error function integral

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April 5th 2009, 11:21 PM
wik_chick88

error function integral

the error function ("erf") is defined as $erf(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} {e^{-u^2} du}$ .

evaluate the definite integral $\int^{\alpha}_0 {erf (x) dx}$ , where $\alpha$ is greater than ${0}$

note: does anyone know the LaTeX symbol for greater than? all i could find was greater than or equal to which is \geq
April 5th 2009, 11:43 PM
mr fantastic

Quote:

Originally Posted by wik_chick88

the error function ("erf") is defined as $erf(x) = \frac{2}{\sqrt{\pi}}\int^{x}_{0} {e^{-u^2} du}$ .

evaluate the definite integral $\int^{\alpha}_0 {erf (x) dx}$ , where $\alpha$ is greater than ${0}$

note: does anyone know the LaTeX symbol for greater than? all i could find was greater than or equal to which is \geq

The latex symbol for greater than is >.

Reverse the order of integration: $\frac{2}{\sqrt{\pi}} \int_{x = 0}^{\alpha} \left( \int_{u = 0}^{u = x} e^{-u^2} \, du \right) dx = \frac{2}{\sqrt{\pi}} \int_{u = 0}^{\alpha} \left( \int_{x = u}^{x = \alpha} e^{-u^2} \, dx \right) du$
April 5th 2009, 11:48 PM
chisigma

Integrating by parts...

$\int erf(x)\cdot dx = x\cdot erf(x) - \frac {e^{-x^{2}}}{\sqrt {\pi}} +c$

... so that...

$\int_{0}^{\alpha} erf(x)\cdot dx = \alpha\cdot erf(\alpha) + \frac {1- e^{-\alpha^{2}}}{\sqrt{\pi}}$

Kind regards

$\chi$ $\sigma$
April 6th 2009, 05:49 AM
wik_chick88

Quote:

Originally Posted by mr fantastic

The latex symbol for greater than is >.

Reverse the order of integration: $\frac{2}{\sqrt{\pi}} \int_{x = 0}^{\alpha} \left( \int_{u = 0}^{u = x} e^{-u^2} \, du \right) dx = \frac{2}{\sqrt{\pi}} \int_{u = 0}^{\alpha} \left( \int_{x = u}^{x = \alpha} e^{-u^2} \, dx \right) du$

i tried that but i keep getting a -2u on the bottom of the fraction which, when I substitute u=0, i get an undefined answer! can someone please explain how to use integration by parts in this question?
April 6th 2009, 06:37 AM
chisigma

Integration by parts...

$\int u(x)\cdot v(x)\cdot dx = U(x)\cdot v(x) + \int U(x)\cdot v^{'}(x)\cdot dx$

... where $U(x)$ is a primitive of $u(x)$ . Setting $u(x)=1$ and $v(x)=erf(x)$ you obtain...

$\int erf(x)\cdot dx= x\cdot erf(x) + \frac{2}{\sqrt{\pi}}\cdot \int x\cdot e^{-x^{2}}\cdot dx = x\cdot erf(x) - \frac{e^{- x^{2}}}{\sqrt{\pi}} + c$

Kind regards

$\chi$ $\sigma$
April 6th 2009, 01:48 PM
mr fantastic

Quote:

Originally Posted by wik_chick88

i tried that but i keep getting a -2u on the bottom of the fraction which, when I substitute u=0, i get an undefined answer! can someone please explain how to use integration by parts in this question?

There should be no trouble here:

When you do the inner integral you obviously get $\frac{2}{\sqrt{\pi}} \int_0^{\alpha} \alpha e^{-u^2} - u e^{-u^2} \, du$ .

When you evaluate this integral you get $\alpha \, \text{Erf} (\alpha) + \frac{1}{\sqrt{\pi}} \left[ e^{-u^2} \right]_{0}^{\alpha} = \, ....$