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Integrating out of the real domain of a function 
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#1
Nov2712, 11:43 AM

P: 232

Check this out:
[tex]\int_{1}^{0} ln(x) dx[/tex] [tex]u=ln(x), dv=dx[/tex] [tex]du=\frac{1}{x},v=x[/tex] [tex]\int_{1}^{0} ln(x) dx= x\space ln(x) \_{1}^{0} \int_{1}^{0} \frac{x}{x} dx[/tex] [tex]=x\space ln(x) \_{1}^{0}x \_{1}^{0}[/tex] [tex]=\lim_{x\to0} x\space ln(x)(ln(1))(0(1))[/tex] [tex]\lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}[/tex] [tex]=\lim_{x\to0} \frac{1/x}{1/x^2}[/tex] [tex]=\lim_{x\to0} x=0[/tex] [tex]\int_{1}^{0} ln(x)dx=ln(1)1[/tex] [tex]e^{i\pi}=1\to ln(1)=i\pi[/tex] [tex]\int_{1}^{0}ln(x)\space dx=i\pi1[/tex] Is this legitimate? P.S. Why don't my limits look right? 


#2
Nov2712, 02:11 PM

P: 232

By the way, this is not a textbook style question, I'm wondering whether or not it makes sense to integrate out of a function's domain.



#3
Nov2712, 02:33 PM

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#4
Nov2712, 03:23 PM

P: 232

Integrating out of the real domain of a function



#5
Nov2712, 05:59 PM

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[itex]\int_{1}^0ln(z)dz = \int_{1}^0(ln(r)+i\pi) d(r) = \left[r ln(r)  r + i\pi r \right]_0^1 = i\pi  1[/itex] 


#6
Nov2712, 06:12 PM

P: 232

Also, is integrating the function without reinterpreting it into the complex plane o.k. (i.e. is my way of arriving at the same answer legitimate)? 


#7
Nov2712, 07:25 PM

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#8
Nov2712, 08:10 PM

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#9
Nov2712, 08:44 PM

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[itex]\int_Pln(z)dz = \int_P(ln(re^{iπ}) dre^{iπ}= e^{iπ}\int_{r=1}^0(ln(r)+iπ) dr = \int_{r=1}^0(ln(r)+iπ) dr[/itex] [itex] = \int_{r=0}^1(ln(r)+iπ) = \left[(r ln(r)r+iπr)\right]_0^1 = 1+iπ [/itex] 


#10
Nov2712, 08:59 PM

P: 232




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