Integral of 1/ln(x) formula

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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
 
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1,073
1
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
Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.
 
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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?
 
1,073
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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."
 
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I guess that was wrong.
 
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Something like [tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex] the first part which could be written as [tex]\int{\frac{e^zdz}{z}}[/tex] 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.
 
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[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]
 
1,073
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[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]
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

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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?
 
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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.
 

MWM

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Ah, you're right. Ok... as if I hadn't said anything :-)
 
95
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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]

??
 
1,073
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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]

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

Gib Z

Homework Helper
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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

268
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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.
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

Homework Helper
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O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
 
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[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.
 
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Gib Z

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

ssd

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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.
 

Gib Z

Homework Helper
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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.
 

ssd

268
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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.
Probably you mean ILATE, for choosing the 1st function for integration by parts.
 

Gib Z

Homework Helper
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O no, im 100% sure i ment LIATE, i've seen ILATE, but I was taught LIATE and i havent had any problems, everyman to his own :)
 

ssd

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Since L and I are have higher preference than others, the preference between L & I does not matter if you have a function containing one of L or I. But when only L & I functions are there, I is first and then comes L...... this is what my class notes of 11th std said. I just have checked a text book after your post and that confirms mine. I really do not know what the truth is.... but had a bias for ILATE as the text books mentioned ..... and never questioned it before today.
 
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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.
Correct.
How do you get the special math text to appear?
Cheers,
 

Gib Z

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3,344
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You have to learn laTex code. Theres a tutorial somehwere, I cant remember...I didnt learn from the tutorial btw, I clicked on other peoples code and learned from that. Click on my code to see what I typed to get that to appear.
 

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