Integrating 1/x: From Undefined to Natural Logs

In summary, the integral of 1/x is ln(x), which can be derived using both the limit definition of a derivative and the fundamental theorem of calculus. The natural logarithm function has two equivalent definitions, one being ln(x) = \int_1^x dt/t and the other being \exp(\ln{x}) = x. This allows us to define the inverse of ln(x) as exp(x), which is just a number (e) to the x power.
  • #1
DiracPool
1,243
516
The integral of 1/x is ln(x). Where does that come from? That always puzzled me. We can continue to take derivatives through x^0 and into the negative integers, and just use the plain old power rule to get the answers. We can do the same for the integral of x all the way from negative exponents through positive exponents with the exception of x^-1. If we try to take the integral here, we get x^0/0, which is 1/0, and is undefined. OK, I get that, but how do we get a natural logarithm out of this undefined expression?
 
Physics news on Phys.org
  • #2
First use the limit definition of a derivative on ln(x) to show that dln(x)/dx=1/x then use the fundamental theorem of calculus to show that the antiderivative of 1/x is ln(x).
 
  • #3
Hi DiracPool! :smile:
DiracPool said:
If we try to take the integral here, we get x^0/0, which is 1/0, and is undefined. OK, I get that, but how do we get a natural logarithm out of this undefined expression?

Let's find [itex]\lim_{n\rightarrow 0}\int_1^{e^a} x^{n-1} dx[/itex]

that's limn->0 (ena - 1)/n

= limn->0 (na + n2a2/2! + …)/n

= a = [ln(ea) - ln(1)] :wink:
 
  • #4
bp_psy said:
First use the limit definition of a derivative on ln(x) to show that dln(x)/dx=1/x then use the fundamental theorem of calculus to show that the antiderivative of 1/x is ln(x).

Agreed. Using the rule for other exponents here gives us an undefined integral. Does this mean the integral of 1/x is undefined? No, it just means we have to find another way to evaluate the integral (FTC, as bp_psy pointed out,) as there's obviously an expression for the area under the curve 1/x, so long as both bounds are on the same side of the y-axis.

Just for a few other examples where applying a rule to find a defined expression results in an undefined answer:

Quotient rule on [itex]\lim\limits_{x\to0}\left(\dfrac xx\right)[/itex]

L'Hôpital twice on [itex]\lim\limits_{x\to0}\left(\dfrac{x\cdot\sin \left(x\right)}{x^2}\right)[/itex]

Both of these result in meaningless answers, but both limits are clearly defined. So we have to find another method to solve them.
 
  • #5
Great, thanks for the insight everyone.
 
  • #6
Many newer Calculus texts handle this by noting, first, that the rule "[itex]\int x^n dx= 1/(n+1) x^{n+1}+ C[/itex]" cannot be used for n= -1 because it would involve dividing by 0 and then defining [itex]ln(x)= \int_1^t dt/t[/itex]
From that definition one can derive all of the properties of ln(x), including the facts that ln(ab)= ln(a)+ ln(b), that [itex]ln(a^b)= bln(a)[/itex], that ln(x) is invertible, and that its inverse is [itex]e^x[/itex] where is the unique number satisfying 1= ln(e).
 
  • #7
The natural logarithm function [itex]\ln{x}[/itex] has two equivalent definitions.
1) [itex]\ln{x}[/itex] is the function such that [itex]\exp(\ln{x})=\ln(e^x)=x[/itex], and
2) [itex]\ln{x}=\int_1^x \frac{dt}{t}[/itex].

Proof. From definition 1 to 2:
The derivative of [itex]\exp(\ln{x})[/itex] is [itex]\exp(\ln{x})\frac{d}{dx}\ln{x}[/itex]. But [itex]\exp(\ln{x})=x[/itex], so [itex]x\frac{d}{dx}\ln{x}=\frac{d}{dx}x=1\implies\frac{d}{dx}\ln{x}= \frac{1}{x}[/itex]. By the Fundamental Theorem of Calculus, [itex]\frac{d}{dx}\int_1^x \frac{dt}{t}=\frac{1}{x}[/itex]. Since [itex]f'(x)=g'(x)\implies f(x)=g(x)+C[/itex] for some [itex]C\in\mathbb{R}[/itex], we have [itex]\ln{x}=\int_1^x \frac{dt}{t}[/itex] (as it turns out, [itex]C=0[/itex]).

From definition 2 to 1:
Use the substitution [itex]t=e^u,\,\,dt=e^udu[/itex] to get the integral [itex]\ln{x}=\int_{g(1)}^{g(x)} \frac{e^udu}{e^u}[/itex] where [itex]g(e^x)=e^{g(x)}=x[/itex]. From this we can surmise that [itex]g(1)=0[/itex]. The integral simplifies to [itex]\ln{x}=\int_0^{g(x)}du=\left. u\right|_0^{g(x)}=g(x)[/itex]. So [itex]\ln{x}[/itex] satisfies [itex]\exp(\ln{x})=\ln(e^x)=x[/itex].
 
Last edited:
  • #8
Since you post that, it is worth noting:
If we define [itex]ln(x)= \int_1^x dt/t[/itex] it is easy to show that, since 1/t is continuous for all t except 0 ln(x) is defined for all positive x. Of course, 1/x, the derivative of ln(x) (by the fundamental theorem of Calculus) is positive for all positive x, ln(x) is an increasing function which means that ln(x) is a "one-to-one" function mapping the set of all positive numbers to the set or all real numbers. That tells us that it has an inverse and we can define "exp(x)" to be that inverse.

Now, the crucial point is this: if y= exp(x) then, since exp(x) is the inverse function to ln(x), x= ln(y). If x is not 0, that is equivalent to [itex]1= (1/x)ln(y)= ln(y^{1/x})[/itex]. Going back to the exponential form, [itex]exp(1)= y^{1/x}[/itex] so that [itex]y= exp(x)= (exp(1))^x[/itex]. That is, that inverse function really is just a number (and we can define e= exp(1)) to the x power.
 

Related to Integrating 1/x: From Undefined to Natural Logs

1. What is the process for integrating 1/x?

The process for integrating 1/x, also known as the natural logarithm, involves using the formula ∫1/x dx = ln|x| + C, where C is the constant of integration. This means that the integral of 1/x is equal to the natural logarithm of the absolute value of x plus a constant.

2. Why is the integral of 1/x undefined at x = 0?

The integral of 1/x is undefined at x = 0 because when x is equal to 0, the denominator becomes 0. This results in a division by 0 error, which is undefined in mathematics. This is why the natural logarithm is considered a discontinuous function.

3. What is the relationship between integrating 1/x and the natural logarithm?

The process of integrating 1/x is closely related to the natural logarithm. In fact, the natural logarithm is the inverse function of e^x, also known as the exponential function. This means that the natural logarithm "undoes" the effect of the exponential function, and is represented by the integral of 1/x.

4. Can the constant of integration be any value?

Yes, the constant of integration (C) can be any real number. This is because when we take the derivative of ln|x|, the constant disappears, so any value of C will result in the same derivative. However, when solving a specific problem or finding the exact value of an integral, the constant should be determined based on the given information.

5. How can integrating 1/x be applied in real life situations?

The natural logarithm and integrating 1/x have many real-life applications, such as in finance, physics, and chemistry. In finance, the natural logarithm is used in compound interest calculations and in modeling population growth. In physics, it is used in radioactive decay and in calculating the half-life of a substance. In chemistry, it is used in the Arrhenius equation to determine the rate of a chemical reaction.

Similar threads

Replies
4
Views
3K
Replies
14
Views
2K
Replies
14
Views
1K
Replies
25
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
3
Views
1K
Replies
10
Views
4K
Back
Top