ x ∞ e − t 2 d t \begin{aligned} 1 - \text{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_0^\infty e^{-t^2}\,dt - \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt \\ &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}\,dt \end{aligned} so that ∫ 0 ∞ ( 1 − erf ( x ) ) d x = ∫ 0 ∞ 2 π ∫ x ∞ e − t 2 d t d x = 2 π ∫ 0 ∞ ∫ 0 t e − t 2 d x d t = 2 π ∫ 0 ∞ t e − t 2 d t = 2 π ∫ 0 ∞ 1 2 e − u d u = 1 π [ − e − u ∣ 0 ∞ = 1 π ≈ 0.56419 \begin{aligned} \int_0^\infty \Big(1 - \text{erf}(x)\Big)\,dx &= \int_0^\infty \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}\,dt \,dx \\ &= \frac{2}{\sqrt{\pi}} \int_0^\infty \int_0^t e^{-t^2}\,dx\,dt \\ &= \frac{2}{\sqrt{\pi}} \int_0^\infty t e^{-t^2}\,dt \\ &= \frac{2}{\sqrt{\pi}} \int_0^\infty \frac{1}{2} e^{-u}\,du \\ &= \frac{1}{\sqrt{\pi}} \left[-e^{-u}\right|_0^\infty \\ &= \frac{1}{\sqrt{\pi}} \approx \boxed{0.56419} \end{aligned}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

In the second line how did you intertwine dx and dt?

Younes Bouhafid - 2 years, 8 months ago

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I changed the order of integration.

First note that the integrand e − t 2 e^{-t^2} is never negative, which means we can change the order of integration (for those interested, this is a sufficient, not a necessary, condition).

Then to find the new limits, note that the iterated integral ∫ 0 ∞ ∫ x ∞ e − t 2 d t d x \int_0^\infty \int_x^\infty e^{-t^2}\,dt\,dx corresponds to integrating over the region 0 ≤ x ≤ t < ∞ 0 \leq x \leq t < \infty in the x t xt -plane. Using this inequality, we can see the limits on x x are x = 0 x=0 and x = t x=t , giving ∫ 0 ∞ ∫ 0 t e − t 2 d x d t \int_0^\infty \int_0^t e^{-t^2}\,dx\,dt as an equivalent iterated integral.

Brian Moehring - 2 years, 8 months ago

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