Integrating out of the real domain of a function

piercebeatz

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?

piercebeatz

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

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.
Use \lim, not lim.

piercebeatz

So one can never integrate out of a function's domain? What about in complex analysis?

haruspex

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]

piercebeatz

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

Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=re^iθ). It's better not to confuse the two.
No, for the reasons given in other posts.

piercebeatz

OK, but what did you do to transform the integral into that form?

haruspex

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^iπ.
[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]

piercebeatz

Alright. What exactly do you mean by a "path"? And why did the limits of integration change?