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Integrating out of the real domain of a function

by pierce15
Tags: domain, function, integrating, real
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pierce15
#1
Nov27-12, 11:43 AM
P: 246
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\pi-1[/tex]

Is this legitimate?

P.S. Why don't my limits look right?
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pierce15
#2
Nov27-12, 02:11 PM
P: 246
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.
Mark44
#3
Nov27-12, 02:33 PM
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Quote Quote by piercebeatz View Post
Check this out:

[tex]\int_{-1}^{0} ln(x) dx[/tex]
Nothing below is legitimate. For the Fundamental Theorem of calculus to apply, the function (ln(x) here) has to be defined on the interval [-1, 0], and continuous on the interior of this interval. ln(x) is defined only for x > 0.
Quote Quote by piercebeatz View Post

[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\pi-1[/tex]

Is this legitimate?

P.S. Why don't my limits look right?
Use \lim, not lim.

pierce15
#4
Nov27-12, 03:23 PM
P: 246
Integrating out of the real domain of a function

Quote Quote by Mark44 View Post
Nothing below is legitimate. For the Fundamental Theorem of calculus to apply, the function (ln(x) here) has to be defined on the interval [-1, 0], and continuous on the interior of this interval. ln(x) is defined only for x > 0.
So one can never integrate out of a function's domain? What about in complex analysis?
haruspex
#5
Nov27-12, 05:59 PM
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Quote Quote by piercebeatz View Post
So one can never integrate out of a function's domain? What about in complex analysis?
Reinterpreting your integral in the complex plane, we need r >= 0, so over the x interval (-1,0) we will have θ=π:
[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]
pierce15
#6
Nov27-12, 06:12 PM
P: 246
Quote Quote by haruspex View Post
Reinterpreting your integral in the complex plane, we need r >= 0, so over the x interval (-1,0) we will have θ=π:
[itex]\int_{-1}^0ln(z)dz = \int_{1}^0(ln(r)+i\pi) d(-r)[/itex]
How did you go to polar coordinates?

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)?
haruspex
#7
Nov27-12, 07:25 PM
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Quote Quote by piercebeatz View Post
How did you go to polar coordinates?
Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=re). It's better not to confuse the two.
is integrating the function without reinterpreting it into the complex plane o.k.
No, for the reasons given in other posts.
pierce15
#8
Nov27-12, 08:10 PM
P: 246
Quote Quote by haruspex View Post
Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=re.
OK, but what did you do to transform the integral into that form?
haruspex
#9
Nov27-12, 08:44 PM
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Quote Quote by piercebeatz View Post
OK, but what did you do to transform the integral into that form?
First, you need to assume a specific path, P (though for the ln() function it won't matter as long as we don't go around the origin to get there). Assume it's along the negative real axis. So at all points in the path, z is of the form re.
[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]
pierce15
#10
Nov27-12, 08:59 PM
P: 246
Quote Quote by haruspex View Post
First, you need to assume a specific path, P (though for the ln() function it won't matter as long as we don't go around the origin to get there). Assume it's along the negative real axis. So at all points in the path, z is of the form re.
[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]
Alright. What exactly do you mean by a "path"? And why did the limits of integration change?


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