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The Natural Logarithm
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34.1 DEFINITION We already know the formula x r+1 +C [r = 1] r+1 There remains the problem of nding an antiderivative of x 1 . Figure 34-1 shows the graph of y = 1/t for t > 0; it is one branch of a hyperbola. For x > 1, the de nite integral x r dx =
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1 dt t
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represents the area under the curve y = 1/t and above the t-axis, between t = 1 and t = x.
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Fig. 34-1 De nition: ln x =
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1 dt t
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The function ln x is called the natural logarithm. By Theorem 31.1, Dx (ln x) = 1 x
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(34.1)
THE NATURAL LOGARITHM
[CHAP. 34
and so the natural logarithm is the desired antiderivative of x 1 , but only on (0, ). An antiderivative for all x = 0 will be constructed in the following section.
PROPERTIES ln 1 = 0.
PROPERTY 1.
This follows from PROPERTY 2.
1 dt = 0. t
If x > 1, ln x > 0.
This is apparent from the area interpretation (Fig. 34-1) or, more rigorously, from Problem 30.11. PROPERTY 3. In fact, and, for 0 < x < 1, If 0 < x < 1, ln x < 0. ln x =
1 dt = t
1 dt t
[reversing limits of integration]
1 dt > 0 t
by Problem 30.11. PROPERTY 4. 1 dx = ln |x| + C x [x = 0].
In other words, ln |x| is an antiderivative of x 1 for all nonzero x. The proof is simple. When x > 0, then |x| = x, and so Dx (ln |x|) = Dx (ln x) = When x < 0, then |x| = x, and so Dx (ln |x|) = Dx (ln ( x)) = Du (ln u)Dx (u) 1 ( 1) u 1 1 = = u x = PROPERTY 5. ln uv = ln u + ln v. [by the chain rule; u = x > 0] [by (34.1)] 1 x [by (34.1)]
For a proof, see Problem 34.2. u PROPERTY 6. ln = ln u ln v. v u Proof: In Property 5, replace u by . v 1 PROPERTY 7. ln = ln v. v Proof: Let u = 1 in Property 6 and use Property 1. PROPERTY 8. For any rational number r, ln x r = r ln x.
See Problem 34.3 for a proof.
CHAP. 34]
THE NATURAL LOGARITHM
PROPERTY 9.
ln x is an increasing function. 1 > 0, ln x must be an increasing function. x ln u = ln v implies u = v. 1 < ln 2 < 1. 2
Proof: Since Dx (ln x) = PROPERTY 10.
This follows from Property 9. Since ln x is increasing, it can never repeat a value. PROPERTY 11.
Proof: The maximum of 1/t on [1, 2] is 1, and the minimum is 1 . Hence, by Problem 30.3(b), 1 (2 1) < 2 2 1(2 1); that is,
(1/t)dt <
< ln 2 < 1. The strict inequalities follow from Problem 30.11.
A more intuitive proof would use the area interpretation of
(1/t) dt.
We shall see later that ln 2 is 0.693 , and we shall assume this value in what follows. PROPERTY 12.
x +
lim ln x = + .
Proof: By Property 9, we need only show that ln x eventually exceeds any given positive integer k. For x > 22k , ln x > ln 22k = 2k ln 2 So PROPERTY 13. Proof: Let u = 1 ln x > 2k 2
x 0+
[by Property 8] [by Property 11]
lim ln x = .
1 . As x 0+ , u + . So, x lim ln x = lim ln
u +
x 0+
1 = lim ln u u u + = lim ln u =
u +
[by Property 7] [by Property 12]
Solved Problems
34.1 Sketch the graph of y = ln x.
We know that ln x is increasing (Property 9), that ln 1 = 0 (Property 1), and that 1 < ln 2 < 1 (Property 11). From 2 the value y = ln 2 = 0.693 we can estimate the y-values at x = 4, 8, 16, and at x = 1 , 1 , 1 , . . . by Property 8, 2 4 8 ln 4 = 2 ln 2 1 ln = ln 2 2 ln 8 = 3 ln 2 1 ln = 2 ln 2 4 ln 16 = 4 ln 2 . . . 1 ln = 3 ln 2 . . . 8
2 Dx (ln x) = Dx (x 1 ) = x 2 = 1/x 2 < 0 and, therefore, the graph is concave downward. There is no horizontal asymptote (by Property 12), but the negative y-axis is a vertical asymptote (by Property 13). The graph is sketched in Fig. 34-2. Notice that ln x assumes all real numbers as values.
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