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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

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