Is this valid for improper integrals.

[tnutty]

note: INT = integral and the above and bottom arguments are the bounds for the integral.


infinity
INT [ f(x) ]
-infinity

=

lim a - >-infinity and lim b-> infinity of :
b
INT [f(x)] = pi
a

I am just wondering if the move having two limits in one integral is valid,
as done above. Or would I have to separate it

 

[Cyosis]

I suppose it is, but usually we write it as:

$$\lim_{a \to \infty} \int_{-a}^a \arctan x dx$$

 

[tnutty]

Oh, even better and more efficient, Thanks. To me its useless to split it into two integrals because there is not need.

 

[HallsofIvy]

I suppose it is, but usually we write it as:

$$\lim_{a \to \infty} \int_{-a}^a \arctan x dx$$

Not, that is NOT how we usually write it! That is the "principal value" of the integral and can be very misleading.

For example, if we write
$$\int_{-\infty}^\infty \frac{1}{x} dx$$
as
$$\lim_{a\rightarrow \infty} \int_{-a}^a \frac{1}{x}dx$$
it becomes
$$\lim_{a\rightarrow \infty} ln(|a|)- ln(|-a|)= 0$$

The standard way to write it is, in fact,
$$\lim_{a\rightarrow -\infty}\lim_{b\rightarrow \infty}\int_a^b f(x) dx$$
which, for f(x)= 1/x, does not exist.

 

[Cyosis]

I'm confused myself now, while I see it going wrong for the limit to infinity this raises a question for the interval (-1,1).

$$\lim_{a \to 1} \int_{-a}^a \frac{1}{x}dx=\lim_{a \to 1} \ln|a|-\ln|-a|=0$$

Now do this again with double limits.
$$\lim_{a \to -1} \lim_{b \to 1} \int_a^b \frac{1}{x} dx=\lim_{a \to -1} \lim_{b \to 1} (\ln|b|-\ln|a|)=\ln|1|-\ln|-1|=0$$

But this integral should diverge should it not? So given that the function has a singularity somewhere within its integration interval should we not also split the integral up in, in this case -1,0 and 0,1?

For example:

$$\lim_{a \to -1} \lim_{\epsilon \uparrow 0} \int_a^\epsilon \frac{1}{x}dx+\lim_{b \to 1} \lim_{\epsilon \downarrow 0} \int_\epsilon^b \frac{1}{x} dx$$