## Thursday, August 22, 2019

### Integrals with the Error (erf) Function

Integrals with the Error (erf) Function

Introduction - Integrals Involving the Erf Function

erf(x) = 2 / √π * ∫( e^-(t^2), t, 0, x)   ** graphing calculator syntax

Problem 1:  ∫( e^-(t^2), t, a, x) with a > 0 and x > a

∫( e^-(t^2), t, 0, x) = ∫( e^-(t^2), t, 0, a)  + ∫( e^-(t^2), t, a, x)

∫( e^-(t^2), t, 0, x) - ∫( e^-(t^2), t, 0, a)  = ∫( e^-(t^2), t, a, x)

√π/2 * 2/√π * (  ∫( e^-(t^2), t, 0, x) - ∫( e^-(t^2), t, 0, a) ) =   √π/2 * 2/√π * ∫( e^-(t^2), t, a, x)

√π/2 * ( 2/√π *  ∫( e^-(t^2), t, 0, x) - 2/√π *  ∫( e^-(t^2), t, 0, a) ) =   ∫( e^-(t^2), t, a, x)

√π/2 * ( erf(x) - erf(a)  ) =   ∫( e^-(t^2), t, a, x)

∫( e^-(t^2), t, a, x) = √π/2 * (erf(x) - erf(a))

Problem 2:  ∫( e^-((t + b)^2), t, 0, x)

∫( e^-((t + b)^2), t, 0, x)

Substitutions:
u = t + b
u - b = t
du = dt

Limits:
t  = x, u = x + b
t = 0, u = b

∫( e^-((t + b)^2), t, 0, x)

= ∫( e^-(u^2), u, b, b+x)

Note √π/2  * 2 /√π  = 1

=  √π/2  * 2 /√π  * ∫( e^-(u^2), u, b, b+x)

Using problem 1 as an aid to help us get:

= √π/2  * (eft(x + b) - erf(x))

∫( e^-((t + b)^2), t, 0, x) = √π/2  * (eft(x + b) - erf(x))

Problem 3:   ∫( e^(-a*t^2 + b), t, 0, x)

∫( e^(-a*t^2 + b), t, 0, x)

= ∫( e^(-a*t^2) * e^b, t, 0, x)

e^b is a constant.

= e^b * ∫( e^(-a*t^2), t, 0, x)

Substitutions:
u = √a * t
u^2 = a * t^2

du = dt * √a
du / √a = dt

Limits:
t  = x, u = √a * x
t = 0, u = 0

= e^b * 1/√a  * ∫( e^(-u^2), u, 0, √a * x)

= e^b * 1/√a  * √π/2  * 2 /√π  * ∫( e^(-u^2), u, 0, √a * x)

= e^b * 1/√a  * √π/2  * erf(√a * x)

∫( e^(-a*t^2 + b), t, 0, x) = e^b * 1/√a  * √π/2  * erf(√a * x)

Problem 4:  ∫( e^-(ln^2 t)/t, t, 1, x)

∫( e^-(ln^2 t)/t, t, 1, x)

Substitutions:
u^2 = ln^2 t
u = ln t
du = 1/t dt
t  du = dt

1/t * t = 1

Limits:
t = x, u = ln x
t = 1, u = ln 1 = 0

= ∫( e^-(u^2), u, 0, ln x)

= √π/2  * 2 /√π  * ∫( e^-(u^2), u, 0, ln x)

= √π/2  * erf(ln x)

∫( e^-(ln^2 t)/t, t, 1, x) = √π/2  * erf(ln x)

Problem 5:  2/√π * ∫(e^(t^2), t, 0, x)

2/√π * ∫(e^(t^2), t, 0, x)

Substitutions:
-u^2 = t^2
u^2 = -t^2
u = i * t   where i = √ -1
du = i dt
- i du = dt  because 1/i = -i

Limits:
t = x, u = i *x
t = 0, u = 0

= 2/√π * ∫(e^-(u^2) * -i, t, 0, i*x)

= 2/√π * -i * ∫(e^-(u^2), t, 0, i*x)

= -i * erf(i*x)

2/√π * ∫(e^(t^2), t, 0, x) = -i * erf(i * x)

Alternative substitution:
-u^2 = t^2
i * u = t
u = -i * t
du = -i dt
i du = dt

Limits:
t = x, u = -i *x
t = 0, u = 0

Then we get

2/√π * ∫(e^(t^2), t, 0, x)

= 2/√π * i * ∫(e^-(u^2), t, 0,- i*x)

= i * erf(-i * x)

In summary....

2/√π * ∫(e^(t^2), t, 0, x) = -i * erf(i * x) = i * erf(-i * x)

These are several of integrals using the erf function. - Eddie

All original content copyright, © 2011-2019.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

### Retro Review: Texas Instruments BA-Solar

Retro Review: Texas Instruments BA-Solar Finance + Solar + 1980s Quick Facts Model:  BA-Solar Company:  Texas Instruments Years:  1986 - 199...