Why Split Improper Integrals?

[Dethrone]

From textbooks, I usually see that when there is an integral like this:
$$\int_{-\infty}^{+\infty} f(x)\,dx$$, they generally split it two, usually by 0.
$$\int_{-\infty}^{0} f(x)\,dx + \int_{0}^{\infty} f(x) \,dx$$
They do the same for points of discontinuity, but if you notice, the number that they use to divide the integral in two, cancels out!

Why can't we do something like this?
$$ \lim_{{u}\to{-\infty}} \lim_{{v}\to{\infty}} \int_{u}^{v} f(x)\,dx$$

Or is there a rule against having two limits? (Now that I think about it, there probably is a rule against it, but nevertheless, I will proceed with my post)

It just seems tedious, where, if you get an integral which is not only bounded by infinities, but also have discontinuities , you would have to split it into at least 4-6 integrals. Take my friend ineedhelpnow's integral.
$$\int_{-\infty}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$, you would divide the integral into 4 parts.

 

[Aryth1]

I'm pretty sure you can do the double limits if there are no discontinuities. I'm not positive though since I never did it that way.

As for the discontinuities, remember the rule for improper integrals:

If $\int_a^t f(x) ~dx$ exists for every $t > a$, then $$\int_a^{\infty} f(x) ~dx = \lim_{{t}\to{\infty}}\int_a^t f(x) ~dx$$ provided the limit exists and is not infinite.

You run into problems if you don't split the integral at discontinuities.

 

[Dethrone]

That is what I'm trying to get at. $$\displaystyle \int_{-\infty}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$
If you need to split the integral for both discontinuities and infinities, then for the above integral, you're going to have to split it 4 times. Do you usually chose a variable to split the integral, or just use a number? I guess it doesn't matter since it ends up cancelling, though.

 

[Aryth1]

That is what I'm trying to get at. $$\displaystyle \int_{-\infty}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$
If you need to split the integral for both discontinuities and infinities, then for the above integral, you're going to have to split it 4 times. Do you usually chose a variable to split the integral, or just use a number? I guess it doesn't matter since it ends up cancelling, though.

Well, as you said, you can technically split it up however you like, but as with most problems there are reasonably good choices you can make.

If you're splitting the interval, convenient choices are at the origin or at discontinuities, and you choose discontinuities because you can be more certain about the limits and the integrals existence (You don't want to accidentally have divergence in your integrals and get nowhere). If you split at infinities, you would use variables since you have to take the limits. If you split at discontinuities, then you use the value at which the discontinuity occurs (of course, provided at least one of either the left or right limits exists and is not infinity).

As for your example integral, I saw that it was covered in the original thread that there is no real value for this integral since the function is not defined in the negative reals... So splitting it doesn't really seem to have a point...

 

[Dethrone]

Let's assume that the integral is possible.

$$\displaystyle \int_{-\infty}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$

Discontinuity at x = 0 and x = -2.

$$=\displaystyle \int_{-\infty}^{-2} \ \frac{dx}{(2+x) \sqrt{x}} + \displaystyle \int_{-2}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$

So would I need to further split these integrals? I have to take the limit as they approach infinity, but I also have to the limit as it approaches the discontinuities.

$$=\displaystyle \int_{-\infty}^{a} \ \frac{dx}{(2+x) \sqrt{x}} + \displaystyle \int_{a}^{-2} \ \frac{dx}{(2+x) \sqrt{x}} +\displaystyle \int_{-2}^{b} \ \frac{dx}{(2+x) \sqrt{x}} + \displaystyle \int_{b}^{0} \ \frac{dx}{(2+x) \sqrt{x}}$$

Then I would have one limit per integral? For example, the first integral would be $$\lim_{{u}\to{-\infty}}$$, and the second integral would be $$\lim_{{v}\to{-2^-}}$$, or am I overcomplicated it? If so, how should it be attempted?

 

[chisigma]

The first thought that comes to me when I am faced with an 'impossible' integral is: take it easy! ... first let's see if the integrand admits a primitive elementary and now we see that it is so. Setting $\displaystyle u = \sqrt{x}$ we obtain ...

$\displaystyle \int \frac{d x}{(2 + x)\ \sqrt{x}} = 2\ \int \frac{d u}{2 + u^{2}} = \sqrt{2}\ \tan^{-1} \frac{u}{\sqrt{2}} + c = \sqrt{2}\ \tan^{-1} \sqrt{\frac{x}{2}} + c\ (1)$

Now we proceed correctly and isolate the singularity x = -2 calculating the limit ...

$\displaystyle I = \lim_{b \rightarrow 0 +} \lim_{a \rightarrow - \infty} \int_{a}^{- 2 - b} \frac{d x}{(2 + x)\ \sqrt{x}} + \int_{- 2 + b}^{0} \frac{d x}{(2 + x)\ \sqrt{x}} = \sqrt{2}\ \lim_{b \rightarrow 0+} \lim_{a \rightarrow - \infty} - \tan^{-1} \sqrt{\frac{a}{2}} + \tan^{-1} \sqrt{\frac{-2 - b}{2}} - \tan^{-1} \sqrt{\frac{-2 + b}{2}} = \sqrt{2}\ i\ \lim_{t \rightarrow \infty} \tanh^{-1} t = \frac{\pi}{\sqrt{2}}\ (2)$

The procedure now seen for the calculation of an improper integral is called 'at principal value' and is written as...

$\displaystyle I = \text{PV}\ \int_{- \infty}^{0} \frac{d x}{(2 + x)\ \sqrt{x}} = \frac{\pi}{\sqrt{2}}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

That's great. I just wanted to know how someone would split the integral. My textbooks split discontinuities and infinity bounded integrals the same way,
i.e for infinity-bounded integrals:
$$\int_{-\infty}^{0} f(x)\,dx = \lim_{{n}\to{-\infty}}\int_{n}^{0} f(x)\,dx$$
i.e for discontinuities:
$$\int_{-5}^{0} \frac{1}{x^2}\,dx = \lim_{{n}\to{0^-}}\int_{-5}^{0} \frac{1}{x^2}\,dx$$

and amalgamating the process, it would look something like this:$$=\displaystyle \lim_{{u}\to{-\infty}}\int_{u}^{a} \ \frac{dx}{(2+x) \sqrt{x}} + \displaystyle \lim_{{v}\to{-2^-}}\int_{a}^{v} \ \frac{dx}{(2+x) \sqrt{x}} +\displaystyle \lim_{{w}\to{-2^+}}\int_{w}^{b} \ \frac{dx}{(2+x) \sqrt{x}} + \displaystyle \lim_{{z}\to{0^-}}\int_{b}^{z} \ \frac{dx}{(2+x) \sqrt{x}}$$

The way you did it, chisigma, is much more efficient when you have discontinuities. And if you're allowed to use double limits like that, wouldn't it be more feasible (assuming that there isn't any discontinuities) to do this?:

$$\displaystyle \lim_{{u}\to{-\infty}} \lim_{{v}\to{\infty}} \int_{u}^{v} f(x)\,dx $$ instead of $$\displaystyle \int_{-\infty}^{0} f(x)\,dx + \int_{0}^{\infty} f(x) \,dx$$?