{"id":4612,"date":"2026-09-15T12:48:04","date_gmt":"2026-09-15T12:48:04","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4612"},"modified":"2026-09-15T14:48:44","modified_gmt":"2026-09-15T14:48:44","slug":"improper-integrals","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/improper-integrals.html","title":{"rendered":"Improper Integrals: Convergence, Divergence, and Examples"},"content":{"rendered":"

Improper integrals extend the concept of the definite integral<\/a> to cases where the interval of integration is infinite or the function becomes unbounded. In such situations, the main question is no longer only how to calculate the integral, but also whether it is possible to obtain a finite result at all.<\/p>\n

This is why convergence and divergence are especially important when working with improper integrals. Next, we will look at the different types of improper integrals, how they are correctly defined using limits, and under what conditions they have a finite value.<\/p>\n

Improper Integrals: Why a Limit Process Is Needed<\/h2>\n

For an ordinary definite integral, we consider the function on a finite interval \\( [a,b] \\). Under appropriate conditions, we can write<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx.
\n\\]<\/p>\n

But what should we do if one of the limits of integration is infinite? Or if the function becomes unbounded near a certain point of the interval? In such cases, the usual definition of a definite integral cannot be applied directly.<\/p>\n

Instead, an improper integral is defined using a limit. To do this, the infinite limit of integration or the point near which the function becomes unbounded is replaced by a variable limit of integration. We then calculate the definite integral over the corresponding finite interval and examine the limit of the resulting expression.<\/p>\n

For example, the symbol \\( +\\infty \\) as the upper limit of integration does not mean that infinity should be substituted into the antiderivative<\/a> instead of a number. Infinity is not a numerical value. It simply indicates that the upper limit of integration increases without bound.<\/p>\n

A similar approach is used when the function becomes unbounded near a certain point. In this case, the variable limit of integration approaches that point from the left or from the right, depending on its position within the interval.<\/p>\n

Therefore, an improper integral is not calculated directly in the same way as an ordinary definite integral. First, it is written using the appropriate limit, and only then is the resulting expression examined. Next, we will see how this approach is applied to improper integrals of the first and second kinds.<\/p>\n

Infinite Interval: Integrals of the First Kind<\/h2>\n

If at least one of the limits of integration is infinite, we have an improper integral of the first kind.<\/p>\n

Let us first consider the case where the upper limit is infinite:<\/p>\n

\\[
\n\\int_a^{+\\infty} f(x)\\,dx.
\n\\]<\/p>\n

To give this expression a precise mathematical meaning, we replace \\( +\\infty \\) with a finite variable limit \\( b \\):<\/p>\n

\\[
\n\\int_a^{+\\infty} f(x)\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\int_a^b f(x)\\,dx.
\n\\]<\/p>\n

First, the function is integrated over the ordinary finite interval \\( [a,b] \\). Then the value of \\( b \\) is increased, and we examine how the value of the resulting definite integral changes.<\/p>\n

If the lower limit is infinite, we use a similar approach:<\/p>\n

\\[
\n\\int_{-\\infty}^{b} f(x)\\,dx
\n=
\n\\lim_{a\\to-\\infty}
\n\\int_a^b f(x)\\,dx.
\n\\]<\/p>\n

Here, the variable limit \\( a \\) decreases without bound.<\/p>\n

Special attention is needed when both limits of integration are infinite:<\/p>\n

\\[
\n\\int_{-\\infty}^{+\\infty} f(x)\\,dx.
\n\\]<\/p>\n

Such an integral is not defined by letting both limits approach infinity at the same time. Instead, we choose any finite point \\( c \\) and split the interval:<\/p>\n

\\[
\n\\int_{-\\infty}^{+\\infty} f(x)\\,dx
\n=
\n\\int_{-\\infty}^{c} f(x)\\,dx
\n+
\n\\int_c^{+\\infty} f(x)\\,dx.
\n\\]<\/p>\n

Then each part is written using its own limit:<\/p>\n

\\[
\n\\begin{gathered}
\n\\int_{-\\infty}^{c} f(x)\\,dx
\n=
\n\\lim_{a\\to-\\infty}
\n\\int_a^c f(x)\\,dx,\\\\[6pt]
\n\\int_c^{+\\infty} f(x)\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\int_c^b f(x)\\,dx.
\n\\end{gathered}
\n\\]<\/p>\n

Why do we need two independent limit processes? Because the behavior of the function as \\( x\\to-\\infty \\) may be different from its behavior as \\( x\\to+\\infty \\). Therefore, the left-hand and right-hand parts must be examined separately.<\/p>\n

Unbounded Function: Integrals of the Second Kind<\/h2>\n

An infinite interval is not the only reason an integral can be improper. The interval may remain finite, while the function itself becomes unbounded near one of its points. In this case, we have an improper integral of the second kind.<\/p>\n

Suppose that the function \\( f(x) \\) becomes unbounded as we approach the right endpoint \\( b \\). Then<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n\\]<\/p>\n

is defined by the formula<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n=
\n\\lim_{t\\to b-}
\n\\int_a^t f(x)\\,dx.
\n\\]<\/p>\n

The notation \\( t\\to b- \\) means that the point \\( t \\) approaches \\( b \\) from the left. This way, when calculating the ordinary definite integral, we do not reach the point near which the function becomes unbounded.<\/p>\n

If the function is unbounded near the left endpoint \\( a \\), we use an approach from the right:<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n=
\n\\lim_{t\\to a+}
\n\\int_t^b f(x)\\,dx.
\n\\]<\/p>\n

Here, the notation \\( t\\to a+ \\) means that \\( t \\) approaches the point \\( a \\) from the right.<\/p>\n

Another important case occurs when the function is unbounded at an interior point of the interval. Let \\( c\\in(a,b) \\), and suppose the function becomes unbounded near \\( c \\). Then the integral is split into two parts:<\/p>\n

\\[
\n\\int_a^b f(x)\\,dx
\n=
\n\\int_a^c f(x)\\,dx
\n+
\n\\int_c^b f(x)\\,dx.
\n\\]<\/p>\n

The first part is defined using a left-hand limit:<\/p>\n

\\[
\n\\int_a^c f(x)\\,dx
\n=
\n\\lim_{t\\to c-}
\n\\int_a^t f(x)\\,dx,
\n\\]<\/p>\n

and the second part is defined using a right-hand limit:<\/p>\n

\\[
\n\\int_c^b f(x)\\,dx
\n=
\n\\lim_{t\\to c+}
\n\\int_t^b f(x)\\,dx.
\n\\]<\/p>\n

Why do we need two different limits here? The function may behave differently on the left and right sides of the point \\( c \\). That is why both parts must be examined independently.<\/p>\n

Sometimes an integral can be improper for more than one reason at the same time. For example, the interval may be infinite while the function is also unbounded at a certain point within it. In this case, the interval is divided into parts so that each reason for the integral being improper can be examined with a separate limit process.<\/p>\n

Convergence and Divergence: Conditions for an Improper Integral to Exist<\/h2>\n

Now that we know how to write an improper integral correctly using one or more limits, the main question arises: does it have a finite numerical value?<\/p>\n

An improper integral is called convergent if the corresponding limits exist and each has a finite value.<\/p>\n

For example, if<\/p>\n

\\[
\n\\lim_{b\\to+\\infty}
\n\\int_a^b f(x)\\,dx=L,
\n\\qquad L\\in\\mathbb{R},
\n\\]<\/p>\n

then<\/p>\n

\\[
\n\\int_a^{+\\infty}f(x)\\,dx=L.
\n\\]<\/p>\n

In this case, the number \\( L \\) is the value of the improper integral.<\/p>\n

If the required limit does not exist or is not finite, the improper integral is called divergent. In particular, this happens when the corresponding expression increases without bound, decreases without bound, or does not approach a finite number.<\/p>\n

If an improper integral consists of several improper parts, it converges only if each part converges separately. If even one part diverges, the entire integral diverges.<\/p>\n

Therefore, when working with an improper integral, it is important to follow a clear sequence. First, identify why the integral is improper. Then write it correctly using the appropriate limits, and only after that examine whether those limits exist. The result of this analysis determines whether the improper integral can be assigned a finite numerical value.<\/p>\n

Improper Integrals in Practice: Step-by-Step Calculation and Convergence Testing<\/h2>\n

Now let’s apply the definitions we have discussed to actual calculations. In each case, we will first identify why the integral is improper, then rewrite it using the appropriate limit, calculate the definite integral, and determine whether the improper integral converges or diverges.<\/p>\n

Example 1. Calculate the improper integral
\n\\[
\n\\int_1^{+\\infty}\\frac{1}{x^2}\\,dx
\n\\]
\nand determine whether it converges.<\/h3>\n

The function<\/p>\n

\\[
\nf(x)=\\frac{1}{x^2}
\n\\]<\/p>\n

is continuous on the interval \\( [1,+\\infty) \\). The integral is improper because its upper limit is infinite.<\/p>\n

Therefore, we replace \\( +\\infty \\) with a variable limit \\( b \\):<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x^2}\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\int_1^b\\frac{1}{x^2}\\,dx.
\n\\]<\/p>\n

Rewrite the integrand in power form:<\/p>\n

\\[
\n\\frac{1}{x^2}=x^{-2}.
\n\\]<\/p>\n

Find the antiderivative:<\/p>\n

\\[
\n\\int x^{-2}\\,dx
\n=
\n-\\frac{1}{x}.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_1^b\\frac{1}{x^2}\\,dx
\n=
\n\\left.-\\frac{1}{x}\\right|_1^b.
\n\\]<\/p>\n

Substitute the limits of integration:<\/p>\n

\\[
\n\\left.-\\frac{1}{x}\\right|_1^b
\n=
\n-\\frac{1}{b}-(-1)
\n=
\n1-\\frac{1}{b}.
\n\\]<\/p>\n

Now find the limit:<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x^2}\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\left(1-\\frac{1}{b}\\right).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\lim_{b\\to+\\infty}\\frac{1}{b}=0,
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x^2}\\,dx=1.
\n\\]<\/p>\n

The limit exists and has a finite value. Therefore, the improper integral converges, and its value is \\( 1 \\).<\/p>\n

Example 2. Determine whether the improper integral
\n\\[
\n\\int_1^{+\\infty}\\frac{1}{x}\\,dx.
\n\\]
\nconverges or diverges.<\/h3>\n

Once again, the integral is improper because its upper limit is infinite. Therefore, we write it using a limit:<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x}\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\int_1^b\\frac{1}{x}\\,dx.
\n\\]<\/p>\n

An antiderivative of \\( 1\/x \\) on the interval \\( x>0 \\) is \\( \\ln(x) \\). Therefore,<\/p>\n

\\[
\n\\int_1^b\\frac{1}{x}\\,dx
\n=
\n\\left.\\ln(x)\\right|_1^b.
\n\\]<\/p>\n

Substitute the limits:<\/p>\n

\\[
\n\\left.\\ln(x)\\right|_1^b
\n=
\n\\ln(b)-\\ln(1).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\ln(1)=0,
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\n\\int_1^b\\frac{1}{x}\\,dx=\\ln(b).
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x}\\,dx
\n=
\n\\lim_{b\\to+\\infty}\\ln(b).
\n\\]<\/p>\n

As \\( b\\to+\\infty \\), the logarithm increases without bound:<\/p>\n

\\[
\n\\lim_{b\\to+\\infty}\\ln(b)=+\\infty.
\n\\]<\/p>\n

Therefore, there is no finite limit, so<\/p>\n

\\[
\n\\int_1^{+\\infty}\\frac{1}{x}\\,dx
\n\\]<\/p>\n

diverges.<\/p>\n

This result shows that simply having an antiderivative is not enough for an improper integral to converge. What matters is the value of the corresponding limit.<\/p>\n

Example 3. Calculate the improper integral
\n\\[
\n\\int_0^1\\frac{1}{\\sqrt{x}}\\,dx
\n\\]
\nand determine whether it converges.<\/h3>\n

Here, the interval of integration is finite, but the function<\/p>\n

\\[
\nf(x)=\\frac{1}{\\sqrt{x}}
\n\\]<\/p>\n

increases without bound as \\( x\\to0+ \\). Therefore, this is an improper integral of the second kind.<\/p>\n

Replace the lower limit \\( 0 \\) with a variable \\( a>0 \\):<\/p>\n

\\[
\n\\int_0^1\\frac{1}{\\sqrt{x}}\\,dx
\n=
\n\\lim_{a\\to0+}
\n\\int_a^1\\frac{1}{\\sqrt{x}}\\,dx.
\n\\]<\/p>\n

Rewrite the integrand:<\/p>\n

\\[
\n\\frac{1}{\\sqrt{x}}
\n=
\nx^{-\\frac12}.
\n\\]<\/p>\n

Find the antiderivative:<\/p>\n

\\[
\n\\int x^{-\\frac12}\\,dx
\n=
\n2\\cdot\\sqrt{x}.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_a^1\\frac{1}{\\sqrt{x}}\\,dx
\n=
\n\\left.2\\cdot\\sqrt{x}\\right|_a^1.
\n\\]<\/p>\n

Substitute the limits:<\/p>\n

\\[
\n\\left.2\\cdot\\sqrt{x}\\right|_a^1
\n=
\n2-2\\cdot\\sqrt{a}.
\n\\]<\/p>\n

Now find the limit:<\/p>\n

\\[
\n\\int_0^1\\frac{1}{\\sqrt{x}}\\,dx
\n=
\n\\lim_{a\\to0+}
\n\\left(2-2\\cdot\\sqrt{a}\\right).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\lim_{a\\to0+}\\sqrt{a}=0,
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\n\\int_0^1\\frac{1}{\\sqrt{x}}\\,dx=2.
\n\\]<\/p>\n

Therefore, even though the function is unbounded near \\( x=0 \\), the corresponding limit is finite. So the integral converges, and its value is \\( 2 \\).<\/p>\n

Example 4. Determine whether the improper integral
\n\\[
\n\\int_{-1}^{1}\\frac{1}{x^2}\\,dx.
\n\\]
\nconverges or diverges.<\/h3>\n

The limits of integration are finite, but the function<\/p>\n

\\[
\nf(x)=\\frac{1}{x^2}
\n\\]<\/p>\n

is not defined at \\( x=0 \\) and increases without bound near this point.<\/p>\n

Since the point \\( x=0 \\), where the function is not defined, lies inside the interval \\( [-1,1] \\), the integral must be split at this point:<\/p>\n

\\[
\n\\int_{-1}^{1}\\frac{1}{x^2}\\,dx
\n=
\n\\int_{-1}^{0}\\frac{1}{x^2}\\,dx
\n+
\n\\int_0^1\\frac{1}{x^2}\\,dx.
\n\\]<\/p>\n

Let’s begin with the left part:<\/p>\n

\\[
\n\\int_{-1}^{0}\\frac{1}{x^2}\\,dx
\n=
\n\\lim_{a\\to0-}
\n\\int_{-1}^{a}\\frac{1}{x^2}\\,dx.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\int\\frac{1}{x^2}\\,dx
\n=
\n-\\frac{1}{x},
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\n\\int_{-1}^{a}\\frac{1}{x^2}\\,dx
\n=
\n\\left.-\\frac{1}{x}\\right|_{-1}^{a}.
\n\\]<\/p>\n

Substitute the limits:<\/p>\n

\\[
\n\\left.-\\frac{1}{x}\\right|_{-1}^{a}
\n=
\n-\\frac{1}{a}-1.
\n\\]<\/p>\n

As \\( a\\to0- \\),<\/p>\n

\\[
\n\\frac{1}{a}\\to-\\infty,
\n\\]<\/p>\n

so<\/p>\n

\\[
\n\\lim_{a\\to0-}
\n\\left(-\\frac{1}{a}-1\\right)
\n=
\n+\\infty.
\n\\]<\/p>\n

Therefore, the left part already diverges. This is enough to conclude that the entire integral diverges.<\/p>\n

However, for completeness, let’s also check the right part:<\/p>\n

\\[
\n\\int_0^1\\frac{1}{x^2}\\,dx
\n=
\n\\lim_{b\\to0+}
\n\\int_b^1\\frac{1}{x^2}\\,dx.
\n\\]<\/p>\n

Calculate the definite integral:<\/p>\n

\\[
\n\\int_b^1\\frac{1}{x^2}\\,dx
\n=
\n\\left.-\\frac{1}{x}\\right|_b^1
\n=
\n-1+\\frac{1}{b}.
\n\\]<\/p>\n

As \\( b\\to0+ \\),<\/p>\n

\\[
\n\\frac{1}{b}\\to+\\infty,
\n\\]<\/p>\n

so<\/p>\n

\\[
\n\\lim_{b\\to0+}
\n\\left(-1+\\frac{1}{b}\\right)
\n=
\n+\\infty.
\n\\]<\/p>\n

Therefore, both improper parts diverge. Thus,<\/p>\n

\\[
\n\\int_{-1}^{1}\\frac{1}{x^2}\\,dx
\n\\]<\/p>\n

diverges.<\/p>\n

This example shows why it is important to check the behavior of the function over the entire interval of integration before applying the Newton-Leibniz formula<\/a>.<\/p>\n

Example 5. Calculate the improper integral
\n\\[
\n\\int_{-\\infty}^{+\\infty}\\frac{1}{1+x^2}\\,dx
\n\\]
\nand determine whether it converges.<\/h3>\n

Here, both limits of integration are infinite. Therefore, we choose the finite point \\( 0 \\) and split the integral into two parts:<\/p>\n

\\[
\n\\int_{-\\infty}^{+\\infty}\\frac{1}{1+x^2}\\,dx
\n=
\n\\int_{-\\infty}^{0}\\frac{1}{1+x^2}\\,dx
\n+
\n\\int_0^{+\\infty}\\frac{1}{1+x^2}\\,dx.
\n\\]<\/p>\n

First, consider the left part:<\/p>\n

\\[
\n\\int_{-\\infty}^{0}\\frac{1}{1+x^2}\\,dx
\n=
\n\\lim_{a\\to-\\infty}
\n\\int_a^0\\frac{1}{1+x^2}\\,dx.
\n\\]<\/p>\n

An antiderivative of the integrand is \\( \\arctan(x) \\). Therefore,<\/p>\n

\\[
\n\\int_a^0\\frac{1}{1+x^2}\\,dx
\n=
\n\\left.\\arctan(x)\\right|_a^0.
\n\\]<\/p>\n

Substitute the limits:<\/p>\n

\\[
\n\\arctan(0)-\\arctan(a)
\n=
\n-\\arctan(a).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\lim_{a\\to-\\infty}\\arctan(a)
\n=
\n-\\frac{\\pi}{2},
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\n\\int_{-\\infty}^{0}\\frac{1}{1+x^2}\\,dx
\n=
\n\\frac{\\pi}{2}.
\n\\]<\/p>\n

Now consider the right part:<\/p>\n

\\[
\n\\int_0^{+\\infty}\\frac{1}{1+x^2}\\,dx
\n=
\n\\lim_{b\\to+\\infty}
\n\\int_0^b\\frac{1}{1+x^2}\\,dx.
\n\\]<\/p>\n

We have<\/p>\n

\\[
\n\\int_0^b\\frac{1}{1+x^2}\\,dx
\n=
\n\\left.\\arctan(x)\\right|_0^b
\n=
\n\\arctan(b).
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\lim_{b\\to+\\infty}\\arctan(b)
\n=
\n\\frac{\\pi}{2},
\n\\]<\/p>\n

we get<\/p>\n

\\[
\n\\int_0^{+\\infty}\\frac{1}{1+x^2}\\,dx
\n=
\n\\frac{\\pi}{2}.
\n\\]<\/p>\n

Both improper parts converge. Therefore, we can add their values:<\/p>\n

\\[
\n\\int_{-\\infty}^{+\\infty}\\frac{1}{1+x^2}\\,dx
\n=
\n\\frac{\\pi}{2}
\n+
\n\\frac{\\pi}{2}
\n=
\n\\pi.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int_{-\\infty}^{+\\infty}\\frac{1}{1+x^2}\\,dx=\\pi.
\n\\]<\/p>\n

Next Steps in Integral Calculus: What to Learn<\/h2>\n

Improper integrals are only one area where integral calculus is applied. A natural next step is to explore topics where integrals help solve geometric problems and work with functions of several variables.<\/p>\n

    \n
  1. Area of a Region Using an Integral: Formulas and Examples<\/a> \u2014 In this article, we will look at how to use the definite integral to find the areas of regions bounded by graphs of functions and given lines.<\/li>\n
  2. Arc Length of a Curve: Formula Using a Definite Integral<\/a> \u2014 We will learn how to calculate the arc length of the graph of a function over a given interval and how to apply the definite integral step by step.<\/li>\n
  3. Double Integrals: Meaning, Region of Integration, and Examples<\/a> \u2014 We will explore the meaning of a double integral, learn how to define the region of integration, and see how the basic calculations are performed.<\/li>\n<\/ol>\n

    Improper Integrals in Code: From a Flowchart to Your Own Program<\/h2>\n

    If you enjoy programming, try putting your knowledge of improper integrals into practice. The flowchart below shows an algorithm that gradually increases the upper limit of integration, compares consecutive approximations, and stops the calculation once the required accuracy is reached. Try implementing this algorithm in your favorite programming language, run the program with different accuracy values, and observe how the calculated result approaches the value of the improper integral.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

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