Integrating out of the real domain of a function

[pierce15]

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?

 

[pierce15]

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]

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.

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.

 

[pierce15]

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]

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

 

[pierce15]

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)?

 

[haruspex]

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^iθ). 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]

Not polar co-ordinates (which would be (r,θ)) but the polar form of complex variables (z=re^iθ.

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

 

[haruspex]

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^iπ.
\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

 

[pierce15]

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.
\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

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