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 2^Oscar Mar19-09 02:06 PM

Quick question about integral of (1/x)

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 Mar19-09 02:45 PM

Re: Quick question about integral of (1/x)

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 Mar19-09 06:04 PM

Re: Quick question about integral of (1/x)

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 Mar19-09 09:49 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by lurflurf (Post 2124327) 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 Mar20-09 07:17 AM

Re: Quick question about integral of (1/x)

Quote:
 Quote by 2^Oscar (Post 2124030) 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 Mar20-09 11:33 AM

Re: Quick question about integral of (1/x)

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:

$$\frac{dy}{dx} = \frac{1}{x}$$

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

$$\frac{dx}{dy} = x$$

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 Mar20-09 12:10 PM

Re: Quick question about integral of (1/x)

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 realise 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 Mar20-09 04:44 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by lanedance (Post 2124557) 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 Mar21-09 12:27 AM

Re: Quick question about integral of (1/x)

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

 nikkor180 Mar21-09 05:06 AM

Re: Quick question about integral of (1/x)

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 Mar21-09 05:47 AM

Re: Quick question about integral of (1/x)

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 Dec21-11 05:29 AM

Re: Quick question about integral of (1/x)

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 text book called Brief Calculus.

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

 checkitagain Dec21-11 03:29 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by 2^Oscar (Post 2124030) 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 Dec21-11 03:38 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by 2^Oscar (Post 2124030) 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 Dec21-11 04:00 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by lavinia (Post 3679007) While the answers already given are correct ...

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

Please see the post above yours regarding this.

 micromass Dec21-11 04:09 PM

Re: Quick question about integral of (1/x)

Quote:
 Quote by checkitagain (Post 3679027) $\text{Not those answers as they concern where the}$ $\text{absolute value symbol is missing, not to mention the "+ C."}$ Please see the post above yours regarding this.

 dextercioby Dec21-11 04:50 PM

Re: Quick question about integral of (1/x)

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 Dec21-11 05:38 PM

Re: Quick question about integral of (1/x)

Hey sbcdave -welcome to PF!

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

Quote:
 Quote by sbcdave (Post 3678387) 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.

Quote:
 Quote by sbcdave (Post 3678387) 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 lets 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 infinte 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 wierd with lmiting processes