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Integral of 1/ln(x) 
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#1
Dec2606, 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 


#2
Dec2606, 04:36 PM

P: 1,076




#3
Dec2606, 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?



#4
Dec2606, 11:24 PM

P: 1,076

Integral of 1/ln(x)



#5
Dec2706, 01:21 AM

PF Gold
P: 1,059

I guess that was wrong.



#6
Dec2706, 01:28 AM

P: 1,076




#7
Dec2706, 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}{xlog m(0)}[/tex] 


#8
Dec2706, 02:03 AM

P: 1,076




#9
Dec2706, 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 loglog(m/m0)+c; isn't it?



#10
Dec2706, 04:14 AM

P: 1,076




#11
Dec2706, 04:19 AM

P: 2

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



#12
Dec2706, 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] ?? 


#13
Dec2706, 04:02 PM

P: 1,076




#14
Dec2806, 06:40 AM

HW Helper
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. 


#15
Dec2906, 02:08 AM

P: 239

ln(m/mo)=t+c, c is the constant of integration Simplifying, ln(m)=t+k, k= c+ ln(mo). 


#16
Dec3006, 12:16 AM

HW Helper
P: 3,352

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



#17
Dec3006, 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}{xlog 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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