Integral of 1/x: Proving Invalidity of Method

[alice22]

Sure you can.
\int_{.4}^{.8} \frac{dt}{t} + \int_{.8}^{1} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t}
\Rightarrow \int_{.4}^{.8} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t} - \int_{.8}^{1} \frac{dt}{t}
= -\int_{1}^{.4} \frac{dt}{t} + \int_{1}^{.8} \frac{dt}{t}
= - ln(.4) + ln(.8) = ln(.8/.4) = ln 2 ~.693
You're quoting someone in this thread, but what is said in the quote is incorrect. For any base b, with b > 0 and b != 1, log_b 1 = 0. This has nothing to do with whatever the base happens to be.


I don't understand what you're asking here. The natural exponential function e^x has all real numbers as its domain and the positive reals as its range. The inverse of this function (the natural log function) has a domain of the positive reals and its range is all real numbers.

What I am saying is the exponential function does not stop at 1, however the integral of 1/x
does not seem to be defined below 1.
You seem to have ignored that bit and it was that bit I was specifically concerned with.

 

[losiu99]

For any base b, with b > 0 and b != 1, log_b 1 = 0. This has nothing to do with whatever the base happens to be.

What I ment was that 1 is an unique zero of natural log. Of course, any other log as well. My point is that antiderivative of 1/x, namely natural log, defined as the inverse of the exponential, must be zero at the lower limit of this integration, for otherwise the integral does not represent natural log.

Area under 1/x between 0 and 1 is indeed infinite.

By messier properties I mean things like this:
Suppose we define
<br /> f(x)=\int_{0.5}^{x} \frac{dx}{x}<br />
Then instead of nice addition formula for log, we get
<br /> f(x)+f(y)=\log2 +f(xy) <br />

So defined function also is not inverse of the exponential. Not because of domain, but because e^0=1, and f(1)=log 2.

Which does not of course mean you cannot integrate 1/x over intervals containing numbers less than one. You cannot start integration from zero, for answer will be undefined, but any other intervals are perfectly ok.

 

[Mark44]

What I am saying is the exponential function does not stop at 1, however the integral of 1/x
does not seem to be defined below 1.
You seem to have ignored that bit and it was that bit I was specifically concerned with.

No, I didn't ignore that, but you must have ignored what I wrote. You asked about
\int_{.4}^{.8} \frac{dt}{t}

and I showed how that could be evaluated using the integral definition of the natural log function. In the example, both .4 and .8 are less than 1.

The natural log function ln(x) is defined for any x > 0.

ln(x) = \int_{1}^{x} \frac{dt}{t}

The only restriction on x in the integral above is that it must be positive.

One other thing, don't confuse "area" and "value of a definite integral." Area is always nonnegative, but a definite integral can have a negative value. In this integral, if x is between 0 and 1,

ln(x) = \int_{1}^{x} \frac{dt}{t} &lt; 0

 

[HallsofIvy]

No they don't and no they don't. Actually the lower limit in the integral is 1, therefore

\ln x=\int_{1}^{x} \frac{1}{t}{}dt.

Ouch! Thanks!

 

[alice22]

No, I didn't ignore that, but you must have ignored what I wrote. You asked about
\int_{.4}^{.8} \frac{dt}{t}

and I showed how that could be evaluated using the integral definition of the natural log function. In the example, both .4 and .8 are less than 1.

The natural log function ln(x) is defined for any x > 0.

ln(x) = \int_{1}^{x} \frac{dt}{t}

The only restriction on x in the integral above is that it must be positive.

One other thing, don't confuse "area" and "value of a definite integral." Area is always nonnegative, but a definite integral can have a negative value. In this integral, if x is between 0 and 1,

ln(x) = \int_{1}^{x} \frac{dt}{t} &lt; 0

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

 

[losiu99]

If we define log to be this integral from zero to x, it will be divergent for every real number. That's not the same answer, I'm afraid
The fact is: if we want to represent natural log understood as an inverse of the exponential function by integral of 1/t with x as the upper limit, lower limit must be 1. For no other number the equality holds.

 

[Bohrok]

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

You know, that's pretty much exactly what I was wondering in post #19 here
https://www.physicsforums.com/showthread.php?t=412403&page=2

If we define log to be this integral from zero to x, it will be divergent for every real number. That's not the same answer, I'm afraid
The fact is: if we want to represent natural log understood as an inverse of the exponential function by integral of 1/t with x as the upper limit, lower limit must be 1. For no other number the equality holds.

But how did we know to choose 1 in the first place? Just luck, or kind of working backwards from what we know the answer to be?

 

[losiu99]

Well, from the definition of natural log as inverse of the exponential, it is easy to show that it's derivative is 1/x. So, by fundamental theorem of calculus,
<br /> \int_{a}^x \frac{dt}{t}=\log x - \log a<br />
Since log 1 = 0 (directly from previously mentioned definition), it is obvious that
<br /> \log x = \int_{1}^x \frac{dt}{t}<br />
Natural log is increasing (once again, definition), so no number other than one in place of a fits.

Edit: of course, if we didn't want log to be inverse of the exponential, it doesn't really matter where do we start integrating from.

 

[HallsofIvy]

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

I can't help but wonder what kind of Calculus course you took!

\int_a^b f(x)dx does NOT mean that b must be larger than a.

Surely one of the first things you learned about the integral is that
\int_a^b f(x)dx= -\int_b^a f(x) dx[/itex].<br /> <br /> The only thing starting the integration at x= 1 causes is that ln(1)= 0.<br /> Since, by the fundamental theorem of calculus, the derivative of \int_1^x (1/T)dt is 1/x&gt; 0 for x&gt; 0 so ln(x) is an increasing function and so if x&lt; 1, ln(x)&lt; 0 and if x&gt; 1, ln(x)&gt; 0.<br /> <br /> The only thing that restricts what x can be is that we cannot integrate across a point where f(x) becomes unbounded- which is why ln(x) is not defined for x&lt; 0.<br /> That is also why you <b>cannot</b> start the integral at 0- 1/t is ubounded in any neighborhood of t= 0.<br /> <br /> &quot;If you defined the integral to be from 0 to x, you would come up with the same answers wouldn&#039;t you?&quot; No, you wouldn&#039;t! Especially because 1/t is not bounded in any neighborhood of 0 and so that integral is impossible. <br /> <br /> <blockquote data-attributes="" data-quote="Bohrok" data-source="post: 2788254" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Bohrok said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You know, that&#039;s pretty much exactly what I was wondering in post #19 here<br /> <a href="https://www.physicsforums.com/showthread.php?t=412403&amp;page=2" class="link link--internal">https://www.physicsforums.com/showthread.php?t=412403&amp;page=2</a><br /> <br /> <br /> <br /> But how did we know to choose 1 in the first place? Just luck, or kind of working backwards from what we know the answer to be? </div> </div> </blockquote> Not &quot;luck&quot; but just because 1 is an easy number (and 0 won&#039;t work as I showed above).<br /> <br /> If we were to define <br /> L_a(x)= \int_a^x \frac{1}{t} dt<br /> then we have<br /> L_a(x)= \int_1^x \frac{1}{t}dt- \int_1^a \frac{1}{t}dt<br /> so that L_a(x)= ln(x)- ln(a)= ln(x/a).<br /> <br /> If y= L_a(x) then we would have y= e^{x/a} which is the same as (e^{1/a})^x. That is, choosing &quot;a&quot; as the lower limit of the integral would just lead to a different base, e^{1/a}- and so, perhaps, to a different choice of numerical value for &quot;e&quot;.<br /> <br /> Historically, of course, it worked the other way. The function f(x)= a^x can be defined, without using logarithms, for any positive number, a. The derivative of a^x would be given by<br /> \lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to a}\frac{a^xa^h- a^x}{h}<br /> = \lim_{h\to a}\left(a^x\right)\frac{a^h- 1}{h}= \left(a^x\right)\lim_{h\to 0}\frac{a^h- a}{h}<br /> that is, the derivative of a^x, for any positive a, is some constant (which depends on a), C_a, times a^x itself.<br /> <br /> Now, it is easy to see that if a= 2,<br /> C_2= \lim_{h\to 0}\frac{2^h- 2}{h}<br /> is close to .69 (take h= .001) and that if a= 3,<br /> C_3= \lim_{h\to 0}\frac{3^h- 3}{h}<br /> is close to 1.099.<br /> <br /> C_2 is less than 1 and C_3 is larger than 1. It doesn&#039;t take much (essentially showing that this limit process is a continuous function of a) to see that there is some number between 2 and 3 such that the limit is 1. We call that number &quot;e&quot; and have the result that (e^x)&amp;#039;= e^x (instead of some weird constant times a^x).<br /> <br /> But either way, our choice of numerical value for &quot;e&quot; and the definition of ln(x) are based on the fact that &quot;1&quot; is a convenient number.<br /> <br /> If <b>you</b> were able to choose a value for a <b>positive</b> constant that would keep popping up in your work, what value would you choose?<br /> <br /> (By the way, I just noticed that alice22 resurrected this from a thread that had been closed three years ago.)

 

[Unit]

Surely one of the first things you learned about the integral is that
\int_a^b f(x)dx= \int_b^a f(x) dx.

I think you meant

\int_a^b f(x) dx = - \int_b^a f(x) dx

with the minus sign :) Other than that, I really like your explanation!

 

[alice22]

(By the way, I just noticed that alice22 resurrected this from a thread that had been closed three years ago.)

Yes I said that it the first line of my post.
The thread was not closed though, it was still open for posting.

 

[HallsofIvy]

I think you meant

\int_a^b f(x) dx = - \int_b^a f(x) dx

with the minus sign :) Other than that, I really like your explanation!

I think that is the second time in a few days I have made that mistake.

Oh, well, I'll go back and edit and then no one will know!

Thanks for the heads-up.

 

[alice22]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?
Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

Maybe I am making a mountain out of a molehill?

 

[Mark44]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

And lots of them don't cross the x-axis anywhere, one of them being y = 1/x, the function that plays a role in this definition of ln(x). What's so special about 1 is that ln(1) = log(1) = log_b(1) = 0, so 1 seems as good a starting point for the lower limit of integration as any.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?

But we don't ordinarily define cos(x) as an integral.

Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

If you ask a very specific question about a definite integral (as you have done), we will naturally assume that you have some knowledge of the notation that is used in calculus. Although it would be possible to explain everything in English without resorting to symbols, the explanations would be much more verbose and likely not as precise.

That's not to say that the best explanation is one that is purely equations and symbols.

Maybe I am making a mountain out of a molehill?

Maybe. The equation
ln(x) = \int_1^x \frac{dt}{t}
is one definition for the natural log function.

 

[Mute]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

Choosing 1 as the lower limit makes this definition of the logarithm correspond exactly with our other definitions.

Even if the integral definition had been the first definition of the logarithm known, I would guess that 1 still would have been chosen as the lower limit initially. Why? Well, zero obviously wouldn't work, so might as well just pick the next natural number, which is one. People like natural numbers much more than fractions, and skipping 1 to go to 2 seems even more arbitrary than picking 1 in this scenario. Even if some other number had been picked, once the function's relation to the exponential function had been realized (for example), the definition probably would have been changed to have the lower limit of 1 so that it corresponded exactly to the inverse of the exponential function.

 

[HallsofIvy]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?
Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

Maybe I am making a mountain out of a molehill?

You have now been told repeatedly that the choice of "1" as the lower limit for the integral defining ln(x) is simply because it is convenient. Choosing that lower limit to be any positive number other than 1 would just change the numerical value of "e".

If you could choose a positive constant that will keep popping up in your calculations, don't you think "1" would be the most reasonable choice?

Do you see why "0" cannot be the lower bound? (Because 1/t is not bounded in any neighborhood of t= 0.)

Finally, Mark44 refers to the fact that ln(1)= 0 but, of course, that is due to the definition of ln(x) as the inverse of e^x and I am referring to it as defined by the integral (and e^x then defined as its inverse).

 

[alice22]

You have now been told repeatedly that the choice of "1" as the lower limit for the integral defining ln(x) is simply because it is convenient. Choosing that lower limit to be any positive number other than 1 would just change the numerical value of "e".

If you could choose a positive constant that will keep popping up in your calculations, don't you think "1" would be the most reasonable choice?

Do you see why "0" cannot be the lower bound? (Because 1/t is not bounded in any neighborhood of t= 0.)

Finally, Mark44 refers to the fact that ln(1)= 0 but, of course, that is due to the definition of ln(x) as the inverse of e^x[/math] and I am referring to it as <b>defined</b> by the integral (and e^x[/math] then defined as &lt;b&gt;its&lt;/b&gt; inverse).

I am looking for a viable explanation not to be &amp;#039;told&amp;#039; this is the answer.&lt;br /&gt; I don&amp;#039;t think convenience is a credible answer.&lt;br /&gt; &lt;br /&gt; I also do not see how the value of &amp;quot;e&amp;quot; can be changed no more than the value of pi can be changed.&lt;br /&gt; &lt;br /&gt; It might be convenient to say pi=3 however, convenience is not what is required, the correct&lt;br /&gt; answer is the only acceptable answer.&lt;br /&gt; &lt;br /&gt; So no I do not see convenience as an acceptable answer.

 

[losiu99]

Okay then, answer to what problem is this 1 here? If the question is "Why is 1 the lower limit of this integral representing natural log (defined as the inverse of the exponential)?", then I believe I answered it in my previous posts (In fact, this integral with other lower limit doesn't define log of any base). If the question is "why log is defined as this integral with 1 as lower limit, and not some other real number", the answer is: someone wanted log to be inverse of exponential. If the question is "Why did he?", I think "He thought it would be nice to find some integral representation for inverse of such a fundamental function as exp". It is perfectly ok to define some function as the same integral with other lower limit, it simply will be different function, satisfying different properties etc. Could you formulate your question once again? I feel a little lost, I no longer understand what are you asking for.

 

[Char. Limit]

Actually, convenience is an acceptable answer. Much of what mathematicians do is based off of convenience. Just think about the table of integrals in any calculus book. A mathematician could solve any of those functions by himself every time he needed them, but why would he when he could just solve the general case and use it?

It's the same general idea of convenience, picking the number that makes things easier.

Also, just out of curiousity, is it possible to define the sine function as an integral of the cosine function? Just a little side question.

 

[alice22]

Okay then, answer to what problem is this 1 here? If the question is "Why is 1 the lower limit of this integral representing natural log (defined as the inverse of the exponential)?", then I believe I answered it in my previous posts (In fact, this integral with other lower limit doesn't define log of any base). If the question is "why log is defined as this integral with 1 as lower limit, and not some other real number", the answer is: someone wanted log to be inverse of exponential. If the question is "Why did he?", I think "He thought it would be nice to find some integral representation for inverse of such a fundamental function as exp". It is perfectly ok to define some function as the same integral with other lower limit, it simply will be different function, satisfying different properties etc. Could you formulate your question once again? I feel a little lost, I no longer understand what are you asking for.

No I have problems with that.
For starters show me the mathematics equation for "someone wanted something"?
I do not recall covering the that in my math course, maybe I missed the maths of "someone wanted".

example x^2 = 3 (because someone wanted it to be).

The inverse of the exponential function does not stop at one so that again is a wholely invalid reason.

 

[alice22]

Actually, convenience is an acceptable answer. Much of what mathematicians do is based off of convenience. Just think about the table of integrals in any calculus book. A mathematician could solve any of those functions by himself every time he needed them, but why would he when he could just solve the general case and use it?

It's the same general idea of convenience, picking the number that makes things easier.

Also, just out of curiousity, is it possible to define the sine function as an integral of the cosine function? Just a little side question.

You can define cos as tan if it makes the problem easier, after all convenience is what it is all about, who cares if it makes any sense?

 

[losiu99]

We can define things the way we want. If we define x as the number whose square is three, "x" will mean all the objects with this property. If such notation proves useful, it may become standard. Just like "pi" is defined to be ratio of circumference to diameter. Why doesn't "pi" mean 3? Because we defined it not to.

Next, I don't imply you have problem. You mentioned "answer" in you post, so I just asked "answer to what question?". You don't want to be just told the answer, that's good. But in order to help you understand the answer, I need to know the question. I'm obviously missing your point, so it would be helpful if you clarified it.

I'm not sure what you mean by "stop at one". If you mean that it's defined for positive reals less than one, you're obviously right. It has nothing to do with lower limit of the integral, though. We don't have to start integrating "from the very left".

Edit: Sure we could define cos as tan. There was no point, however. As long as convenience doesn't collide with validity, there's nothing wrong about it. We want definitions to be as useful and comfortable as it's possible, don't we? The whole point of defining is about convenience. We call some specific subsets of X\times Y a function, so that we don't have to describe this desired properties every time.

 

[Mute]

You can define cos as tan if it makes the problem easier, after all convenience is what it is all about, who cares if it makes any sense?

As I said in the post at the top of the page, choosing the lower limit to be 1 makes the integral definition coincide with all of the other definitions of the logarithm. It would otherwise differ by a constant. Even if the integral definition came first, once the other definitions became known (and were found to be more useful), someone would have modified the definition to have the constant 1 so as to make the integral definition match up with the other ones.

If you don't find that to be a satisfactory answer then I don't know what answer you could possibly be looking for.

 

[Mark44]

No I have problems with that.
For starters show me the mathematics equation for "someone wanted something"?
I do not recall covering the that in my math course, maybe I missed the maths of "someone wanted".

There is no mathematics equation for "someone wanted something" but there are many equations that arose because someone wanted to find out something.

An equation such as x - 2 = 0 is very simple to solve, but a similar equation, x + 2 = 0 is impossible to solve if you don't understand the concept of negative numbers, which came along only about a thousand years ago.

Also, if you know that 3*3 = 9, you can guess at a solution to x² = 9, but if you don't have the concept of irrational numbers, you'll have a tough time solving the equation x² = 3.
[/quote]

example x^2 = 3 (because someone wanted it to be).

The inverse of the exponential function does not stop at one so that again is a wholely invalid reason.

What does that have to do with anything? Numerous people in this very long thread have given you perfectly valid reasons for the integral definition of the natural log function to be as it is. If you have a question about what they said, ask it, but do not wave your hand and say that these explanations are "wholely (sic) invalid" merely because you don't understand them.