Jim Lambers
MAT 280
Fall Semester 2016-17
Lecture 1 Example
Consider the problem of evaluating the improper integral
Z ∞
2
e−x dx.
−∞
2
It has been proven that the function e−x does not have an antiderivative in terms of elementary
functions. Therefore, to evaluate this integral, we must take a nonstandard approach. From
the following steps,
Z
∞
−x2
e
"Z
∞
−x2
dx =
e
−∞
2 #1/2
dx
−∞
∞
Z
=
−x2
e
−∞
Z ∞
=
Z
=
Z
dx
e
2
−∞
Z ∞
Z
∞
e−x dx
−∞
∞
−x2
e
=
e
Z Z
=
−x2 −y 2
e
e−x
−∞ −∞
∞ Z ∞
−∞
−y 2
e
Z
=
1/2
dx
2
e−y dy
1/2
1/2
dy
dx
−∞
e
−∞ −∞
Z ∞ Z ∞
−x2
−∞
−∞
Z ∞ Z ∞
=
∞
2 −y 2
1/2
dy dx
1/2
dy dx
−(x2 +y 2 )
1/2
dy dx
−∞
−(x2 +y 2 )
e
1/2
dA
,
R2
2
2
we have now expressed our original integral as the square root of a double integral of e−(x +y )
over all of two-dimensional space, instead of the entire real number line. It might appear that
this makes the original problem even more difficult, but in fact it does the opposite, for the
switch to two dimensions allow us to use polar coordinates (r, θ) instead of (x, y), which are
related by
x = r cos θ, y = r sin θ.
From these relations, we have
x2 + y 2 = (r cos θ)2 + (r sin θ)2 = r2 (cos2 θ + sin2 θ) = r2 ,
2
so our integrand becomes e−r . But can the element of area dA = dx dy be expressed in polar
coordinates?
To answer this, suppose that a polar rectangle r1 ≤ r ≤ r2 , θ1 ≤ θ ≤ θ2 is mapped
to (x, y)-space, or rectangular coordinates, using x = r cos θ and y = r sin θ. Then, if the
1
Figure 1: Mapping a polar rectangle to an approximate parallelogram in rectangular coordinates
rectangle has dimensions dr and dθ that are infinitesimally small, the rectangle is mapped to
an approximate parallelogram in rectangular coordinates, as can be seen in Figure 1. The area
of this parallelogram, which is dA, can be expressed in terms of dr and dθ using the formula
for the area of a parallelogram
A = ku × vk
where u and v are vectors describing adjacent edges of the parallelogram, and kxk is the
magnitude, or length, of a vector x.
How can we get the vectors u and v? For example, one edge of the parallelogram, the bottom
edge in the figure, corresponds to the side of the rectangle r1 ≤ r ≤ r2 , θ = θ1 . Therefore, the
edge of the parallelogram is given by the vector
u = hx(r2 , θ1 ) − x(r1 , θ1 ), y(r2 , θ1 ) − y(r1 , θ1 )i,
where x(r, θ) = r cos θ and y(r, θ) = r sin θ are the functions that perform the mapping from
polar to rectangular coordinates. Now, if we keep in mind that along this edge, only r is varying
while θ is kept constant, we can use the Mean Value Theorem
f (b) − f (a) = f 0 (c)(b − a),
where c is some point satisfying a ≤ c ≤ b. Applying the Mean Value Theorem only along the
r-direction yields
x(r2 , θ1 ) − x(r1 , θ1 ) =
∂x ∗
(r , θ1 )(r2 − r1 ) = xr (r∗ , θ1 ) dr
∂r
where ∂x/∂r, also written as xr , is the partial derivative of x with respect to r. That is, x is
treated as a function of only r, and is differentiated with respect to r while θ is held constant.
The point r∗ at which xr is evaluated is some value satisfying r1 ≤ r∗ ≤ r2 .
Applying the Mean Value Theorem to both the x- and y-coordinates of two adjacent edges
u and v of the parallelogram, we obtain
u = hxr dr, yr dri,
v = hxθ dθ, yθ dθi.
To compute the area of the parallelogram, we need to compute the cross-product of these
vectors. Since they are only two-dimensional, we add a z-coordinate of zero to both. We then
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have
dA = dx dy
≈ khx(r2 , θ1 ) − x(r1 , θ1 ), y(r2 , θ1 ) − y(r1 , θ1 ), 0i ×
≈
=
=
=
=
hx(r1 , θ2 ) − x(r1 , θ1 ), y(r1 , θ2 ) − y(r1 , θ1 ), 0ik
∂x
∂y
∂x
∂y
∂r dr, ∂r dr, 0 × ∂θ dθ, ∂θ dθ, 0 ∂x ∂y
∂x ∂y
,
,
0
dr
×
,
,
0
dθ
∂r ∂r
∂θ ∂θ
∂x ∂y
∂x ∂y
∂r , ∂r , 0 × ∂θ , ∂θ , 0 dr dθ
0, 0, ∂x ∂y − ∂y ∂x dr dθ
∂r ∂θ
∂r ∂θ ∂x ∂y ∂y ∂x ∂r ∂θ − ∂r ∂θ dr dθ.
Then, recalling that x = r cos θ and y = r sin θ, we obtain
∂(r cos θ) ∂(r sin θ) ∂(r sin θ) ∂(r cos θ) dr dθ
dA = −
∂r
∂θ
∂r
∂θ
= |(cos θ)(r cos θ) − (sin θ)(−r sin θ)| dr dθ
= |r cos2 θ + r sin2 θ| dr dθ
= |r(cos2 θ + sin2 θ)| dr dθ
= |r| dr dθ
= r dr dθ.
Note the factor of r that is introduced; this will be very helpful when we return to our integral.
Now, we can express our double integral in terms of polar coordinates. Using the fact that
R2 , the entire 2-D space, can be described in both rectangular and polar coordinates as
R2 = {(x, y)| − ∞ < x < ∞, −∞ < y < ∞}
= {(r, θ)|0 ≤ r < ∞, 0 ≤ θ ≤ 2π}
we can reduce this double integral to a single integral as follows:
Z 2π Z ∞
Z Z
2
−(x2 +y 2 )
e
dA =
e−r r dr dθ
2
R
0Z 2π 0 Z ∞
−r2
=
dθ
e r dr
0
0
Z ∞
2π
−r2
=
θ|0
e r dr
Z ∞ 0
2
= 2π
e−r r dr.
0
2
Here, we use the fact that the integrand e−r r only depends on r, not θ; therefore, the double
integral, when expressed in polar coordinates, can actually be viewed as a product of two single
3
integrals with respect to different variables. Therefore, they can be evaluated separately, and
then multiplied.
Finally, we can evaluate this integral using u-substitution:
Z Z
e
−(x2 +y 2 )
b
Z
dA = 2π lim
b→∞ 0
R2
2
e−r r dr
Z 2
1 b −u
= 2π lim
e du,
b→∞ 2 0
Z b2
e−u du
= π lim
u = r2 ,
du = 2r dr
b→∞ 0
b2
= π lim −e−u 0
b→∞
h
i
2
= π lim −e−b − (−e0 )
b→∞
h
i
2
= π lim 1 − e−b
b→∞
−b2
= π 1 − lim e
b→∞
= π.
We then conclude that our original integral is
sZ Z
Z
∞
−∞
2
e−x dx =
e−(x2 +y2 ) dA =
√
π.
R2
Thus, an apparent complication of changing a one-dimensional integral to a two-dimensional
integral, also changed an integrand that was impossible to anti-differentiate into one that was
relatively simple.
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