Parametric Differentiation and Integration Hongwei Chen Department of Mathematics Christopher Newport University Newport News, VA 23606
[email protected] October 20, 2008 To appear in the Int. J. Math. Ed. Sci. & Tech. Abstract Parametric differentiation and integration under the integral sign constitutes a powerful technique for calculating integrals. However, this topic is generally not included in the undergraduate mathematics curriculum. In this note, we give a comprehensive review of this approach, and show how it can be systematically used to evaluate most of the famous improper integrals in analysis. Students in mathematics and science would benefit from an exposure to this wonderful method.
In this note, we present an integration method that evaluates integrals through differentiation and integration with respect to a parameter. This approach has been a favorite tool of applied mathematicians and theoretical physicists. In his autobiography, “Surely You’re Joking, Mr. Feynman!”, eminent physicist Feynman mentioned how he frequently used this approach when confronted with difficult integrations associated with mathematics and physical problems. He referred to this approach as “a different box of tools”. However, most modern texts either ignore this subject or provide only few examples. Though [2, 5, 6] addressed this approach, they focused on either proper integrals or indefinite integrals. In the following, we illustrate this method with the help of some selected examples, most of them are improper integrals. Those examples will show that the parametric differentiation and integration technique only requires the mathematical maturity of calculus, and often provides a straightforward method to evaluate difficult integrals for which one commonly uses the method of contour integration. Students in mathematics and science will benefit from an exposure to this wonderful approach. In my view, parametric differentiation and integration should be 1
included in the undergraduate mathematics curriculum
To illustrate the basic idea, we begin with two examples.
Example 1. Show that Z ∞
sin x π dx = . x 2
0
This integral is generally evaluated by using contour integration and thus requites the theory of complex functions. Here we consider the parametric integral I(p) =
Z ∞
sin x dx, x
e−px
0
(p > 0).
Differentiating under the integral sign yields 0
I (p) = −
Z ∞
e−px sin x dx = −
0
1 . 1 + p2
Hence I(p) = − arctan p + C. Since limp→∞ I(p) = 0, it follows that I(p) =
π − arctan p. 2
Setting p = 0 leads to the desired value.
Example 2. Evaluate the Gauss normal integral G=
Z ∞
2
e−x dx.
0
One clever way of evaluating this integral resides in the introduction of polar coordinates to 2
G =
Z ∞Z ∞ 0
e−(x
2 +y 2 )
dxdy.
0
Here we offer a modification that borrows the double integrals but applies the parametric integration. Notice that if p > 0, the change of variable x = pt yields G=p
Z ∞ 0
2
e−p
2 t2
dt.
2
Multiplying both sides by e−p and then integrating p from 0 to ∞ leads to G·
Z ∞
−p2
e
2
dp = G =
Z ∞
0
−p2
pe
dp
0
Z ∞
e−p
2 t2
dt.
0
Exchanging the order of the right-hand side integration gives G2 =
Z ∞
dt
Z ∞
0
2 )p2
pe−(1+t
dp =
0
Thus,
1 Z ∞ dt π = . 2 2 0 1+t 4
√
π . 2
G=
After these two examples, a natural questions arises: How should one introduce a parameter within the integrand? In many integrals, especially in the formulas of integrals, parameters are already present. For example, differentiating the formula Z 1
xp dx =
0
1 , p+1
(p 6= −1)
with respect to p k times gives Z 1
xp (ln x)k dx =
0
(−1)k k! (p + 1)k+1
as a new integral formula. However, there exist integrals containing no parameter, like examples 1 and 2. In such cases, a parameter is generally introduced either by substitution, as we did in Example 2, or by specialization, where the given integral is viewed as a special case of a parametric integral. See Example 3 below. The above two examples also require the permission of three operations: 1. Differentiation with respect to a parameter under the integral sing 2. Integration with respect to a parameter under the integral sign 3. Exchange the limit and integral operations Many theorems have been devoted to determining when the order of these processes may be interchanged. For the proper integrals, our discussion and illustrations will be based on Leibniz’s rule for differentiation and Fubini’s theorems for interchanging orders of integration (See [1, 4] for details). Let I be an integral whose integrand f (x, p) contains a parameter p: I(p) =
Z b
f (x, p) dx.
a
3
Leibniz’s Rule. If f and its partial derivative fp are continuous in the rectangle R = {(x, p) : x ∈ [a, b], p ∈ [p1 , p2 ]}, then I(p) is differentiable in (p1 , p2 ) and Z b
0
I (p) =
a
fp (x, p) dx.
Fubini’s Theorem. If f is continuous in the rectangle R = {(x, p) : x ∈ [a, b], p ∈ [p1 , p2 ]}, then for any t ∈ [p1 , p2 ], lim
Z b
p→t
Z b
f (x, p)dx =
a
a
lim f (x, p) dx p→t
and Z t
I(p) dp =
Z t
dp
Z b
Z b
a
p1
p1
f (x, p) dx =
dx
a
Z t
f (x, p) dp.
p1
Example 3. To evaluate Z 1 0
arctan x √ dx, x 1 − x2
we introduce a parametric integral T (p) =
Z 1 0
arctan(px) √ dx. x 1 − x2
Differentiate with respect to p by Leibniz’s rule to get 0
T (p) =
Z 1 0
(1 +
dx √ . 1 − x2
p 2 x2 )
The substitution x = sin θ yields T 0 (p) =
Z π/2 0
q dθ 1 1 π 2 tan θ)|π/2 = √ √ arctan( . = 1 + p 0 2 2 1 + p2 1 + p2 sin θ 1 + p2
Since T (0) = 0, it follows that T (p) = Thus, Z 1 0
q π ln(p + 1 + p2 ). 2
√ arctan x π √ dx = T (1) = ln(1 + 2). 2 x 1 − x2
Example 4. To find I=
Z π/2 0
!
ln
a + b sin x dx · a − b sin x sin x 4
(a > b > 0),
observing that a + b sin x a − b sin x
1 ln sin x
!
Z 1
= 2ab
dy
0
a2
−
b2 y 2
−
b2 y 2
sin2 x
,
we get I = 2ab
Z π/2
dx
0
Z 1
dy a2
0
sin2 x
.
For a > b > 0, 1/(a2 − b2 y 2 sin2 x) is continuous on [0, π/2] × [0, 1]. By Fubini’s theorem, interchanging the order of the integration, we obtain K = 2ab
Z 1
dy
0
Z π/2
dx a2
0
b2 y 2
−
sin2 x
Z 1
πdy √ 2 0 2a a − b2 y 2 ! b . = π arcsin a
= 2ab
The following example shows what can go wrong when interchanging the order of integration without justification. Example 5. Consider y 2 − x2 (x2 + y 2 )2
f (x, y) =
on [0, 1] × [0, 1].
We have Z 1
f (x, y)dx =
0
x2
x 1 |10 = 2 +y 1 + y2
and so Z 1
dy
Z 1
f (x, y) dx =
0
0
π . 4
However, Z 1
dx
Z 1 0
0
π f (x, y) dy = − . 4
Here the discontinuity of f (x, y) at (0, 0) fails the interchange of integral orders. For the improper integral I(p) =
Z ∞
f (x, p) dx,
a
to avoid the absurdity in Example 5, the uniformly converges and absolute integrability are usual requirements. We say that I(p) converges uniformly on [p1 , p2 ] if for every > 0 and p ∈ [p1 , p2 ] there is a constant A ≥ a, independent of p, such that x ≥ A implies Z ∞ a
f (x, p) dx −
Z A a
f (x, p) dx
5
Z ∞
=
A
f (x, p) dx < .
There is a very convenient test for uniform convergence, due to Weierstrass. M-Test. If f (x, p) is integrable on every finite interval and there is an integrable function φ(x) on [0, ∞) such that |f (x, p)| ≤ φ(x),
for all x ≥ a, p ∈ [p1 , p2 ]
then I(p) converges uniformly on [p1 , p2 ]. When the integrand is in product form, the following criterion is sometimes useful. Dini Test. If Z A a
f (x, p) dx
is uniformly bounded in A and p, g(x, p) is monotonic in x and g(x, p) → 0 uniformly in p as x → ∞, then Z ∞
f (x, p)g(x, p) dx
a
converges uniformly on [p1 , p2 ]. As an immediate consequence of the Dini test, if Z ∞
e
−xp
f (x) dx and
R∞
Z ∞
a
f (x) dx converges, then 2
e−x p f (x) dx
a
a
are uniform convergent for p > 0. When the range of parameter p also becomes infinite, to switch the order of integration, besides the uniform convergence of Z ∞
f (x, p) dx and
Z ∞
f (x, p) dp,
b
a
Fubini’s theorem requires that one of double integrals Z ∞ b
dp
Z ∞
|f (x, p)| dx and
a
Z ∞ a
dx
Z ∞
|f (x, p)| dp
b
exists. Combining with the M-test, we have the following theorem to justify the desired steps in our examples. Extended Fubini’s Theorem. Let f (x, y) be continuous on (a, ∞) × (b, ∞). Suppose the improper integrals Z ∞
f (x, y) dy and
b
Z ∞ a
6
f (x, y) dx
exist and converge uniformly for x and y restricted to every finite interval, respectively. In addition, suppose that for s, t > a Z t
f (x, y) dx
s
and
R∞ b
≤ M (y)
M (y) dy exists. Then Z ∞
dx
Z ∞
a
f (x, y) dy =
Z ∞
b
dy
Z ∞
b
f (x, y) dx.
a
Now, we turn to more selected examples, in which most of the justifications of applications of Leibniz’s rule or Fubini’s theorem are straight forward. We will skip such a demonstration and leave the verification to the reader.
Example 6. Let α, β > 0. Evaluate Z ∞
e−α
2 2 x2 − β x2
dx.
0
Let y = αx, p = αβ. Then Z ∞
2
e
−α2 x2 − β2 x
0
1 Z ∞ −y2 − yp22 dx = e dy. α 0
Set J(p) =
Z ∞ 2 −y 2 − p2 y
e
dy.
0
Then p −y2 − yp22 e dy. y2
Z ∞
J 0 (p) = −2
0
The substitution z = p/y yields J 0 (p) = −2
Z ∞
e−z
2 2− p z2
dz = −2 J(p).
0
Thus, −2p
J(p) = J(0)e
Z ∞
=
e
−y 2
−2p
dy e
0
and so Z ∞
2
e
−α2 x2 − β2 x
0
√ π −2p = e 2
√ π −2αβ dx = e . 2α
Remark. In general, we can establish that Z ∞ 0
"
f
B Ax − x
2 #
dx =
1 Z∞ f (y 2 ) dy, A 0 7
(A > 0, B > 0),
as long as the right hand side integral converges. Example 7. Let α, β > 0. Show that H(β) =
Z ∞
e
−αx2
0
1 cos βx dx = 2
r
π −β 2 /4α e . α
Differentiating with respect to β and then integrating by parts gives 0
H (β) = −
Z ∞
2
xe−αx sin βx dx
0
1 Z∞ 2 = sin βx d(e−αx ) 2α 0 β Z ∞ −αx2 = − e cos βx dx 2α 0 β = − H(β). 2α Moreover, by Example 2, H(0) =
Z ∞
−αx2
e
0
r 1 Z ∞ −x2 1 π e dx = √ dx = . α 0 2 α
The desired integral follows by solving the initial value problem. It is interesting to see that the analogous integral Z ∞
2
e−αx sin βx dx
0
has no elementary closed form.
Example 8. Let α, β > 0. Evaluate the Laplace integral L1 (β) =
Z ∞ 0
Z ∞ cos βx x sin βx dx and L2 (β) = dx. 2 2 α +x α 2 + x2 0
Applying the Leibniz rule gets L01 (β)
=−
Z ∞ 0
x sin βx dx = − L2 (β). α 2 + x2
Here differentiating L01 (β) under integral sign once more yields −
Z ∞ 0
Z π/2β 2 Z (2n+1)π/2β 2 ∞ X x2 cos βx x cos βx x | cos βx| n+1 dx = − dx + (−1) dx. 2 2 2 2 α +x α +x α 2 + x2 0 (2n−1)π/2β n=1
Since Z (2n+1)π/2β (2n−1)π/2β
x2 | cos βx| 2((2n − 1)π/2β)2 2 dx ≥ → , as n → ∞, α 2 + x2 (α2 + ((2n − 1)π/2β)2 )β β 8
the series, and so the corresponding integral, is divergent. However, observing that sin βx x sin βx α2 sin βx = − 2 2 2 x(α + x ) x α + x2 and Z ∞
Z ∞ sin βx sin x π dx = dx = , x x 2 0
0
we have L01 (β) = −
π Z ∞ α2 sin βx + dx. 2 x(α2 + x2 ) 0
Then L001 (β) =
Z ∞ 0
α2 cos βx dx = α2 L1 (β). α 2 + x2
So, L1 (β) = c1 eαβ + c2 e−αβ , where c1 and c2 are arbitrary constants. Since L1 (β) ≤
Z ∞ 0
α2
dx π = , 2 +x 2α
L1 (β) is uniformly bounded. Therefore, c1 = 0, c2 = L1 (0) =
π 2α
and consequently L1 (β) =
π −αβ π e and L2 (β) = −L01 (β) = e−αβ . 2α 2
L1 also can be recaptured by using parametric integration. Indeed, noticing Z ∞ 1 2 2 = e−t(α +x ) dt, 2 2 α +x 0
we have L1 =
Z ∞
cos βx dx
0
Z ∞
e−t(α
2 +x2 )
dt.
0
Interchanging the integral order yields L1 =
Z ∞ 0
−α2 t
e
dt
Z ∞ 0
9
2
e−tx cos βx dx.
Example 7 implies the inner integral Z ∞
e
−tx2
0
1 cos βx dx = 2
r
π −β / 4t e . t
Finally, by Example 6, we arrive at √ Z∞ π dt 2 2 L1 = e−α t−β /4t √ 2 0 t √ Z ∞ −α2 s2 −β 2 /4s2 = π e ds 0 π −αβ = e . 2α
(let t = s2 )
Example 9. Evaluate the Fresnel integrals Fc =
Z ∞ 0
cos x2 dx and Fs =
Z ∞
sin x2 dx.
0
It is standard to substitute x for x2 , so 1 Z ∞ cos x 1 Z ∞ sin x √ dx and Fs = √ dx. Fc = 2 0 x 2 0 x Recall that 1 2 Z ∞ −xy2 √ =√ e dy. x π 0 Then Z ∞ 1 Z∞ 2 e−xy dy cos x dx Fc = √ π 0 0 Z ∞ 1 Z∞ 2 = √ dy e−xy cos x dx π 0 0 r Z ∞ dy 1 π 1 1 π =√ √ = . = √ π 0 1 + y4 π 2 2 2 2
Similarly, 1 Fs = 2
r
π . 2
Example 10. Evaluate F =
Z ∞ 0
e−ax − e−bx dx x
(a, b > 0)
Recall that Z b a
e−yx dy =
e−ax − e−bx . x
10
This yields F =
Z ∞
dx
0
Z b
e
−yx
dy =
a
Z b
dy
Z ∞
a
−yx
e
dx =
0
Z b a
1 b dy = ln . y a
Remarks. F belongs to the family of Frullani integrals, which are defined by Z ∞ 0
f (ax) − f (bx) dx x
(a, b > 0).
In general, if f (x) is continuous on [0, ∞) and f (+∞) = lim f (x) x→∞
exists, then Z ∞ 0
f (ax) − f (bx) b dx = (f (0) − f (+∞)) ln . x a
If limx→∞ f (x) has no finite limit, but Z ∞ A
f (x) dx x
exists, then Z ∞ 0
f (ax) − f (bx) b dx = f (0) ln . x a
Similarly, if f fails to be continuous at x = 0, but Z A 0
f (x) dx x
exists, then Z ∞ 0
f (ax) − f (bx) b dx = f (∞) ln . x a
Finally, we present two examples of what can go wrong when differentiating under the integral sign or interchanging the order of integration without justification. In 1851 Cauchy obtained the result Z ∞ 0
1 sin(x ) cos(px) dx = 2 2
r
π p2 cos 2 4 "
!
p2 − sin 4
!#
.
He then differentiated under the integral sign with respect to p yielding Z ∞ 0
p x sin(x ) sin(px) dx = 4 2
r
11
π p2 sin 2 4 "
!
p2 + cos 4
!#
.
(1)
This formula has subsequently been reproduced and still appears in standard tables today. However, [3] recently proved that the integral in (1) is divergent! Next, the direct calculation shows Z ∞ 1
dy
Z ∞ 1
Z ∞ Z ∞ y 2 − x2 y 2 − x2 π π and dx dx = − dy = . 2 2 2 2 2 2 (x + y ) 4 (x + y ) 4 1 1
Since both Z ∞
dy
1
Z ∞ 1
Z ∞ Z ∞ |y 2 − x2 | |y 2 − x2 | dx and dx dy (x2 + y 2 )2 (x2 + y 2 )2 1 1
diverge, Fubini’s theorem does not apply to the given function. In conclusion, to give the reader some practice in applying this method to evaluate integrals, we list a series of problems in no particular order in the Appendix. Acknowledgment. The author wish to thank the referees and the editor for their thoughtful comments and suggestions, which resulted in significant improvements of this article. References 1. T. M. Apostol, Mathematics Analysis, 2nd edition, Addison-Wesley, 1974. 2. L. R. Bragg, Parametric Differentiation Revisited, American Mathematical Monthly, 98(1991) 259-262. 3. E. Talvila, Some divergent trigonometric integrals, American Mathematical Monthly, 108(2001) 432-436. 4. E. Talvila, Necessary and sufficient conditions for differentiating under the integral sign, American Mathematical Monthly, 108(2001) 544-548. 5. J. Wiener, Differentiation with respect to a parameter, College Mathematics Journal, 32(2001) 180-184. 6. A. J. Zajta & S. K. Goel, Parametric integration techniques, Mathematics Magazine, 62(1989) 318-322.
12
Appendix 1. Let α, β, k > 0. Evaluate (a) (b)
R ∞ 1−cos αx −kx e dx 0 x R ∞ sin αx sin βx −kx 0
x
·
x
e
dx
2. Let a, b > 0. Prove that (a) (b)
R ∞ ln(1+a2 x2 )
0 b2 +x2 R ∞ arctan ax 0 x(1+x2 )
dx =
dx =
π 2
π b
ln(1 + ab).
ln(1 + a).
R a+b bx dx = π2 ln (a+b) (c) 0∞ arctan ax·arctan . x2 aa ·bb
3. For s, t > 0, evaluate
Z ∞ −sx e cos ax − e−tx cos bt
t
0
4. Evaluate
Z ∞
I(p) =
−1
x−1/2 e−ax−px
dt.
dx.
0
5. Evaluate
Z ∞
e
−x2
0
α2 cos 2 dx x
6. Define I(p) =
Z ∞ 0
Z ∞ −p(1+x2 ) e
1 + x2
−∞
Show that I 0 (p) = −
2
e−x sin
and
Z ∞
2
e−x dx
α2 dx. x2
dx.
p−1/2 e−p .
−∞
Using the solution recaptures that Z ∞
2
e−x dx =
√
π.
−∞
7. Prove that
2 Z ∞ xe−x dx
0
8. Evaluate
Z ∞
Z ∞ −x2 e dx
a √ =√ π x2 + a2 2
(a > 0).
x2 + a2
0
e−ax cos(px2 ) dx and
Z ∞
2
e−ax sin(px2 ) dx.
0
0
9. Monthly Problem 11001. Evaluate Z ∞
a arctan √
0
b 2 a + x2
√
dx + x2
a2
in closed form for a, b, > 0. 10. Monthly Problem 11113. Evaluate
√
Z ∞ Z ∞ −k x2 +y2 e sin(ax) sin(bx) p Ik (a, b) = dxdy 0
xy x2 + y 2
0
in closed form for a, b, k > 0.
13
