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200 Chapter 2 The Heat Equation

And finally, the error function supplies the integral

∫b

a

e−y^2 dy=

√

π

2

(

erf(b)−erf(a)

)

. (6)

The reason for the choice of the constant in front of the integral in Eq. (4) is
to make

z→∞limerf(z)=^1. (7)
To see that this is true, define

A=

∫∞

0

e−y^2 dy.

We are going to show thatA=

√

π/2. First writeA^2 as the product of two
integrals,

A^2 =

∫∞

0

e−y^2 dy

∫∞

0

e−x^2 dx.

Remember that the name of the variable of integration in a definite integral
is immaterial. This expression forA^2 can be interpreted as an iterated double
integral over the first quadrant of thex,y-plane, equivalent to

A^2 =

∫∞

0

∫∞

0

e−(x^2 +y^2 )dx dy.

Now change to polar coordinates. The first quadrant is described by the in-
equalities 0<r<∞,0<θ<π/2, and the element of area in polar coordi-
nates isrdrdθ.Thus,wehave

A^2 =

∫π/ 2

0

∫∞

0

e−r^2 rdrdθ. (8)

This integral, which can be evaluated by elementary means (see Exercise 2),
has valueπ/4. HenceA=√π/2 and Eq. (7) is validated.
Many workers also use thecomplementary error function,erfc(z),definedas

erfc(z)=√^2

π

∫∞

z

e−y^2 dy. (9)

By using Eq. (7) we obtain the identity

erfc(z)= 1 −erf(z). (10)

Some properties of the complementary error function are found in Exercise 3.