Can Integrals Ever Blow Up and What Does That Mean?

[eep]

Integrals that "blow up"

Hi, I'm trying to figure out if there's any way I can integrate a function over a range which includes a value that the integral "blows up" at, for example 1/x integrated from -1 to 1.

 

[Hurkyl]

Yes; it's called an improper integral. In this case, you split your integral into two pieces, so that the singularity is at the boundary. Then, you replace the bound with a variable, and take the limit as the bound goes to zero.

(In this particular case, the result is that the integral does not exist)

 

[eep]

so 1/x cannot be integrated over any range that passes through 0?

 

[Hurkyl]

Not even a range that just touches zero!

But for fun, try integrating $1/\sqrt{x}$.

 

[eep]

What about the intergral of 1/|x|? (absolute value of 1/x)

I don't have a problem with $\frac{1}{\sqrt{x}}$ because $x^\frac{1}{2}$ is defined everywhere (even though you might get imaginary numbers)

 

[Hurkyl]

Oh, my bad, I meant $1 / \sqrt{ |x| }$.

You can most easily answer your question by working it out. You already know the answer for the integral of 1/x, right? And what is the definition of |x|?

 

[eep]

Well, what I want to try and do for 1/|x| from say -1 to 1 is to change the integral into the integral of -1/x from -1 to 0 + the integral of 1/x from 0 to 1 but this doesn't work either. I tried replaced the 0's with h's and taking the limit but you still end up with the same problem. Grr. Can it just not be evaluated at all? |x| is just the absolute value of x. It gets rid of any sign information.

 

[eep]

A thought... suppose we are trying to do the integral of 1/|x| from -1 to 1. We make the substitution u^2 = x, therefore...

$$2udu = dx$$

$$\int_{-1}^{1} \frac{1}{|x|} dx = \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2u}{|u^2|} du<br /> <br /> = \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2}{u}<br /> <br /> = 2 (\ln{|1|} - \ln{|1|})<br /> <br /> = 0<br />$$
I can't immediately see what's wrong with this but I'm sure this can't be right as there is obviously only positive area under the graph of 1/|x|.

 

[HallsofIvy]

A thought... suppose we are trying to do the integral of 1/|x| from -1 to 1. We make the substitution u^2 = x, therefore...

$$2udu = dx$$

$$\int_{-1}^{1} \frac{1}{|x|} dx = \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2u}{|u^2|} du<br /> <br /> = \int_{\sqrt{-1}}^{\sqrt{1}} \frac{2}{u}<br /> <br /> = 2 (\ln{|1|} - \ln{|1|})<br /> <br /> = 0<br />$$
I can't immediately see what's wrong with this but I'm sure this can't be right as there is obviously only positive area under the graph of 1/|x|.

In your first post, you asked how to do an integral like this and, in his first response, Hurkyl told you how. Perhaps, "what's wrong with this" is that you are simply ignoring what he said! In general, if there is a singularity in the middle of an interval, you cannot just ignore it and evaluate the anti-derivative at the two endpoints.

Since 1/|x| is not defined at 0, integrate from -1 to some negative $\alpha$, integrate from some positive $\beta$ and add. Then take the limit, if it exists, as $\alpha$ and $\beta$ go to 0. That is:
$$\int_{-1}^\alpha \frac{dx}{|x|}+ \int_\beta^1 \frac{dx}{|x|}$$
Since |x|= -x for x< 0 and |x|= x for x> 0, this is just
$$-\int_{-1}^\alpha \frac{dx}{x}+ \int_\beta^1 \frac{dx}{x}$$
$$= -ln|x|\left|_{-1}^\alpha+ ln|x|left|_\beta^1$$
$$= -ln|\alpha|+ ln|-1|+ ln|\beta|- ln|1|= ln|\beta|- ln|\alpha|$$
Finally, take the limit as $\alpha$ and $\beta$ go to 0 independently. In this case, that limit does not exist so the integral does not exist..

 

[reilly]

When you use complex variable theory, the integral of 1/x or 1/z can be defined in terms of what's called a principal part integral or principal-value integral. And then the integral of 1/x from -R to R can be shown to be zero -- it is, after all, an odd function.

This approach goes back to the turn of the 20th century when much research was being done on solutions of the wave equations in terms of Green's functions. You'll find discussions in most texts on advanced E&M, or on complex variable theory and, so-called Hilbert Transforms. Byron and
Fuller's Mathematics of Classical and Quantum Physics (Dover) gives a detailed discussion

Regards,
Reilly Atkinson

 

[eljose]

By symmetry the integral $$\int_{-1}^{1}dx/x$$ is 0 taking Cauchy,s principal value for integral.