**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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