Integral of 1/ln(x)

NEWO

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

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

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

d_leet

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

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?

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."

robert Ihnot

I guess that was wrong.

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.

robert Ihnot

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

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

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.

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?

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.

MWM

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

NEWO

Ok thanks guys for your replies,

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

surely the integral would be

[tex]\int{\frac{1}{m_{0}ue^{u}}}du[/tex]

??

d_leet

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

Gib Z

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

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

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).

Gib Z

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

robert Ihnot

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

What was wrong with substiting X=Log m?

[tex]\int{\frac{dx}{x-log m(0)}[/tex]

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