As you know, finding the integral can be a rather difficult task. It would be a great disappointment to take up the calculation of an improper integral and find out at the end of the path that it diverges. Therefore, methods are of interest that allow one to draw a conclusion about the convergence or divergence of an improper integral without serious calculations for one type of functions. The first and second comparison theorems, which will be discussed below, help to a large extent to investigate improper integrals for convergence.
Let f(x)?0. Then the functions
are monotonically increasing in the variables t or -g (since we take q > 0, -g tends to zero from the left). If, as the arguments increase, the functions F 1 (t) and F 2 (-d) remain bounded from above, this means that the corresponding improper integrals converge. This is the basis of the first comparison theorem for integrals of non-negative functions.
Let the following conditions be satisfied for the function f(x) and g(x) at x?a:
- 1) 0?f(x)?g(x);
- 2) The functions f(x) and g(x) are continuous.
Then the convergence of the integral implies the convergence of the integral, and the divergence of the integral implies the divergence
Since 0?f(x)?g(x) and the functions are continuous, then
By assumption, the integral converges, i.e., has a finite value. Therefore, the integral also converges.
Now let the integral diverge. Suppose that the integral converges, but then the integral must converge, which contradicts the condition. Our assumption is wrong, the integral diverges.
Comparison theorem for improper integrals of the 2nd kind.
Let for the functions f(x) and g(x) on the interval , increase indefinitely for x>+0. For it, for x>+0, the inequality<. Несобственный интеграл есть эталонный интеграл 2-го рода, который при p=<1 сходится; следовательно, по 1-й теореме сравнения для несобственных интегралов 2-го рода интеграл сходится также.
Comparison theorem for improper integrals of the first kind.
Let the integral diverge for the function f(x) and g(x) on the interval.
This means that the integral also diverges on the segment.
Thus, this integral diverges on the entire segment [-1, 1]. Note that if we began to calculate this integral, not paying attention to the discontinuity of the integrand at the point x = 0, we would get an incorrect result. Really,
, which is impossible.
So, to study the improper integral of a discontinuous function, it is necessary to "split" it into several integrals and investigate them.
If the integrand has a discontinuity of the second kind on the (finite) interval of integration, one speaks of an improper integral of the second kind.
10.2.1 Definition and basic properties
Let us denote the interval of integration $\left[ a, \, b \right ]$, both of these numbers are assumed to be finite below. If there is only 1 gap, it can be either at the point $a$, or at the point $b$, or inside the interval $(a,\,b)$. Let us first consider the case when there is a discontinuity of the second kind at the point $a$, and the integrand is continuous at other points. So we are discussing the integral
\begin(equation) I=\int _a^b f(x)\,dx, (22) \label(intr2) \end(equation)
where $f(x) \rightarrow \infty $ when $x \rightarrow a+0$. As before, the first thing to do is to give meaning to this expression. To do this, consider the integral
\[ I(\epsilon)=\int _(a+\epsilon)^b f(x)\,dx. \]
Definition. Let there be a limit
\[ A=\lim _(\epsilon \rightarrow +0)I(\epsilon)=\lim _(\epsilon \rightarrow +0)\int _(a+\epsilon)^b f(x)\,dx. \]
Then the improper integral of the second kind (22) is said to converge and the value $A$ is assigned to it, the function $f(x)$ itself is said to be integrable on the interval $\left[ a, \, b\right]$.
Consider the integral
\[ I=\int ^1_0\frac(dx)(\sqrt(x)). \]
The integrand $1/\sqrt(x)$ for $x \rightarrow +0$ has an infinite limit, so at the point $x=0$ it has a discontinuity of the second kind. Let's put
\[ I(\epsilon)=\int ^1_(\epsilon )\frac(dx)(\sqrt(x))\,. \]
In this case, the antiderivative is known,
\[ I(\epsilon)=\int ^1_(\epsilon )\frac(dx)(\sqrt(x))=2\sqrt(x)|^1_(\epsilon )=2(1-\sqrt( \epsilon ))\rightarrow 2 \]
for $\epsilon \rightarrow +0$. Thus, the original integral is a convergent improper integral of the second kind, and it is equal to 2.
Let us consider the variant when there is a discontinuity of the second kind of the integrand at the upper limit of the integration interval. This case can be reduced to the previous one by changing the variable $x=-t$ and then rearranging the limits of integration.
Let us consider the case when the integrand has a discontinuity of the second kind inside the integration interval, at the point $c \in (a,\,b)$. In this case, the original integral
\begin(equation) I=\int _a^bf(x)\,dx (23) \label(intr3) \end(equation)
presented as a sum
\[ I=I_1+I_2, \quad I_1=\int _a^cf(x)\,dx +\int _c^df(x)\,dx. \]
Definition. If both integrals $I_1, \, I_2$ converge, then the improper integral (23) is called convergent and it is assigned a value equal to the sum of the integrals $I_1, \, I_2$, the function $f(x)$ is called integrable on the interval $\left [a, \, b\right]$. If at least one of the integrals $I_1,\, I_2$ is divergent, the improper integral (23) is said to be divergent.
Converging improper integrals of the 2nd kind have all the standard properties of ordinary definite integrals.
1. If $f(x)$, $g(x)$ are integrable on the interval $\left[ a, \,b \right ]$, then their sum $f(x)+g(x)$ is also integrable on this interval, and \[ \int _a^(b)\left(f(x)+g(x)\right)dx=\int _a^(b)f(x)dx+\int _a^(b)g (x)dx. \] 2. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$, then for any constant $C$ the function $C\cdot f(x)$ is also integrable on this interval , and \[ \int _a^(b)C\cdot f(x)dx=C \cdot \int _a^(b)f(x)dx. \] 3. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$ and $f(x)>0$ on this interval, then \[ \int _a^(b ) f(x)dx\,>\,0. \] 4. If $f(x)$ is integrable on the interval $\left[ a, \, b \right ]$, then for any $c\in (a, \,b)$ the integrals \[ \int _a^ (c) f(x)dx, \quad \int _c^(b) f(x)dx \] also converge, and \[ \int _a^(b)f(x)dx=\int _a^(c ) f(x)dx+\int _c^(b) f(x)dx \] (additivity of the integral over the interval).
Consider the integral \begin(equation) I=\int _0^(1)\frac(1)(x^k)\,dx. (24) \label(mod2) \end(equation) If $k>0$, the integrand tends to $\infty$ as $x \rightarrow +0$, so the integral is improper of the second kind. We introduce the function \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx. \] In this case, the antiderivative is known, so that \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k )|_(\epsilon)^1= \frac(1)(1-k)-\frac(\epsilon ^(1-k))(1-k). \] for $k \neq 1$, \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x)\,dx\,=lnx|_(\epsilon)^1= -ln \epsilon. \] for $k = 1$. Considering the behavior for $\epsilon \rightarrow +0$, we conclude that the integral (20) converges for $k
10.2.2 Criteria for the convergence of improper integrals of the 2nd kind
Theorem (the first sign of comparison). Let $f(x)$, $g(x)$ be continuous for $x\in (a,\,b)$, and $0 1. If the integral \[ \int _a^(b)g(x)dx \] converges, then the integral \[ \int _a^(b)f(x)dx also converges. \] 2. If the integral \[ \int _a^(b)f(x)dx \] diverges, then the integral \[ \int _a^(b)g(x)dx also diverges. \]
Theorem (the second sign of comparison). Let $f(x)$, $g(x)$ be continuous and positive for $x\in (a,\,b)$, and let there be a finite limit
\[ \theta = \lim_(x \rightarrow a+0) \frac(f(x))(g(x)), \quad \theta \neq 0, \, +\infty. \]
Then the integrals
\[ \int _a^(b)f(x)dx, \quad \int _a^(b)g(x)dx \]
converge or diverge at the same time.
Consider the integral
\[ I=\int _0^(1)\frac(1)(x+\sin x)\,dx. \]
The integrand is a positive function on the integration interval, the integrand tends to $\infty$ as $x \rightarrow +0$, so our integral is improper of the second kind. Further, for $x \rightarrow +0$ we have: if $g(x)=1/x$, then
\[ \lim _(x \rightarrow +0)\frac(f(x))(g(x))=\lim _(x \rightarrow +0)\frac(x)(x+\sin x)=\ frac(1)(2) \neq 0,\, \infty \, . \]
Applying the second criterion of comparison, we come to the conclusion that our integral converges or diverges simultaneously with the integral
\[ \int _0^(+1)\frac(1)(x)\,dx . \]
As shown in the previous example, this integral diverges ($k=1$). Therefore, the original integral also diverges.
Calculate the improper integral or establish its convergence (divergence).
1. \[ \int _(0)^(1)\frac(dx)(x^3-5x^2)\,. \] 2. \[ \int _(3)^(7)\frac(x\,dx)((x-5)^2)\,. \] 3. \[ \int _(0)^(1)\frac(x\,dx)(\sqrt(1-x^2))\,. \] 4. \[ \int _(0)^(1)\frac(x^3\,dx)(1-x^5)\,. \] 5. \[ \int _(-3)^(2)\frac(dx)((x+3)^2)\,. \] 6. \[ \int _(1)^(2)\frac(x^2\,dx)((x-1)\sqrt(x-1))\,. \] 7. \[ \int _(0)^(1)\frac(dx)(\sqrt(x+x^2))\,. \] 8. \[ \int _(0)^(1/4)\frac(dx)(\sqrt(x-x^2))\,. \] 9. \[ \int _(1)^(2)\frac(dx)(xlnx)\,. \] 10. \[ \int _(1)^(2)\frac(x^3\,dx)(\sqrt(4-x^2))\,. \] 11. \[ \int _(0)^(\pi /4)\frac(dx)(\sin ^4x)\,. \]
