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Integral of 1/ln(x)

by NEWO
Tags: 1 or lnx, integral
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NEWO
#1
Dec26-06, 03:38 PM
P: 97
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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d_leet
#2
Dec26-06, 04:36 PM
P: 1,076
Quote Quote by NEWO View Post
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.
Dbjergaard
#3
Dec26-06, 11:04 PM
P: 14
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
#4
Dec26-06, 11:24 PM
P: 1,076
Integral of 1/ln(x)

Quote Quote by Dbjergaard View Post
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."
robert Ihnot
#5
Dec27-06, 01:21 AM
PF Gold
P: 1,059
I guess that was wrong.
d_leet
#6
Dec27-06, 01:28 AM
P: 1,076
Quote Quote by robert Ihnot View Post
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.
robert Ihnot
#7
Dec27-06, 01:55 AM
PF Gold
P: 1,059
[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
#8
Dec27-06, 02:03 AM
P: 1,076
Quote Quote by robert Ihnot View Post
[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
#9
Dec27-06, 04:10 AM
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?
d_leet
#10
Dec27-06, 04:14 AM
P: 1,076
Quote Quote by MWM View Post
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
#11
Dec27-06, 04:19 AM
P: 2
Ah, you're right. Ok... as if I hadn't said anything :-)
NEWO
#12
Dec27-06, 06:04 AM
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

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

??
d_leet
#13
Dec27-06, 04:02 PM
P: 1,076
Quote Quote by NEWO View Post
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
#14
Dec28-06, 06:40 AM
HW Helper
Gib Z's Avatar
P: 3,352
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
#15
Dec29-06, 02:08 AM
P: 239
Quote Quote by Gib Z View Post
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
#16
Dec30-06, 12:16 AM
HW Helper
Gib Z's Avatar
P: 3,352
O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
robert Ihnot
#17
Dec30-06, 04:27 PM
PF Gold
P: 1,059
[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.
Gib Z
#18
Dec30-06, 09:46 PM
HW Helper
Gib Z's Avatar
P: 3,352
Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.


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