1. Improper integrals with infinite limits
Let us recall the definition of an integral as the limit of integral sums: 
The definition assumes that the integration interval is finite and the function f(x) is continuous within it. Violation of these assumptions leads to improper integrals.
Definition. If the integral tends to a finite limit as it increases indefinitely "b", then this limit is called an improper integral with an infinite upper bound of the function f (x) and is denoted by the symbol 
In this case, the improper integral is said to exist or converge.
If the specified limit does not exist or exists but is infinite, then the integral is said to not exist or to diverge.
An improper integral with an infinite lower bound is defined similarly:
An improper integral with two infinite boundaries is given by:
where c is any fixed point on the Ox axis.
So, improper integrals can have an infinite lower bound, an infinite upper bound, and also two infinite bounds.
Signs of convergence. Absolute and conditional convergence
An integral exists only if each of the integrals exists: and .
Example. Examine the convergence of the integral
Assuming c = 0, we get:
those. the integral converges.
Sometimes there is no need to calculate an improper integral, but it is enough to just know whether it converges or diverges by comparing it with another integral.
Comparison theorem for improper integrals.
Let the function f (x) in the interval have several (finite number) discontinuity points of the first kind, this “obstacle” can be easily eliminated by dividing the segment into several segments with discontinuity points, calculating definite integrals on each individual section and adding up the results.
Let's consider definite integral from a function that is unlimited when approaching one of the ends of the segment, for example, 
.
(In such cases they usually say: ''The function has an infinite discontinuity at the right end of the interval of integration.''.)
It is clear that the usual definition of an integral loses its meaning here.
Definition. An improper integral of the function f(x), continuous for a £ x< b и неограниченной при x ® b - 0, называется предел:
The improper integral of a function that has an infinite discontinuity at the left end of the segment is defined similarly:
Consequently, in the section [-1, 0] the integral diverges.
This means that the integral also diverges in the section.
Thus, this integral diverges over the entire interval [-1, 1]. Note that if we began to calculate this integral without 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 study them.
Examples of studying improper integrals for convergence
Example 1
.
Thus, this integral converges for a>1 and diverges for a £1.
Example 2
Examine for convergence. Let's calculate the integral by definition: 
.
Thus, this integral converges at a<1 и расходится при a³1.
Example 3
Examine for convergence 
.
<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два
.
We study the convergence of the first integral I1 using the equivalent function: (since n>0), and the integral converges for m>-1 (example 2). Similarly, for the integral I2:
And the integral converges at m+n<-1 (пример2). Следовательно, исходный интеграл сходится при выполнении одновременно двух условий m>-1 and m+n<-1, и будет расходится при нарушении хотя бы одного из них.
Example 4 Examine for convergence.
The integrand can be infinitely large (if m<0) при x стремящемся к 0, поэтому разобьем исходный интеграл на два:
Since arctgx »x at x®0, the integral I1 is equivalent to the integral, which converges for m+1>-1, i.e. for m>-2 (example 1).
For the integrand in the improper integral of the first kind I2, we select an equivalent one:
since arctgx » p/2 at x® ¥. Consequently, by the second comparison criterion, the integral I2 will converge for m+n<-1, и расходится в противном случае.
Combining the conditions for the convergence of the integrals I1 and I2, we obtain the conditions for the convergence of the original integral: m>-2 and m+n<-1 одновременно.
Comment. In examples 2-4, the 2nd comparison criterion was used, which provides necessary and sufficient conditions for convergence, which allows, having established convergence under some condition on the parameter values, not to prove the divergence of the integral if the obtained convergence conditions are violated.
Example 5 Examine for convergence.
This integral contains a singular point 0, at which the integrand can go to infinity as p<0, поэтому снова разобьем исходный интеграл на два:
.
The integral I1 is an improper integral of the second kind, and the integrand is equivalent at x®0 to the function xp (e-x ®1 at x®0), i.e. I1 converges at p>-1 (example 1).
The integral I2 is an improper integral of the first kind. It is not possible to select a function equivalent to the integrand such that it does not contain an exponential function. Therefore, it is impossible to use comparison attribute 2, as in the previous examples. Let's apply the first comparison criterion, for which we use the following well-known fact:
For a>0 and any p. From this, and the fact that the function xpe-ax is continuous, it follows that this function is bounded, i.e. there is a constant M>0 such that xpe-ax< M. Возьмем, например, a=1/2, и оценим интеграл I2 сверху:
That is, the integral I2 converges for any p.
Thus, the original integral converges at p>-1.
Example 6 Examine for convergence.
Let's change the variable: t = lnx, and we get
The partition of the integral into two is carried out similarly to Example 5. The integral I1 is completely equivalent to the integral I1 from Example 5 and, therefore, converges at q<1.
Let's consider the integral I2. Subject to 1-p<0 этот интеграл полностью эквивалентен интегралу I2 в примере 5 (доказательство сходимости аналогично, а условие 1-p<0 нужно для выполнения 
and a=(1-p)/2.).
So, I2 converges for p>1. However, the study of the convergence of this integral is not completed here, since the used convergence criterion provides only sufficient conditions for convergence. Therefore, we need to study convergence for 1-p £ 0.
Let's consider the case p=1. Then the integral I2 is equivalent to , which converges for q>1 (note that in this case the integral I1 diverges) and diverges otherwise.
At p<1 оценим интеграл I2 и покажем его расходимость. Для этого вспомним, что 
For 1-p>0, and, therefore, starting from some A>1, T-
QE(1-
P)
T³ M=const>0. Then the integral I2 satisfies the estimate
,
Where the integral on the right side diverges, which proves the divergence of the integral I2.
Summing up the results obtained, we find that the original integral converges at q<1 и p>1, otherwise the integral diverges.
Example 6 Investigate for absolute and conditional convergence.
Let's split the original integral into two:
.
Convergence. The integral I1 is equivalent 
, i.e. converges at p<2 (пример 1) , причем абсолютно, так как подынтегральная функция положительна на отрезке интегрирования.
The integral I2 converges according to the Dirichlet-Abel test for p>0 since the antiderivative sin(x) is bounded, and the function 1/xp monotonically tends to zero as x tends to infinity.
Let us show that for p £ 0 the integral diverges. Let us use the Cauchy criterion for this, or rather its negation
.
Let us take the following values as R1 and R2: R1=2pk and R2=2pk+p/2, then
, for p>0.
Thus, the integral converges at 0 Absolute convergence The absolute convergence of the integral I1 has already been established; let us consider the absolute convergence of I2. Let us estimate the integral from above: To prove the divergence for p £ 1, we estimate the integral from below Let us divide the last integral of the difference of functions into the difference of integrals If both integrals converge, then the integral of the difference converges; if one of the integrals diverges and the other converges, then the integral of the difference diverges. In the case of divergence of both integrals, the convergence of the integral of the difference is subject to further investigation. We are interested in the second of the described cases. Diverges (example 1) at p<1. сходится по признаку Дирихле-Абеля при 1>p>0 (see Convergence), therefore the integral is estimated from below by a divergent integral, i.e. it diverges. The case p³1 does not interest us, since for these values of the parameter the integral diverges. Thus, the original integral converges absolutely at 0 If the integrand has a discontinuity of the second kind on the (finite) interval of integration, we speak of an improper integral of the second kind. Let us denote the integration interval by $\left[ a, \, b \right ]$; both of these numbers are assumed to be finite below. If there is only 1 discontinuity, it can be located either at point $a$, or at point $b$, or inside the interval $(a,\,b)$. Let us first consider the case when there is a discontinuity of the second kind at point $a$, and at other points the integrand function is continuous. So we are discussing the integral \begin(equation) I=\int _a^b f(x)\,dx, (22) \label(intr2) \end(equation) and $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 finite 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 function $1/\sqrt(x)$ at $x \rightarrow +0$ 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, \[ I(\epsilon)=\int ^1_(\epsilon )\frac(dx)(\sqrt(x))=2\sqrt(x)|^1_(\epsilon )=2(1-\sqrt( \epsilon ))\rightarrow 2\] at $\epsilon \rightarrow +0$. Thus, the original integral is a convergent improper integral of the second kind, and it is equal to 2. Let us consider the option when there is a discontinuity of the second kind in the integrand function at the upper limit of the integration interval. This case can be reduced to the previous one by making the change of variable $x=-t$ and then rearranging the limits of integration. Let us consider the option when the integrand function has a discontinuity of the second kind inside the integration interval, at 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 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 called divergent. Convergent 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 on this interval $f(x)>0$, 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. Let's introduce the function \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx. \] In this case the antiderivative is known, so \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k )|_(\epsilon)^1= \frac(1)(1-k)-\frac(\epsilon ^(1-k))(1-k). \] for $k \neq 1$, \[ I(\epsilon)=\int _(\epsilon)^(1)\frac(1)(x)\,dx\,=lnx|_(\epsilon)^1= -ln \epsilon. \] for $k = 1$. Considering the behavior at $\epsilon \rightarrow +0$, we come to the conclusion that integral (20) converges at $k 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 \[ \int _a^(b)f(x)dx converges. \] 2. If the integral \[ \int _a^(b)f(x)dx \] diverges, then the integral \[ \int _a^(b)g(x)dx diverges. \] Theorem (second comparison criterion). 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 simultaneously. 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 an improper integral 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 comparison criterion, we come to the conclusion that our integral converges or diverges simultaneously with the integral \[ \int _0^(+1)\frac(1)(x)\,dx . \] As was shown in the previous example, this integral diverges ($k=1$). Consequently, 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. \[ \int _(0)^(\pi /4)\frac(dx)(\sin ^4x)\,. \] As you know, finding the integral can be a rather difficult task. It would be a great disappointment to start calculating an improper integral and find at the end of the path that it diverges. Therefore, of interest are methods that allow, without serious calculations based on one type of function, to draw a conclusion about the convergence or divergence of an improper integral. The first and second comparison theorems, which will be discussed below, greatly help to study improper integrals for convergence. Let f(x)?0. Then the functions are monotonically increasing in the variables t or -g (since we take g>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 functions f(x) and g(x) at x?a satisfy the following conditions: Then from the convergence of the integral follows the convergence of the integral, and from the divergence of the integral follows the divergence Since 0?f(x)?g(x) and the functions are continuous, then By condition, the integral converges, i.e. has a finite value. Therefore, the integral also converges. Now let the integral diverge. Let us assume that the integral converges, but then the integral must converge, which contradicts the condition. Our assumption is incorrect, 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 without limit for x>+0. For x>+0, the following inequality holds:<. 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 1st kind. Let for the function f(x) and g(x) on the interval )
, i.e. the integral converges for p>1.
.
.10.2.1 Definition and basic properties
10.2.2 Tests for the convergence of improper integrals of the 2nd kind
