DIFFERENTIATION OF INTEGRALS; LEIBNIZ’ RULE

12. DIFFERENTIATION OF INTEGRALS; LEIBNIZ’ RULE

According to the definition of an integral as an antiderivative, if

f (t) dt = F (t) = F (x) − F (a),

where a is a constant. If we differentiate (12.2) with respect to x, we have

[F (x) − F (a)] =

by (12.1). Similarly,

f (t) dt = F (a) − F (x),

= −f(x).

dx x

dx

234 Partial Differentiation Chapter 4

Example 1. Find

sin t dt.

dx π/4 By (12.3), we find immediately that the answer is sin x. We can check this by

finding the integral and then differentiating. We get

and the derivative of this is sin x as before. By replacing x in (12.3) by v, and replacing x in (12.4) by u, we can then write

f (t) dt = −f(u).

du u

Suppose u and v are functions of x and we want dI/dx where

I=

f (t) dt.

When the integral is evaluated, the answer depends on the limits u and v. Finding dI/dx is then a partial differentiation problem; I is a function of u and v, which are functions of x. We can write

But ∂I/∂v means to differentiate I with respect to v when u is a constant; this is just (12.5), so ∂I/∂v = f (v). Similarly, ∂I/∂u means that v is constant and we can use (12.6) to get ∂I/∂u = −f(u). Then we have

d v(x)

− f(u)

dx u(x)

Example 2. Find dI/dx if I =

t 2 dt.

By (12.8) we get

3 3 We could also integrate first and then differentiate with respect to x:

3 This last method seems so simple you may wonder why we need (12.8). Look

0 3 0 3 dx

at another example.

Section 12 Differentiation of Integrals; Leibniz’ Rule 235

Example 3. Find dI/dx if

Here the indefinite integral cannot be evaluated in terms of elementary functions; however, we can find dI/dx by using (12.8). We get

dI sin(sin −1 x)

a f (x, t) dt, where a and b are constants. Under not too restrictive conditions,

Finally, we may want to find dI/dx when I = b

that is, we can differentiate under the integral sign. [A set of sufficient conditions for this to be correct would be that b

a f (x, t) dt exists, ∂f /∂x is continuous and |∂f(x, t)/∂x| ≤ g(t), where b

a g(t) dt exists. For most practical purposes this means that if both integrals in (12.9) exist, then (12.9) is correct.] Equation (12.9) is often

useful in evaluating definite integrals. Example 4. Find ∞ 0 t n

e 2 −kt dt for odd n, k > 0.

First we evaluate the integral

te −kt 2 1 dt = − 2 e −kt

I=

0 2k Now we calculate successive derivatives of I with respect to k. dI ∞

0 2k 2 0 2k 2 Repeating the differentiation with respect to k, we get

0 k 4 Continuing in this way (Problem 17), we can find the integral of any odd power of 2

t times e −kt :

t 2n+1 e −kt 2 dt = n!

0 2k n+1

Your computer may give you this result in terms of the gamma function (see Chapter

11, Sections 1 to 5). The relation is n! = Γ(n + 1).

236 Partial Differentiation Chapter 4

Example 5. Evaluate

First we differentiate I with respect to a, and evaluate the resulting integral.

0 a+1 0 a+1 Now we integrate dI/da with respect to a to get I back again (plus an integration

= ln(a + 1) + C.

a+1

If a = 0, (12.11) gives I = 0 and (12.12) gives I = C, so C = 0 and we have from (12.12), I = ln(a + 1).

It is convenient to collect formulas (12.8) and (12.9) into one formula known as Leibniz’ rule :

du v ∂f (12.13)

d v(x)

dv

+ dt. dx u(x)

f (x, t) dt = f (x, v)

− f(x, u)

dx

dx u ∂x

Example 6. Find dI/dx if

By (12.13) we get

dI e x·2x

Although you can do problems like this by computer, in many cases you can just write down the answer using (12.13) in less time than it takes to type the problem into the computer.

PROBLEMS, SECTION 12

Z √ x 1. If y =

sin t 2 dt, find dy/dx.

2. If s = 1−e dt, find ∂s/∂v and ∂s/∂u and also their limits as u and v tend to

zero.

Section 12 Differentiation of Integrals; Leibniz’ Rule 237 Z cos x sin t

dz

3. If z = dt, find

4. Use L’Hˆ opital’s rule to evaluate lim

5. If u = dt, find

, and

at x = π/2, y = π.

∂x ∂y

∂x

Hint: Use differentials. Z 2x−3y du

« 7. If e −t dt = x and u v = y, find „ ∂u , „ ∂u , and „ ∂y at u = 2, v = 0.

8. 2 If e −s ds = u, find dx .

0 du

Z π 9. If y =

sin xt dt, find dy/dx (a) by evaluating the integral and then differenti-

ating, (b) by differentiating first and then evaluating the integral.

e xu −1 du.

10. Find dy/dx explicitly if y =

d Z x 2 11. Find dx

(x − t) dt by evaluating the integral first, and by differentiating first.

3−x

12. Find d Z x 2 du

dx x ln(x + u) d Z 2/x sin xt

14. Given that

, differentiate with respect to y and so evaluate

15. Given that

differentiate with respect to a to show that

and differentiate with respect to k to show that

Z ∞ 16. In kinetic theory we have to evaluate integrals of the form I =

t n e −at 2 dt. Given Z ∞

that e −at dt =

pπ/a , evaluate I for n = 2, 4, 6, · · · , 2m. 17. Complete Example 4 to obtain (12.10).

238 Partial Differentiation Chapter 4

f (t) dt

18. Show that u(x, y) =

(x − t) −∞ 2 +y 2 satisfies u xx +u yy = 0. Z x 19. Show that y =

f (u) sin(x − u) du satisfies y ′′ + y = f (x).

20. (a) Show that y = f (x − t) dt satisfies (dy/dx) = f(x). (Hint: It is helpful to

make the change of variable x − t = u in the integral.)

(b) Show that y = (x − u)f(u) du satisfies y ′′ = f (x).

(c) Show that y = (x − u) (n) n−1 f (u) du satisfies y = f (x).

(n − 1)! 0