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.
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.
