Quick question about integral of (1/x)

[2^Oscar]

I know that the integral of 1/x is ln(x)

but if you follow the conventional method of integrating you raise the power and multiply by the new power, and you get an answer that is obviously wrong

Would someone mind explaining to me why you cannot integrate this function by normal means?



Thanks,
Oscar

 

[jdougherty]

One way to think of it is the following: If we could find the antiderivative by "normal" means, then we would get
\int \frac{1}{x}~dx = x^{0} = \textrm{constant}
which is clearly not true from looking at the graph of 1/x

Another way to look at it: the reason we can do the reverse power rule usually for two reasons: 1) the fundamental theorem of calculus, and 2) the power rule. In other words, it usually works because we know if we find the function F(x) such that x^{n} is the derivative, then we know the integral is equal to F(x). But the power rule only works when n is not zero, since D[x^{0}] = D[1] = 0.

 

[lurflurf]

This is silly.
You are maybe thinking of
Integral[x^a]=x^(a+1)/(a+1) (* a!=-1)
This cannot hold when a=-1 because it would require division by zero.
We can however use limits to extend the result.
Integral[x^a]=lim_{a->a} x^(a+1)/(a+1)
in particular
Integral[x^-1]=lim_{a->-1} x^(a+1)/(a+1)=log(x)

 

[lanedance]

This is silly.
You are maybe thinking of
Integral[x^a]=x^(a+1)/(a+1) (* a!=-1)
This cannot hold when a=-1 because it would require division by zero.
We can however use limits to extend the result.
Integral[x^a]=lim_{x->a} x^(a+1)/(a+1)
in particular
Integral[x^-1]=lim_{x->-1} x^(a+1)/(a+1)=log(x)

hmmm.. do you mean

\int x^a = ^{lim}_{b\rightarrow a} \frac{x^{b+1}}{b+1}
then
\int \frac{1}{x}= ^{lim}_{b\rightarrow -1} \frac{x^{b+1}}{b+1} =ln(x)

 

[HallsofIvy]

I know that the integral of 1/x is ln(x)

but if you follow the conventional method of integrating you raise the power and multiply by the new power, and you get an answer that is obviously wrong

Would someone mind explaining to me why you cannot integrate this function by normal means?



Thanks,
Oscar

It is not clear what you are asking. The obvious answer is what lurflurf said. You cannot use the formula
\int x^n dx= \frac{1}{n+1} x^{n+1}+ C
when n= -1 because then you would be dividing by 0. I'm not sure why you think of that as "normal means".

 

[confinement]

Think of it this way, the antiderivative of 1/x is the function whose inverse is exactly equal to its own derivative. Let y(x) be the antiderivate of 1/x. Then we have:

<br /> <br /> \frac{dy}{dx} = \frac{1}{x}<br />

Inverting the Liebniz notation in the way that he intended yields:

<br /> \frac{dx}{dy} = x<br />

The last equation says that x' = x, i.e. the function x(y) is equal to its own derivative. This means that x(y) cannot be a polynomial or rational function, since all of those functions change when you differentiate them. It is this perfect property, that the antiderivative of 1/x is the inverse function of the function who is exactly equal to its own derivative, that puts in a class of its own, a special case.

 

[2^Oscar]

Hi,

Sorry for being unclear.

My question was supposed to ask why you cannot do the following:

\int x^n dx= \frac{1}{n+1} x^{n+1}+ C

for the special case of x^-1, I realize that it cannot be done from looking at the graph of y=1/x and because the fraction requiring division by 0 cannot be defined. my query was as to why you could not do this for x^-1 (which confinement answered).

Once again appologies for the lack of clarity in my question, all your responses have cleared some of the clouds from my head :)

thanks very much,

Oscar

 

[lurflurf]

hmmm.. do you mean

\int x^a = ^{lim}_{b\rightarrow a} \frac{x^{b+1}}{b+1}
then
\int \frac{1}{x}= ^{lim}_{b\rightarrow -1} \frac{x^{b+1}}{b+1} =ln(x)

Yes, though lim a->a is valid it is confusing to some, I also typoed a into x. Also omiting dx is valid, but confusing to some. Most people denote their favorite base as log, what then is you favorite base if not e?

 

[lanedance]

yeah its got to be e, though i like the ln just to avoid any confusion

 

[nikkor180]

Greetings:

d/dx [ln(x)] = lim(h-->0) [(ln(x+h) - ln(x)) / h]

= lim(h-->0) [(1/h)*ln[(x+h)/x]

= (1/x) lim(h-->0) [(x/h) ln(1+h/x)]

= (1/x) lim(h-->0) [ln(1+h/x)^(x/h)]

= (1/x) ln [lim(h-->0) [(1+(h/x))^(x/h)]] {from lim[f(g(x))] = f(lim[g(x)])}

= (1/x) ln e

= 1/x.

Since d/dx [ln(x)] = 1/x, it follows that intgrl[1/x] = ln x.

Regards,

Rich B.
rmath4u2@aol.com

 

[arildno]

You might prefer the following argument for why the anti-derivative should be ln(x) plus some constant C:

1. We define the arbitrary (non-natural) power of some (positive) variable as follows:
x^{a}=e^{a\ln(x)}

2. In terms of natural powers, we define the exponential function itself as follows:
e^{y}=1+\sum_{n=1}^{\infty}\frac{y^{n}}{n!}

3. Let us utilize the following definite integral to prove our point:
We have:
\int_{a}^{b}x^{-1+\epsilon}dx=\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon}),
where a,b,\epsilon are all considered greater than 0.

4. Now, let us utilize definitions from 1, 2 upon the expression in 3. We gain:
\frac{1}{\epsilon}(b^{\epsilon}-a^{\epsilon})=\sum_{n=1}^{\infty}\frac{\epsilon^{n-1}((\ln(b))^{n}-(\ln(a))^{n}}{n!})

5. Thus, in the limit \epsilon\to{0}, all but the first double-term of this sum goes to zero, and we retain:
\lim_{\epsilon\to{0}}\int_{a}^{b}x^{-1+\epsilon}dx=ln(b)-ln(a)
This is in accordance with what we should have!

6. Note that this argument shows that the definite integral of some power of x can be defined as a continuous function of the power variable, even in the "special case" where the power variable has the value -1.

 

[sbcdave]

I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x

In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come on quickly and almost stop increasing.

Can anyone shed light on what I'm missing.

I apologize if this is elementary. I took college algebra 10 years ago, recently took a precalc class at a community college and just finished reading a textbook called Brief Calculus.

Love math and physics by the way and am glad I found this site.

 

[checkitagain]

I know that the integral of 1/x is ln(x)

To 2^Oscar and the rest of this thread's users:



--------> \int{\dfrac{1}{x}}dx\ = \ \ln|x| \ + \ C



Other than missing the arbitrary constant, "C," there must
be absolute value bars around the "x."



*** Edit ***
\text{Good catch by micromass concerning the missing dx term.}

 

[lavinia]

I know that the integral of 1/x is ln(x)

but if you follow the conventional method of integrating you raise the power and multiply by the new power, and you get an answer that is obviously wrong

Would someone mind explaining to me why you cannot integrate this function by normal means?
Thanks,
Oscar

While the answers already given are correct here is a way of looking at it that has always mystified me.

1/x has the remarkable property that the area under it from 1 to some number is the sum of the areas from 1 to any two numbers whose product equals the number. So if x = yz then the area from 1 to x is the sum of the areas from 1 to y and 1 to z. It is easy to show this using the integral definition of area though I would love to see an elementary proof not using calculus.

It is not hard to convince yourself that a rational function (quotient of two polynomials) can not have this property. So the logarithm is something new. It's derivative is a rational function but it is not. It is like a portal from the rational to the transcendental.

A function that compares two laws of addition and/or multiplication is called a homomorphism. The logarithm is a homomorphism from the positive numbers under multiplication to all of the numbers under addition. The exponential function is the reverse homomorphism taking the numbers under addition to the positive numbers under multiplication.

 

[checkitagain]

While the answers already given are correct ...

\text{Not those answers as they concern where the}
\text{absolute value symbol is missing, not to mention the &quot;+ C.&quot;}


Please see the post above yours regarding this.

 

[micromass]

\text{Not those answers as they concern where the}
\text{absolute value symbol is missing, not to mention the &quot;+ C.&quot;}Please see the post above yours regarding this.

You forgot the dx in your post above. So your answer is also incorrect.

 

[dextercioby]

Somehow (to me at least) related to lavinia's post is the interesting fact discovered by Euler

\mbox{for x very large} ~ \ln x \simeq \frac{1}{1} + \frac{1}{2} +\frac{1}{3} +...+\frac{1}{x}

a result which can be recast rigorously in terms of limits.

 

[lanedance]

Hey sbcdave -welcome to PF!

you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads.

I'm new to calculus, but my impression is that an integral should be a function that represents the area under the curve of the function being integrated, which f(x)=ln(x) does not do for f(x)=1/x

The reminann integral (there are other more complex defintions) is interpreted as the area under of a well behaved function

With this is mind and considering it as a definite integral (over a given integral where the function is well behaved
\int_a^bdx \frac{1}{x} = ln(x)|_a^bdx =ln(b)-ln(a)

This is equivalent to the area between the function and the horizontal axis. Notice ln(x) is a a "monotonically" increasing function (ln(b)>ln(a) for all 0<a<b), so it always gives a positive area.

In the graph of 1/x you can see that from 0 toward infinity the area under the curve would come on quickly and almost stop increasing.

Can anyone shed light on what I'm missing.

I don't totally undertstand your question..

But, at x = 0, f(x) = 1/x is not well defined, so to calculate the area you must use limits.

Similarly to deal with taking the integral to infinity, you need to use a limiting process.

Things get subtle when you consder lmiting process, so its important to be rigrorous. First lest split the integral into 2 portions, which we can do when the function is well behaved:
\int_a^bdx f(x) = \int_a^cdx f(x) + \int_c^bdx f(x)

In this case let's choose c=1 and the integral becomes
\lim_{a \to 0^+} \lim_{b \to +\infty} \int_a^b dx \frac{1}{x} = \lim_{a \to 0^+} \int_a^1 dx \frac{1}{x} + \lim_{b \to +\infty}\int_1^bdx \frac{1}{x} = \lim_{a \to 0^+}(ln(1)-ln(a)) + \lim_{b \to +\infty}(ln(b)-ln(1)) = \lim_{a \to 0^+}{-ln(a)} + \lim_{b \to +\infty}ln(b)

As both these diverge (become infinite) we actually find there is inifinite area below 1/x in both the intervals (0,1) and (1,inf). In fact in some repsects there is saimialr amount of infinite area in each.

Though in both cases the curve compresses against the axis, it doesn't do so quickly enough to result in a finite area - things can get weird with lmiting processes

Hopefully this helped answer your question and didn't confuse things more!

I apologize if this is elementary. I took college algebra 10 years ago, recently took a precalc class at a community college and just finished reading a textbook called Brief Calculus.

Love math and physics by the way and am glad I found this site.

can be a nice way to help, learn and keep up with some math skills

Mod note: fixed LaTeX

 

[sbcdave]

Hey sbcdave -welcome to PF!

Thanks

you're generally best off posting question as new threads, gets less confusing that way, particularly for old threads.

Roger

The reminann integral (there are other more complex defintions) is interpreted as the area under of a well behaved function

With this is mind and considering it as a definite integral (over a given integral where the function is well behaved
\int_a^bdx \frac{1}{x} = ln(x)|_a^bdx =ln(b)-ln(a)

This is equivalent to the area between the function and the horizontal axis. Notice ln(x) is a a "monotonically" increasing function (ln(b)>ln(a) for all 0<a<b), so it always gives a positive area.

But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite.

Example of what I'm struggling with: \int_{0.1}^1dx \frac{1}{x}

Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from.

 

[sbcdave]

I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive.

Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that?

Thanks again for humoring me once already I'll understand if the thread dies here...lol

 

[lavinia]

I was looking at a graph of ln(x) and 1/x together on my ti-83 and assumed that because the ln(x) went negative with x<1 that that meant the area was negative, now I see that because the -y value grows as x gets smaller that your subtracting a larger negative from a negative and end up with a positive.

Now I'm a little confused why the left hand method was larger than the ln(x)-ln(x) method instead of smaller though if you have any ideas on that?

Thanks again for humoring me once already I'll understand if the thread dies here...lol

By definition,

ln(x) = \int^{x}_{1}dx/x

This integral is negative when x < 1

It is not the area under the curve when x < 1. It is minus the area.

 

[lanedance]

Thanks

Roger

But the natural log of anything less than 1 is negative, which seems to say that the area under the curve from 0 to anything less than x=1 on the graph of f(x)=1/x should be a negative area. The graph of f(x)=1/x from x=0 to 1 is above the x-axis and as you say in the next quote is infinite.

Example of what I'm struggling with: \int_{0.1}^1dx \frac{1}{x}

Should be a definable positive integer. Approximately equal to 9(0.1) increments with heights 1/0.9 + 1/0.8 + ... 1/0.1 (left hand method) or approximately equal to 2.829. However, if you do ln(1)-ln(0.1) you get approximately 2.3 (I expected a negative number and need to think about this a little more, there's clearly a flaw in my logic...haha) Sorry. Posting anyways so that you might see where I was coming from.

ok so let's try the two bounding methods, dividing it into 9 blocks

Where each is greater than 1/x
= (1/0.1+1/0.2+1/0.3+1/0.4+1/0.5+1/0.6+1/0.7+1/0.8+1/0.9)*0.1~2.829

Where each is less than 1/x
= (1/0.2+1/0.3+1/0.4+1/0.5+1/0.6+1/0.7+1/0.8+1/0.9+1/1)*0.1~1.929

which both bound your exact answer of ln(1)-ln(0.1)~2.3

Note you can also use the useful property of logarithms to see that it is always positive
ln(b)-ln(a) = ln(b/a)

as b>a>0, then b/a>1 and it is always a positive number

 

[sbcdave]

Thanks for the replies. Very interesting subject to me. Sorry for reviving the dead thread, hopefully some others were entertained by it.

 

[Poopsilon]

Im sorry but how can this:

\int \frac{1}{x}dx = lim_{b\rightarrow -1}\frac{x^{b+1}}{b+1} = ln(x)

possibly be correct?

 

[dimension10]

...Also omiting dx is valid, but confusing to some...

That is very unintuitive however,...

\frac{\mbox{d}y}{\mbox{d}x}=f(x)
\mbox{d}y=f(x)\mbox{d}x
y=\int f(x)\mbox{d}x