Non elementary Integral Question?

Question

Im a calc2 student but Im trying to do a little more math on my own. A number theory book I have is talking of prime numbers and in this section they say you need to integrate 1/lnx from 0 to x. This is a non elementary function so parts, substitution, etc will not work. I know I have to express this as an infinite series but I dont know how to do this unless its a very simple function. (I can do geometric series expansions and the like but I dont know how to do this as 1/lnx)

Can anyone integrate 1/lnx from 0 to x and explain it detailed steps how they did everything. I want to understand how the function was expressed as a series and what function it gives you. (Since the upper bound is x it should give a function, not a number...right?) Thanks in advance.

Thanks for the info so far...That link didnt really help...I have no idea what theyare talking about. Im assuming its not THAT difficult to integrate this...My number theory book said gauss did it when he was 15. Theres got to be somone on here who can solve this...

To pessimist_atheist:

Thanks so much for all the information. I brought this to my math teacher before I posted this and he gave me a basic idea how to sovle this although I didnt really understand. You said exactly what he did except I can understand what youre saying. He said to do the same thing including the substitution and the maclaurin series. Thanks so much for the info. (same goes toeveryone)

Answer 1

Ok, follow me on this one.
First, the problem with this function is that it is undefined at one of the limits, so you have to chose an arbitrary value "a" and integrate from x=a to x=x (redundant, I know) and then setup a limit outside the integral that such that a->0.

Second, yes, it is not exactly easy to do in terms of elementary functions and not necessarily substitutions (to an extent, wait for it). What you have to do is a little algebraic trick. So here is what the problem looks like after you set up a limit:

Limit as a->0 of (Int(from x=a to x=x of (1/ln(x))dx))

So far so good. Now how do you actually approach this thing. I decided to do a simple u substitution:

Let u=ln(x)
du=dx/x
dx=x*du

Also, using the substitution, solve for x in terms of u:
x=e^u
so that:
dx=e^u*du

Resubstitute:

Limit as a->0 of (Int(from x=a to x=x of (e^u/u))du))
[Notice I did not change the limits even though I changed variables. I plan on back substituting.]

So you're probably wondering, why did you do that, you're still stuck with a function which you can't integrate using traditional methods. Yes, you are right, but now you can use a McClaurin series for e^u/u and actually do an integration!

e^u=sum((u^n)/n!)
(e^u)/u=sum((u^(n-1))/n!)

Now integrate wrt u:

Limit as a->0 of (Int(from x=a to x=x of (sum((u^(n-1))/n!)*du)

Limit as a->0 of sum((u^n)/(n*n!) evaluated at x=a and x=x

Resubstitute u for x:

Limit as a->0 of sum((ln(x))^n)/(n*n!) evaluated at x=a and x=x

So now we evaluate!

Limit as a->0 of sum((ln(x))^n)/(n*n!) - sum((ln(a))^n)/(n*n!)

This is as far as I go. I think from here you're going to have major problems, mainly because you have to evaluate a bunch of ln(x) as x goes to zero, in which the function tends to negative infinity. Unless there is some trick I'm missing here to get rid of the second sum (say for example, it goes to zero. That would be nice, but I don't think it's happening.) this is as clean an answer you're going to get, well from me anyway.

The only thing you might do is use the limit comparison test to find out how/if the second sum converges or diverges (since it goes from n=0 to infinity). This, I feel, is probably a herculean feat, but I'm sure someone is up for the challenge.

:)

Answer 2

Writing at least a few terms of the series for 1/ln x should not be a problem. However where you choose to expand your integral is important. Also the integral you are talking about is improper if x is greater than 1, as lim (x->1) (1/lnx)=+-infty. Now my guess is if the integral is known to be non-elemantary then getting a formula for its taylor expansion should be very difficult if not impossible. You may want to take a look at numeric integration books for methods to handle questions like this.

Answer 3

You need Fourier serie to solve this and I'm not propose to do it by yourself and it realy take my time to do so I just thinking about giving you a hint.

Answer 4

http://en.wikipedia.org/wiki/Logarithmic_integral_function

x should be t and you're integrating from 0 to x.