Невласні інтеграли с особливістю на нескінченності

One may initially think that any area that is unbounded is infinite. This is a common misconception. While some regions certainly do have an infinite area, think about it this way: as the upper bound increases or the lower bound decreases, the area of the region increases, up until one of the bounds ceases to exist. This state of boundlessness is approached in the form of a limit. Many of the integrals that represent the areas of such regions, known as improper integrals, have an infinite upper bound and are written as

$\lim_{b \rightarrow \infty}\int^{b}_{a}f(x)dx$

However, in some cases the lower bound becomes progressively smaller until it ceases to exist. In this case, the improper integral is

$\lim_{a \rightarrow -\infty}\int^{b}_{a}f(x)dx$

Example: The integral $\int^{b}_{a}e^{-x}dx$ represents the area under the curve $f(x) = e^{-x}$ from $a$ to $b$. Let's say that the lower bound is $0$, then

$A = \int^{b}_{0}e^{-x}dx$

Now let's say the upper bound continually increases until it is infinite. So the area is a limit of this upper bound as it approaches infinity:

$A = \lim_{b \rightarrow \infty}\int^{b}_{0}e^{-x}dx$

Always find the indefinite integral in terms of the bounds first:

$A = \lim_{b \rightarrow \infty}[-e^{-x}]^{b}_{0} =$

$A = \lim_{b \rightarrow \infty}(-e^{-b} -(-e^{-0})) =$

$A = \lim_{b \rightarrow \infty}(-e^{-b} + 1) = 0 + 1 = 1$

The calculations when the lower bound is boundless work in the same general fashion.

If both bounds do not exist, split up the integral at a constant; in other words, pick a real constant $c$. It can be any constant you wish as by the additive rule of integrals:

$\int^{\infty}_{-\infty}f(x)dx = \int^{\infty}_{c}f(x)dx + \int^{c}_{-\infty}f(x)dx$

Previously, we found out that unbounded regions can, in fact, have finite area! So, a natural next step would be to try and calculate that area. This leads us into the notion of improper integrals.

There are two different kinds of improper integrals: in this section, we deal with integrals in which one or both limits of integration are infinity. Geometrically, this involves regions that are unbounded horizontally.

But before we begin, let's look at a couple of examples:

Great! How about the next problem?

Do you notice a pattern, as the higher limit of integration increases?

If we let $t$ be the upper limit of integration, the we find that $$\int_1^t \frac{1}{x^2}\, dx=-\frac{1}{x}\bigg|_1^t=-\frac{1}{t}-\left(-\frac{1}{1}\right)=1-\frac{1}{t}.$$ So, as $t$ increases ``to infinity,'' the value of the integral seems to increase and converge to $1$. Our goal is to formalize what we just discovered.

To continue, let's look at what we've noticed graphically:

So, for instance, the value of the integral $$\int_0^t \frac{1}{x^2}\, dx$$ can be seen as the area of the region, $R$, bounded by the $x$-axis, the graph of $y=\frac{1}{x^2}$, and the lines $x=1$ and $x=t$.

Aha! We can now see that as $t$ increases to infinity, the value of the integral becomes the area of this eventually unbounded region! This is how we can calculate the area of unbounded regions with finite area: by taking the limit of the integral as $t\to\infty$.

Formally, we can state the following:

1. Suppose that the integral $$\int_a^t f(x)\, dx$$ exists for all $t>a$. Then $$\int_a^{\infty} f(x)\, dx=\lim_{t\to\infty} \int_a^t f(x)\, dx,$$ as long as the limit exists.

2. Similarly, if the integral $$\int_{-\infty}^a f(x)\, dx$$ exists for all $t<a$, then $$\int_{-\infty}^a f(x)\, dx=\lim_{t\to-\infty} \int_t^a f(x)\, dx,$$ as long as the limit exists.

3. If $\int_{-\infty}^a f(x)\, dx$ and $\int_a^{\infty} f(x)\, dx$ are both finite, then $$\int_{-\infty}^{\infty} f(x)\, dx=\int_{-\infty}^a f(x)\, dx+\int_a^{\infty} f(x)\, dx.$$

Lastly, if the limit for the improper integral exists and is finite, we call it convergent. Otherwise, we call it divergent.

Now, let's look at some examples: first, let's consider the previous example: $$\int_1^{\infty} \frac{1}{x^2}\, dx.$$ In the beginning of this section, we indirectly tried to solve this integral! We found earlier that $$\int_1^t \frac{1}{x^2}\, dx=1-\frac{1}{t}.$$ So, by the definition of the improper integral, $$\int_1^{\infty} \frac{1}{x^2}\, dx=\lim_{t\to\infty} \int_1^t \frac{1}{x^2}\, dx=\lim_{t\to\infty} \left(1-\frac{1}{t}\right)=1,$$ which is what we expected. By definition, therefore, this integral converges. (Isn't it still remarkable that the area of that infinite, unbounded region is just $1$?)

Let's look at one more problem: $$\int_1^{\infty} \frac{1}{x}\, dx.$$ By the definition of improper integrals listed above, $$\int_1^{\infty} \frac{1}{x}\, dx=\lim_{t\to\infty} \int_1^t \frac{1}{x}\, dx=\lim_{t\to\infty} \ln{x}\bigg|_1^t$$ $$=\lim_{t\to\infty} \left(\ln{t}-\ln{0}\right)=\lim_{t\to\infty} \ln{t}=\infty.$$ In this case, the area of the unbounded region (between the $x$-axis, the line $x=1$, and the curve of $y=\frac{1}{x}$) has infinite area, and therefore the integral diverges.

Now you try!

Let's use the definition of improper integrals, as before: $$\int_{\pi}^{\infty} \cos{x}\, dx=\lim_{t\to\infty} \int_{\pi}^t \cos{x}\, dx\\=\lim_{t\to\infty} \sin{x}\bigg|_{\pi}^t=\lim_{t\to\infty} (\sin{t}-\sin{\pi})=\lim_{t\to\infty} \sin{t}.$$ Since $\sin{x}$ fluctuates between $-1$ and $1$ forever, the limit does not exist! Therefore, the integral diverges.

Now, try the next problem!

Since we have infinities as both limits of integration, we should choose a value of $a$ for which $\int_{-\infty}^a xe^{-x^2}\, dx$ and $\int_a^{\infty} xe^{-x^2}\, dx$ both converge. For this problem, we'll set $a=0$ for simplicity.

So now, we need to check whether the two integrals converge for the chosen value of $a$. We have $$\int_{-\infty}^0 xe^{-x^2}\, dx=\lim_{t\to-\infty} \int_t^0 xe^{-x^2}\, dx.$$ If we let $u=-x^2$, then $du=-2x\, dx$, and so $$\lim_{t\to-\infty} \int_t^0 xe^{-x^2}\, dx=\lim_{t\to-\infty} -\frac{1}{2}\int_{-t^2}^0 e^u\, du=\lim_{t\to-\infty} -\frac{1}{2}e^u\bigg|_{-t^2}^0$$ $$=\lim_{t\to-\infty} -\frac{1}{2}\left(e^0-e^{-t^2}\right)=\lim_{t\to-\infty} \left(-\frac{1}{2}+e^{-t^2}\right)=-\frac{1}{2}.$$ In almost the exact same way, we can also see that $$\int_0^{\infty} xe^{-x^2}\, dx=\frac{1}{2}.$$ Therefore, by definition $$\int_{-\infty}^{\infty} xe^{-x^2}\, dx=\int_{-\infty}^0 xe^{-x^2}\, dx+\int_0^{\infty} xe^{-x^2}\, dx =-\frac{1}{2}+\frac{1}{2}=0.$$

Finally, now that you know enough about improper integrals with infinity, can you answer this challenge question?

For what values of $p$ does the integral $$\int_1^{\infty} \frac{1}{x^p}\, dx$$ converge?

We consider three cases: when $p>1$, when $p=1$, and when $p<1$.

1. Suppose $p>1$. Then $$\int_1^{\infty} \frac{1}{x^p}\, dx=\lim_{t\to\infty} \int_1^t x^{-p}\, dx=\lim_{t\to\infty} \frac{1}{-p+1}x^{-p+1}\bigg|_1^t$$ $$=\lim_{t\to\infty} \frac{1}{1-p} \left(t^{1-p}-1^{1-p}\right)=\lim_{t\to\infty} \frac{1}{1-p}\left(\frac{1}{t^{p-1}}-1\right).$$ Since $p-1>0$, we have $$\lim_{t\to\infty} \frac{1}{t^{p-1}}=0.$$ Therefore, the integral converges to $$\int_1^{\infty} \frac{1}{x^p}\, dx=-\frac{1}{1-p}.$$

2. If $p=1$, by the work done above we know that $$\int_1^{\infty} \frac{1}{x}\, dx=\lim_{t\to\infty} \ln{t}=\infty.$$ So, if $p=1$ the integral diverges.

3. Suppose $p<1$. By the same steps as from case 1, $$\int_1^{\infty} \frac{1}{x^p}\, dx=\lim_{t\to\infty} \frac{1}{1-p}\left(t^{1-p}-1\right).$$ Since $1-p>0$, we have $$\lim_{t\to\infty} t^{1-p}=\infty.$$ Therefore, the integral becomes $$\int_1^{\infty} \frac{1}{x^p}\, dx=\frac{1}{1-p}(\infty-1)=\infty.$$ So, in this case the integral diverges.

Therefore, by the casework we just did, we conclude that the integral concludes for all $p>1$.

This last example will become very important in the context of converging infinite series.