Why is \displaystyle\int \frac{\mathrm{d}x}{x} = \ln x ?

One essential feature of logarithms is that they make a multiplication problem equivalent to an addition problem, by which I mean
ln(ab)=ln(a)+ln(b) ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b)

Meanwhile, ∫1x dx ∫1xdx\int\frac{1}{x}\mathrm{d}x is usually thought of geometrically as the area underneath a curve.  The problem, then, is to try to see visually what an area under a curve has to do with turning multiplication into addition.

Here's a graph of 1/x 1/x1/x, and we're finding, as an example, the area under it from 1 to 2.

Let's say now that we multiply the limits of integration by two, so we're now finding the area from 2 to 4.  Here's what that looks like.

The two portions are actually very similar to each other in their overall shape.  The orange one is twice as wide as the green one, but also half as tall.  Here they are overlaid.

If you take the green shape and first squash it down vertically by a factor of two then stretch it out horizontally by a factor of two, you get the orange shape exactly. (If you don't believe this, convince yourself it works!)  This means the areas of these shapes are exactly the same, even though we don't know what that area is.  (Pro tip: it turns out to be 0.69)

This result is general.  The area under 1/x 1/x1/x from a aa to b bb is the same as that from a∗c a∗ca*c to b∗c b∗cb*c.

What, then, is the area from 1 to 6?  We can break it into two parts – the area from 1 to 2 and the area from 2 to 6.  But the area from 2 to 6 is the same as the area from 1 to 3, by the above reasoning. 

Thus, the area from 1 to 6 is the same as the sum of the areas from 1 to 2 and from 1 to 3.  Note that 6 = 3*2.   Again, this is general.  The area under 1/x 1/x1/x from 1 11 to a∗b a∗ba*b is the same as the sum of the areas from 1 11 to a aa and from 1 11 to b bb.

That's pretty good motivation for the definition

ln(x)=∫ x 1 1t dt ln⁡(x)=∫1x1tdt\ln(x) = \int_1^x\frac{1}{t}\mathrm{d}t

Note that this is being taken as a definition of the natural logarithm, not a proof of the relationship.  However, we have established

ln(ab)=ln(a)+ln(b) ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b)

and it is evident that

ln(1)=0 ln⁡(1)=0\ln(1) = 0

This definition implies that the logarithm grows without bound because if we continually multiply the argument of the logarithm by two, we continually add ln(2) ln⁡(2)\ln(2) to the value.  (i.e. ln(2x)=ln(x)+ln(2) ln⁡(2x)=ln⁡(x)+ln⁡(2)\ln(2x) = \ln(x) + \ln(2)).  Since we can multiply any number by two over and over, we can add ln(2) ln⁡(2)\ln(2) to the logarithm as many times as we want.  That means we can make the logarithm arbitrarily big.

This also means that starting the integral from 1 rather than from zero was a good idea.  If we start from zero, the integral is infinite.  We can see this because 1/x 1/x1/x is symmetric about the line y=x y=xy = x.

This implies that the area to the left of the curve is the same as the area under the curve, like this.

We just showed that the area under the curve diverges as we move the right hand side of the integral out to infinity, so the area to the left of the curve diverges, too.  If we started the integral at zero, it would be infinite.

What about taking the logarithm of numbers less than one?  I think the discussion so far is enough that the reader can work out ln(1/x)=−ln(x) ln⁡(1/x)=−ln⁡(x)\ln(1/x) = – \ln(x).

Since the area under 1/x 1/x1/x starts at zero when x=1 x=1x=1 and goes up infinitely, it is clear that there must be some number such that ln(x)=1 ln⁡(x)=1\ln(x) = 1.  Let's choose to call that number e ee.  We don't know what it is yet, but it certainly exists.  Thus

ln(e)=1 ln⁡(e)=1\ln(e) = 1

It is immediately apparent that, for example, ln(e 5 )=ln(e∗e∗e∗e∗e)=5ln(e)=5 ln⁡(e5)=ln⁡(e∗e∗e∗e∗e)=5ln⁡(e)=5\ln(e^5) = \ln(e*e*e*e*e) = 5\ln(e) = 5. That makes e ee a pretty handy number.  It shows us that the logarithm of a number x xx is how many times you need to multiply e ee to itself in order to get x xx.

How about ln(e 3/2 ) ln⁡(e3/2)\ln(e^{3/2})?  That is ln([e 1/2 ] 3 )=3ln(e 1/2 ) ln⁡([e1/2]3)=3ln⁡(e1/2)\ln\left([e^{1/2}]^3\right) = 3\ln(e^{1/2}).  So in order to understand logarithms of rational numbers,  we need to understand roots of e ee.

That's not so hard, though.

 ln(e 1/2 ∗e 1/2 )=ln(e)=1 ln⁡(e1/2∗e1/2)=ln⁡(e)=1\ln(e^{1/2}*e^{1/2}) = \ln(e) = 1. 

On the other hand,

ln(e 1/2 ∗e 1/2 )=ln(e 1/2 )+ln(e 1/2 ) ln⁡(e1/2∗e1/2)=ln⁡(e1/2)+ln⁡(e1/2)\ln(e^{1/2}*e^{1/2}) = \ln(e^{1/2}) + \ln(e^{1/2})

From this we deduce ln(e 1/2 )=1/2 ln⁡(e1/2)=1/2\ln(e^{1/2}) = 1/2.  Returning to the unfinished example, ln(e 3/2 )=3∗(1/2)=3/2 ln⁡(e3/2)=3∗(1/2)=3/2\ln(e^{3/2}) = 3*(1/2) = 3/2.  Thus, for any rational number x xx, we have

ln(e x )=x ln⁡(ex)=x\ln(e^x) = x

Irrational numbers are squeezed in between the rational ones, and since the definition of the logarithm as the area under a curve is evidently smooth, the above relation holds for all positive numbers.  (The logarithm of a negative number or of zero isn't defined, at least not in the real numbers.  You should make sure you understand why not.)

Finally, we would like some way of determining what e e e is.  Here is one way to do it.  For small values of x xx, we can see that

ln(1+x)≈x ln⁡(1+x)≈x\ln(1+x) \approx x

This follows from the simple approximation below.

The red box is an approximation to the area of the green integral.  The red box clearly has area x xx while the green integral is ln(1+x) ln⁡(1+x)\ln(1+x).  Thus

ln(1+x)≈x ln⁡(1+x)≈x\ln(1+x) \approx x

It's crude, but it works better and better as x xx becomes tiny.  Multiplying both sides of the approximation by 1/x 1/x1/x we get

1x ln(1+x)≈1 1xln⁡(1+x)≈1\frac{1}{x}\ln(1+x) \approx 1

We know how to rewrite the left hand side.  It gives

ln([1+x] 1/x )≈1 ln⁡([1+x]1/x)≈1\ln\left([1+x]^{1/x}\right) \approx 1

Since we have defined e ee by ln(e)=1 ln⁡(e)=1\ln(e) = 1, we finally see

e=lim x→0 (1+x) 1/x  e=limx→0(1+x)1/x e = \lim_{x\to 0} (1+x)^{1/x}

This has been a purposely non-rigorous answer.  You might have fun making it more rigorous if you like.
Updated Nov 14, 2012 • View Upvotes

Answer by Mark Eichenlaub:

One essential feature of logarithms is that they make a multiplication problem equivalent to an addition problem, by which I mean
[math]\ln(ab) = \ln(a) + \ln(b)[/math]

Meanwhile, [math]\int\frac{1}{x}\mathrm{d}x[/math] is usually thought of geometrically as the area underneath a curve.  The problem, then, is to try to see visually what an area under a curve has to do with turning multiplication into addition.

Here's a graph of [math]1/x[/math], and we're finding, as an example, the area under it from 1 to 2.

Let's say now that we multiply the limits of integration by two, so we're now finding the area from 2 to 4.  Here's what that looks like.

The two portions are actually very similar to each other in their overall shape.  The orange one is twice as wide as the green one, but also half as tall.  Here they are overlaid.

If you take the green shape and first squash it down vertically by a factor of two then stretch it out horizontally by a factor of two, you get the orange shape exactly. (If you don't believe this, convince yourself it works!)  This means the areas of these shapes are exactly the same, even though we don't know what that area is.  (Pro tip: it turns out to be 0.69)

This result is general.  The area under [math]1/x[/math] from [math]a[/math] to [math]b[/math] is the same as that from [math]a*c[/math] to [math]b*c[/math].

What, then, is the area from 1 to 6?  We can break it into two parts – the area from 1 to 2 and the area from 2 to 6.  But the area from 2 to 6 is the same as the area from 1 to 3, by the above reasoning. 

Thus, the area from 1 to 6 is the same as the sum of the areas from 1 to 2 and from 1 to 3.  Note that 6 = 3*2.   Again, this is general.  The area under [math]1/x[/math] from [math]1[/math] to [math]a*b[/math] is the same as the sum of the areas from [math]1[/math] to [math]a[/math] and from [math]1[/math] to [math]b[/math].

That's pretty good motivation for the definition

[math]\ln(x) = \int_1^x\frac{1}{t}\mathrm{d}t[/math]

Note that this is being taken as a definition of the natural logarithm, not a proof of the relationship.  However, we have established

[math]\ln(ab) = \ln(a) + \ln(b)[/math]

and it is evident that

[math]\ln(1) = 0[/math]

This definition implies that the logarithm grows without bound because if we continually multiply the argument of the logarithm by two, we continually add [math]\ln(2)[/math] to the value.  (i.e. [math]\ln(2x) = \ln(x) + \ln(2)[/math]).  Since we can multiply any number by two over and over, we can add [math]\ln(2)[/math] to the logarithm as many times as we want.  That means we can make the logarithm arbitrarily big.

This also means that starting the integral from 1 rather than from zero was a good idea.  If we start from zero, the integral is infinite.  We can see this because [math]1/x[/math] is symmetric about the line [math]y = x[/math].

This implies that the area to the left of the curve is the same as the area under the curve, like this.

We just showed that the area under the curve diverges as we move the right hand side of the integral out to infinity, so the area to the left of the curve diverges, too.  If we started the integral at zero, it would be infinite.

What about taking the logarithm of numbers less than one?  I think the discussion so far is enough that the reader can work out [math]\ln(1/x) = – \ln(x)[/math].

Since the area under [math]1/x[/math] starts at zero when [math]x=1[/math] and goes up infinitely, it is clear that there must be some number such that [math]\ln(x) = 1[/math].  Let's choose to call that number [math]e[/math].  We don't know what it is yet, but it certainly exists.  Thus

[math]\ln(e) = 1[/math]

It is immediately apparent that, for example, [math]\ln(e^5) = \ln(e*e*e*e*e) = 5\ln(e) = 5[/math]. That makes [math]e[/math] a pretty handy number.  It shows us that the logarithm of a number [math]x[/math] is how many times you need to multiply [math]e[/math] to itself in order to get [math]x[/math].

How about [math]\ln(e^{3/2})[/math]?  That is [math]\ln\left([e^{1/2}]^3\right) = 3\ln(e^{1/2})[/math].  So in order to understand logarithms of rational numbers,  we need to understand roots of [math]e[/math].

That's not so hard, though.

 [math]\ln(e^{1/2}*e^{1/2}) = \ln(e) = 1[/math]

On the other hand,

[math]\ln(e^{1/2}*e^{1/2}) = \ln(e^{1/2}) + \ln(e^{1/2})[/math]

From this we deduce [math]\ln(e^{1/2}) = 1/2[/math].  Returning to the unfinished example, [math]\ln(e^{3/2}) = 3*(1/2) = 3/2[/math].  Thus, for any rational number [math]x[/math], we have

[math]\ln(e^x) = x[/math]

Irrational numbers are squeezed in between the rational ones, and since the definition of the logarithm as the area under a curve is evidently smooth, the above relation holds for all positive numbers.  (The logarithm of a negative number or of zero isn't defined, at least not in the real numbers.  You should make sure you understand why not.)

Finally, we would like some way of determining what [math] e[/math] is.  Here is one way to do it.  For small values of [math]x[/math], we can see that

[math]\ln(1+x) \approx x[/math]

This follows from the simple approximation below.

The red box is an approximation to the area of the green integral.  The red box clearly has area [math]x[/math] while the green integral is [math]\ln(1+x)[/math].  Thus

[math]\ln(1+x) \approx x[/math]

It's crude, but it works better and better as [math]x[/math] becomes tiny.  Multiplying both sides of the approximation by [math]1/x[/math] we get

[math]\frac{1}{x}\ln(1+x) \approx 1[/math]

We know how to rewrite the left hand side.  It gives

[math]\ln\left([1+x]^{1/x}\right) \approx 1[/math]

Since we have defined [math]e[/math] by [math]\ln(e) = 1[/math], we finally see

[math] e = \lim_{x\to 0} (1+x)^{1/x}[/math]

This has been a purposely non-rigorous answer.  You might have fun making it more rigorous if you like.

Why is \displaystyle\int \frac{\mathrm{d}x}{x}  = \ln x ?

Advertisements

Leave a comment

Filed under Life

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s