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.