# Integral of 1/ln(x)

1. Dec 26, 2006

### NEWO

I have a formula to solve, which I am having trouble solving, it is a differential equation fo sorts

$$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$

could someone help me solve the indefinite integral of such an equation!!

the m multiply of the right handside is suppose to be dm and likewise with the t on the left hand side sorry for any confusion!!!
Thanks for any help

N

Last edited: Dec 26, 2006
2. Dec 26, 2006

### d_leet

Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.

3. Dec 26, 2006

### Dbjergaard

How can this work? Its an indefinite integral and the answer's definate. One's with respect to m and the other t. I'm sure it doesn't change the u substitution, but shouldn't everything be noted as rigorously as possible?

4. Dec 26, 2006

### d_leet

I think everything is as already as rigorous as possible, at least for all that the original poster has given, however, I have no idea what you mean by "Its an indefinite integral and the answer's definate."

5. Dec 27, 2006

### robert Ihnot

I guess that was wrong.

Last edited: Dec 27, 2006
6. Dec 27, 2006

### d_leet

How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.

7. Dec 27, 2006

### robert Ihnot

$$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$

Let m=e^x then we get $$\int{\frac{dx}{x-log m(0)}$$

8. Dec 27, 2006

### d_leet

You're missing a factor of e-x which then gives you something you can't integrate, but if you make the substitution u=log(m/m0) then you end up with the integral of 1/u with respect to u.

9. Dec 27, 2006

### MWM

We may just notice that 1/m is the 1st derivative of (log(m/m0)) and that, then, the first integral contains an expression dividing its derivative: integrate[D(log(m/m0))]/log(m/m0)]dm. But this is just log|log(m/m0)|+c; isn't it?

10. Dec 27, 2006

### d_leet

Yes that's it, and this is the same solution you arrive at using the substitution I mentioned, and is pretty much the sam process.

11. Dec 27, 2006

### MWM

Ah, you're right. Ok... as if I hadn't said anything :-)

12. Dec 27, 2006

### NEWO

Ok thanks guys for your replies,

the thing is I don't get where the m has gone!!

surely the integral would be

$$\int{\frac{1}{m_{0}ue^{u}}}du$$

??

13. Dec 27, 2006

### d_leet

How do you get to that? Try using the substitution u=log(m/m0) the integral will turn out very nicely.

14. Dec 28, 2006

### Gib Z

Let $u=\log(m/mo)$
The integral:$$\int{\frac{1}{m\log(m/mo)}{dm}$$ becomes $$\int{\frac{1}{m\cdot u}{dm}$$. Whats the derivative of log x? 1/x of course.
$$u=\log (m/mo)$$
$$\frac {du}{dm} =\frac{1}{m}$$
Substituting that in, our integral becomes $$\int \frac{1}{u} \frac {du}{dm} dm$$.
Dm's cancel out, thats why your m goes away!!
And obviously $$\int \frac{1}{u} du = \log u$$.

Now that looks the same form as the right hand side of your first post. Therefore $t=\log (m/mo)$. Hopefully that was what you are looking for.

15. Dec 29, 2006

### ssd

You are right but I want to add,
ln(m/mo)=t+c, c is the constant of integration
Simplifying, ln(m)=t+k, k= c+ ln(mo).

16. Dec 30, 2006

### Gib Z

O yes sorry, forgot about the Constant, my bad. Ty for correcting me.

17. Dec 30, 2006

### robert Ihnot

$$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$

What was wrong with substiting X=Log m?

$$\int{\frac{dx}{x-log m(0)}$$

Remember, if e^x=m then, e^xdx =dm; so ms in the top and bottem cancel out.

Last edited: Dec 30, 2006
18. Dec 30, 2006

### Gib Z

Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.

19. Dec 31, 2006

### ssd

My natural instinct was to substitute u= ln(m/mo) , for the probable shortest way, I saw later 'Gib Z' also did the same substitution.

20. Dec 31, 2006

### Gib Z

Theres a special order to choose your u. Its very helpful: LIATE. Logs, Inverse Trig, Algebraic, Trig, and Exponentials. Go down the list and pick the first one as your u. In that case its obvious this time its the Log.