# Integral of 1/ln(x)

1. Mar 26, 2008

### rock.freak667

1. The problem statement, all variables and given/known data

Find

$$\int \frac{1}{lnx} dx$$

3. The attempt at a solution

Let $t=lnx \Rightarrow \frac{dt}{dx}=\frac{1}{x} \Rightarrow dx=e^t dt$

$$\int \frac{1}{lnx} dx \equiv \int \frac{e^t}{t} dt$$

and well

$$e^t= \sum _{n=o} ^{\infty} \frac{t^n}{n!}$$

$$\frac{e^t}{t}=\sum _{n=o} ^{\infty} \frac{t^{n-1}}{n!}$$

$$\int \frac{e^t}{t}=\int \sum _{n=o} ^{\infty} \frac{t^{n-1}}{n!}$$

$$=\sum _{n=o} ^{\infty} \frac{t^{n}}{(n+1)!}$$

Is there any easier closed form solution for this?

2. Mar 26, 2008

### hexphreak

That would be, I guess, the http://mathworld.wolfram.com/LogarithmicIntegral.html" [Broken].

Last edited by a moderator: May 3, 2017
3. Mar 26, 2008

### Schrodinger's Dog

Apparently it's.

$$Ei\:(1,-1\ln(x))$$

This seems a little unfair though unless you knew about such an integral?

Last edited: Mar 26, 2008
4. Mar 26, 2008

### rock.freak667

Nope,never even heard of it. But my usual way is, if you can't get it out directly, make a series and approximate and hopefully it might turn into something nicer.

5. Mar 26, 2008

### Schrodinger's Dog

Probably but with the exponential integral and imaginary numbers there it will only be equivalent to the actual integral in a series. Although I think in that case they are asking you to show the series rather than the actual integral as it is in distinct maths language. Like to see how to do this, sorry but I have no idea, so can't help, I just thought the answer my maths program spat out was quite interesting, good luck.

At least you know that the answer involves an exponentiation, and imaginary numbers I suppose.

Last edited: Mar 26, 2008
6. Mar 26, 2008

### tiny-tim

erm … $$=\sum _{n=o} ^{\infty} \frac{t^{n}}{n.n!}\,.$$

7. Mar 26, 2008

### rock.freak667

Ah silly me...forgot that n-1+1 is n and not n+1..thanks