# Integral of 1/ln(x)

by NEWO
Tags: 1 or lnx, integral
 P: 97 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
P: 1,076
 Quote by 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
Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.
 P: 16 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?
P: 1,076

## Integral of 1/ln(x)

 Quote by 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?
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."
 PF Patron P: 1,059 I guess that was wrong.
P: 1,076
 Quote by robert Ihnot Something like $$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$ the first part which could be written as $$\int{\frac{e^zdz}{z}}$$ I don't suppose has a definite integral in closed form.
How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.
 PF Patron P: 1,059 $$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$ Let m=e^x then we get $$\int{\frac{dx}{x-log m(0)}$$
P: 1,076
 Quote by robert Ihnot $$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$ Let m=e^x then we get $$\int{\frac{dx}{x-log m(0)}$$
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.
 P: 2 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?
P: 1,076
 Quote by 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?
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.
 P: 2 Ah, you're right. Ok... as if I hadn't said anything :-)
 P: 97 Ok thanks guys for your replies, the thing is I don't get where the m has gone!! surely the integral would be $$\int{\frac{1}{m_{0}ue^{u}}}du$$ ??
P: 1,076
 Quote by NEWO Ok thanks guys for your replies, the thing is I don't get where the m has gone!! surely the integral would be $$\int{\frac{1}{m_{0}ue^{u}}}du$$ ??
How do you get to that? Try using the substitution u=log(m/m0) the integral will turn out very nicely.
 HW Helper P: 3,353 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.
P: 233
 Quote by 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.
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).
 HW Helper P: 3,353 O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
 PF Patron P: 1,059 $$\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}$$ 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.
 HW Helper P: 3,353 Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.

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