## Integrating out of the real domain of a function

Check this out:

$$\int_{-1}^{0} ln(x) dx$$

$$u=ln(x), dv=dx$$
$$du=\frac{1}{x},v=x$$

$$\int_{-1}^{0} ln(x) dx= x\space ln(x) \|_{-1}^{0} -\int_{-1}^{0} \frac{x}{x} dx$$
$$=x\space ln(x) \|_{-1}^{0}-x \|_{-1}^{0}$$
$$=\lim_{x\to0} x\space ln(x)-(-ln(-1))-(0-(-1))$$
$$\lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}$$
$$=\lim_{x\to0} \frac{1/x}{-1/x^2}$$
$$=\lim_{x\to0} -x=0$$

$$\int_{-1}^{0} ln(x)dx=ln(-1)-1$$
$$e^{i\pi}=-1\to ln(-1)=i\pi$$

$$\int_{-1}^{0}ln(x)\space dx=i\pi-1$$

Is this legitimate?

P.S. Why don't my limits look right?

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 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.

Mentor
 Quote by piercebeatz Check this out: $$\int_{-1}^{0} ln(x) dx$$
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 by piercebeatz $$u=ln(x), dv=dx$$ $$du=\frac{1}{x},v=x$$ $$\int_{-1}^{0} ln(x) dx= x\space ln(x) \|_{-1}^{0} -\int_{-1}^{0} \frac{x}{x} dx$$ $$=x\space ln(x) \|_{-1}^{0}-x \|_{-1}^{0}$$ $$=lim_{x\to0} x\space ln(x)-(-ln(-1))-(0-(-1))$$ $$lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}$$ $$=lim_{x\to0} \frac{1/x}{-1/x^2}$$ $$=lim_{x\to0} -x=0$$ $$\int_{-1}^{0} ln(x)dx=ln(-1)-1$$ $$e^{i\pi}=-1\to ln(-1)=i\pi$$ $$\int_{-1}^{0}ln(x)\space dx=i\pi-1$$ Is this legitimate? P.S. Why don't my limits look right?
Use \lim, not lim.

## Integrating out of the real domain of a function

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

Recognitions:
Homework Help
 Quote by piercebeatz 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:
$\int_{-1}^0ln(z)dz = \int_{1}^0(ln(r)+i\pi) d(-r) = \left[r ln(r) - r + i\pi r \right]_0^1 = i\pi - 1$

 Quote by haruspex Reinterpreting your integral in the complex plane, we need r >= 0, so over the x interval (-1,0) we will have θ=π: $\int_{-1}^0ln(z)dz = \int_{1}^0(ln(r)+i\pi) d(-r)$
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)?

Recognitions:
Homework Help
 Quote by piercebeatz 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.

 Quote by haruspex Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=reiθ.
OK, but what did you do to transform the integral into that form?

Recognitions:
Homework Help
$\int_Pln(z)dz = \int_P(ln(re^{iπ}) dre^{iπ}= e^{iπ}\int_{r=1}^0(ln(r)+iπ) dr = -\int_{r=1}^0(ln(r)+iπ) dr$
$= \int_{r=0}^1(ln(r)+iπ) = \left[(r ln(r)-r+iπr)\right]_0^1 = -1+iπ$
 Quote by 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 reiπ. $\int_Pln(z)dz = \int_P(ln(re^{iπ}) dre^{iπ}= e^{iπ}\int_{r=1}^0(ln(r)+iπ) dr = -\int_{r=1}^0(ln(r)+iπ) dr$ $= \int_{r=0}^1(ln(r)+iπ) = \left[(r ln(r)-r+iπr)\right]_0^1 = -1+iπ$