Double Limits When Defining Improper Integral

Question

In my textbook , an improper integral is defined as shown below, where $F$ is a primitive function of a function $f$.

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\lim_{b\to+\infty}F(b)-\lim_{a\to-\infty}F(a) \tag{1}$$

Could I rewrite it in one of the following ways?

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\lim_{a\to+\infty}\int_{-a}^af(x)\mathrm{\ d}x \tag{2}$$

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\lim_{a\to-\infty}\lim_{b\to+\infty}\int_a^bf(x)\mathrm{\ d}x \tag{3}$$

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\lim_{b\to+\infty}\lim_{a\to-\infty}\int_a^bf(x)\mathrm{\ d}x \tag{4}$$

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\underset{b\to+\infty}{\lim_{a\to-\infty}}\int_a^bf(x)\mathrm{\ d}x \tag{5}$$

$$\int_{-\infty}^{+\infty}f(x)\mathrm{\ d}x\mathrel{\overset{\text{def}}{=\mathrel{\mkern-3mu}=}}\lim_{(a,b)\to(-\infty,+\infty)}\int_a^bf(x)\mathrm{\ d}x \tag{6}$$

What would be the differences between each one of these notations (talking about the limits)? Actually, are any of them even valid?

Answer 1

Example 1 :

Consider $f(x)=x+c$ , where $c$ is a Constant.

When we Integrate it with (1) , we will get $[b^2/2+cb]-[a^2/2+ca]$ which is $\infty-\infty$ & (1) will not Exist , even when $c$ is Zero.

When we Integrate it with (2) , we will get $[a^2/2+ca]-[(-a)^2/2+c(-a)]=2ca$ which will not Exist when $c$ is Non-Zero , though it will Exist when $c$ is Zero , in which Case the Negative Area "cancels" the Positive Area.

With Current function , (3) & (4) are almost Equivalent to (1) , though there are other functions where these are not Equivalent.

With (5) & (6) ( which are entirely Equivalent to each other ) , we have Single limit , hence we can control the variables , eg : We can make $b=-a+c$ , ensuring that $(a,b) \to (\infty,-\infty)$ : then Integral will be $0$ , the Negative Area "cancelling" the Positive Area.
We might alternately make $b=-a^2$ , ensuring that $(a,b) \to (\infty,-\infty)$ : then Integral will tend to $-\infty$ , because we are making the Negative Area larger than the Positive Area.
In Case we make it $b=-a+c+1$ , then Integral will tend to $+\infty$ , because we are making the Positive Area larger than the Negative Area.
Let $c=0$ , then $b=-a+D/a$ ( where $D$ is a Constant ) will make the Positive Area & Negative Area "almost" Equal , except for a thin Part with Area $D$ : Overall Integral will be $D$.
We get varying values : This indicates that (5) & (6) will not Exist.

Example 2 :

Consider $f(x)=\sin(x)$ , where Area across 1 cycle is $0$ , though the Integral will not Exist in the limit.
(1) , (3) , (4) will not Exist
(2) will be Zero.
(5) & (6) may be Zero when $b=-a+2\pi$ , thought Integral Value can vary with $b=-a+D$ , & it will not Exist when $b=-a^2$

Example 3 :

Consider $f(x)=e^{-(x-c)^2}$ , which has Same value with all the given Integrals.

Example 4 :

Consider $f(x)=\frac{(x-c)}{((x-c)^2+1)}$ , which has Positive Area beyond $x=c$ , though that is "cancellable" with the Negative Area.

ADDENDUM :

(1) When taking limits with Multivariable Calculus , where we have 2 or more variables in the limit , we say that the limit Exists when we get Exact Same Value no matter which way the limit is approached.

(1A) Similarly , (5) & (6) will Exist when we get Exact Same Value no matter which way the variables tend to $\infty$ , otherwise we have to say that the Integral will not Exist.

(1B) With the limit Concept , let us say we get (3) & (4) like $a/b$ : then , Overall Integral will be $a/\infty=0$ with (3) , while it will be $-\infty/b=-\infty$ with (4) , hence Order will matter.

(2) We will have Equivalence between the Integrals when there is Positive Area all over. Equivalence will be lost when we want "cancelation" , which will require that all Possible ways to "cancel" will give that Exact Same Value.

SUMMARY :

Each Integral [ Except (6) ] is unique here , not necessarily Equivalent , in general.

Answer 2

$$I_1=\lim_{b\to+\infty}F(b)-\lim_{a\to-\infty}F(a)$$ $$I_2=\lim_{a\to+\infty}\Big(F(a)-F(-a)\Big)$$ $$I_3=\lim_{a\to-\infty}\bigg(\lim_{b\to+\infty}\Big(F(b)-F(a)\Big)\bigg)$$ $$I_4=\lim_{b\to+\infty}\bigg(\lim_{a\to-\infty}\Big(F(b)-F(a)\Big)\bigg)$$ $$I_5=\lim_{a\to-\infty\\b\to+\infty}\Big(F(b)-F(a)\Big)$$

Note that $I_1$ exists if and only if the two limits

$$L_+=\lim_{x\to+\infty}F(x)$$ $$L_-=\lim_{x\to-\infty}F(x)$$

both exist.

Supposing $I_1$ exists, it follows that all the others exist and have the same value:

$$I_2=\bigg(\lim_{a\to+\infty}F(a)\bigg)-\bigg(\lim_{-a\to-\infty}F(-a)\bigg)=L_+-L_-$$ $$I_3=\lim_{a\to-\infty}\Bigg(\bigg(\lim_{b\to+\infty}F(b)\bigg)-F(a)\Bigg)=\lim_{a\to-\infty}\bigg(L_+-F(a)\bigg)$$ $$=L_+-\bigg(\lim_{a\to-\infty}F(a)\bigg)=L_+-L_-$$ $$I_4=\lim_{b\to+\infty}\Bigg(F(b)-\bigg(\lim_{a\to-\infty}F(a)\bigg)\Bigg)=\lim_{b\to+\infty}\bigg(F(b)-L_-\bigg)$$ $$=\bigg(\lim_{b\to+\infty}F(b)\bigg)-L_-=L_+-L_-$$ $$I_5=\Bigg(\lim_{a\to-\infty\\b\to+\infty}F(b)\Bigg)-\Bigg(\lim_{a\to-\infty\\b\to+\infty}F(a)\Bigg)=L_+-L_-$$

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It has been noted in comments that $I_2$ may exist while $I_1$ does not. For example, take $f(x)=x/(x^2+1)$, so $F(x)=\tfrac12\ln(x^2+1)$, which gives $I_2=0$ but $L_+=L_-=\infty$.

On the other hand, excluding $I_2$, all of them are equivalent.

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Suppose $I_3$ exists:

$$I_3=\lim_{a\to-\infty}\bigg(\lim_{b\to+\infty}\Big(F(b)-F(a)\Big)\bigg)$$

In order for the outer limit to exist, the inner limit must exist for every $a$ (though we only need it for one value of $a$):

$$\lim_{b\to+\infty}\Big(F(b)-F(a)\Big)$$

Any constant can be added to a limit, so we find that $L_+$ exists:

$$L_+=\lim_{b\to+\infty}F(b)=\lim_{b\to+\infty}\Big(F(b)-F(a)\Big)+F(a)$$

Then we have

$$I_3=\lim_{a\to-\infty}\bigg(L_+-F(a)\bigg)$$

and again by adding a constant (and negating) we find that $L_-$ exists:

$$L_-=\lim_{a\to-\infty}F(a)=L_+-\lim_{a\to-\infty}\bigg(L_+-F(a)\bigg)$$

Therefore $I_1=L_+-L_-$ also exists.

By similar reasoning, $I_4$ implies $I_1$.

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And $I_5$ implies $I_3$ and $I_4$. But we need more than those simple limit laws (e.g. "limit of sum is sum of limits"). Consider something like $\cos(a)/b$, which has a limit as $(a,b)\to(\infty,\infty)$ but no limit as only $a\to\infty$. Of course, that doesn't have the form $F(b)-F(a)$.

The definition of $I_5$ is this:

$$\forall\varepsilon>0,\;\exists D>0,\;\forall a<-D,\;\forall b>D,\;|F(b)-F(a)-I_5|<\varepsilon$$

(In words: For any neighbourhood of the point $I_5\in\mathbb R$, there is some neighbourhood of $\infty$ (that is a half-infinite interval) such that, if both $b$ and $-a$ are large enough to be in the latter neighbourhood, then $F(b)-F(a)$ is close enough to $I_5$ to be in the former neighbourhood.)

Based on that, we want to show that $L_+$ exists. We can use Cauchy's criterion to prove that it exists without knowing what it is exactly. The criterion is

$$\forall\varepsilon'>0,\;\exists D'>0,\;\forall x_1>D',\;\forall x_2>D',\;|F(x_1)-F(x_2)|<\varepsilon'$$

(In words: For any distance $\varepsilon'$, there is some neighbourhood of $\infty$ such that all points in that neighbourhood have their $F$ values close to each other, closer than $\varepsilon'$.)

So, given any $\varepsilon'>0$, we can plug in $\varepsilon=\tfrac12\varepsilon'$ to the definition of $I_5$, which will give us some $D$ with the above properties. Then it suffices to take $D'=D$. Indeed, for any $x_1>D$ and $x_2>D$,

$$|F(x_1)-F(x_2)|=\bigg|\Big(F(x_1)-F(-x_1)-I_5\Big)-\Big(F(x_2)-F(-x_1)-I_5\Big)\bigg|$$ $$\leq\Big|F(x_1)-F(-x_1)-I_5\Big|+\Big|F(x_2)-F(-x_1)-I_5\Big|$$ $$<\varepsilon+\varepsilon\quad=\varepsilon'$$

Therefore $L_+$ exists. By similar reasoning, $L_-$ exists, and thus $I_1$.