The Improper Integral Calculator displays the integrated value of the improper integral online for free. STUDYQUERIES’s online improper integral calculator tool makes calculation faster, and it displays a value in a fraction of a second.

**How to Use the Improper Integral Calculator?**

To use the improper integral calculator, follow these steps:

**Step 1:**Enter the function and limits in the appropriate input fields**Step 2:**Click “Integrate” to get the results**Step 3:**The integrated value will be displayed in the new window

Improper Integral Calculator

**What are improper integrals?**

Improper integrals are definite integrals that cover an unbounded area. One type of improper integrals are integrals where at least one of the endpoints is extended to infinity. For example, \(\int_{1}^{\infty}\frac{1}{x^2}dx\) is an improper integral. It can be viewed as the limit \(\lim_{b\to \infty}\int_{1}^{b}\frac{1}{x^2}dx\)

Another type of improper integrals is integrals whose endpoints are finite, but the integrated function is unbounded at one (or two) of the endpoints. For example, \(\int_{0}^{1}\frac{1}{\sqrt{x}}dx\) is an improper integral. It can be viewed as the limit \(\lim_{a\to 0^+}\int_{a}^{1}\frac{1}{\sqrt{x}}dx\)

Is there an unbounded area that isn’t infinite? Is that possible? Of course! Some improper integrals have a finite value, but not all. When the limit exists, we call the integral convergent, and when it doesn’t, we call it divergent.

In other words, we can say that Recall that the Fundamental Theorem of Calculus says that if \(f\) is a continuous function on the closed interval \(\left[a,b \right]\), then

$$\int_{a}^{b}F(x)dx=\left[F(x) \right]_{a}^{b}=F(b)-F(a)$$

where \(F\) is any antiderivative of \(f\). Both the continuity condition and closed interval must hold to use the Fundamental Theorem of Calculus, and in this case,

$$\int_{a}^{b}f(x)dx$$

represents the net area under \(f(x)\) from \(a\) to \(b\).

We begin with an example where blindly applying the Fundamental Theorem of Calculus can give an incorrect result.

**Explain why is \(\int_{-1}^{1}\frac{1}{x^2}dx\) not equal to \(-2\).**

Here is how one might proceed: $$\int_{-1}^{1}\frac{1}{x^2}dx$$

$$=\int_{-1}^{1}{x^{-2}}dx$$

$$=\left[-{x}^{-1} \right]_{-1}^{1}$$

$$=\left[-\frac{1}{x} \right]_{-1}^{1}$$

$$=\left[-\frac{1}{1} \right]-\left[-\frac{1}{(-1)} \right]$$

$$=-2$$

However, the above answer is WRONG! Since \(f(x)=\frac{1}{x^2}\) is not continuous on \([−1,1]\), we cannot directly apply the Fundamental Theorem of Calculus. Intuitively, we can see why \(−2\) is not the correct answer by looking at the graph of \(f(x)=\frac{1}{x^2}\) on \([−1,1]\).

The shaded area appears to grow without bound as seen in the figure below.

Formalizing this example leads to the concept of an improper integral. There are two ways to extend the Fundamental Theorem of Calculus. One is to use an infinite interval, i.e., \([a,\infty),(−\infty,b]\) or \((−\infty,\infty)\).

The second is to allow the interval \([a,b]\) to contain an infinite discontinuity of \(f(x)\). In either case, the integral is referred to as an improper integral. Probability distributions are one of the most significant applications of this concept, as determining quantities such as the cumulative distribution or expected value typically require integrals on infinite intervals.

**Improper Integral: One Infinite Limit of Integration**

To compute improper integrals, we use the concept of limits along with the Fundamental Theorem of Calculus.

**Definition:** One Infinite Limit of Integration. If \(f(x)\) is continuous on \([a,\infty)\), then the improper integral of \(f\) over \([a,\infty)\) is $$\int_{a}^{\infty}f(x)dx=\lim_{R\to \infty}\int_{a}^{R}f(x)dx$$

If f(x) is continuous on(−\infty,b],then the improper integral of f over (−\infty,b] is

$$\int_{-\infty}^{b}f(x)dx=\lim_{R\to -\infty}\int_{R}^{b}f(x)dx$$

Because we are dealing with limits, we are interested in convergence and divergence of the improper integral. When a limit exists and is a finite number, we say the improper integral converges. Alternatively, we say that the improper integral diverges, as captured in the following definition.

**Convergence and Divergence**

If the limit exists and is a finite number, we say the improper integral converges. If the limit is \(\pm{\infty}\) or does not exist, we say the improper integral diverges.

To get an intuitive (though not completely correct) interpretation of improper integrals, we attempt to analyze \(\int_{a}^{\infty}f(x)dx\) graphically. Here assume f(x) is continuous on \([a,∞)\):

We let \(R\) be a fixed number in \([a,∞)\).Then by taking the limit as \(R\) approaches \(\infty\), we get the improper integral: $$\int_{a}^{\infty}f(x)dx=\lim_{R\to \infty}\int_{a}^{R}f(x)dx$$

We can then apply the Fundamental Theorem of Calculus to the last integral as \(f(x)\) is continuous on the closed interval \([a, R]\). We next define the improper integral for the interval \((−\infty,\infty)\).

**Improper Integral: One Infinite Limit of Integration Example**

**Determine \(\int_{1}^{\infty}\frac{1}{x}dx\) whether is convergent or divergent.**

Using the definition for improper integrals we write this as:

$$\int_{1}^{\infty}\frac{1}{x}dx=\lim_{R\to \infty}\int_{1}^{R}\frac{1}{x}dx$$

$$=\lim_{R\to \infty}[\log_e |{x}|]_{1}^{R}$$

$$=\lim_{R\to \infty}[\log_e |{R}|]-[\log_e |{1}|]$$

$$=\lim_{R\to \infty}[\log_e |{R}|]$$

$$=+\infty$$

Therefore, the integral is divergent.

**Improper Integrals: Two Infinite Limits of Integration**

If both \(\int_{-\infty}^{a}f(x)dx\) and \(\int_{a}^{\infty}f(x)dx\) are convergent, then the improper integral of \(f\) over \((-\infty,\infty)\) is

$$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx$$

The above definition requires both of the integrals \(\int_{-\infty}^{a}f(x)dx\) and \(\int_{a}^{\infty}f(x)dx\) to be convergent for \(\int_{-\infty}^{\infty}f(x)dx\) to also be convergent. If either of \(\int_{-\infty}^{a}f(x)dx\) or \(\int_{a}^{\infty}f(x)dx\) is divergent, then so is \(\int_{-\infty}^{\infty}f(x)dx\).

**Improper Integrals: Two Infinite Limits of Integration Example**

**Determine \(\int_{-\infty}^{\infty}x\sin(x^2)dx\) whether is convergent or divergent.**

We must compute both \(\int_{-\infty}^{0}x\sin(x^2)dx\) and \(\int_{0}^{\infty}x\sin(x^2)dx\).

Note that we don’t have to split the integral up at 0, any finite value \(a\) will work. First, we compute the indefinite integral. Let \(u=x^2\), then \(du=2xdx\) and hence,

$$\int{x\sin(x^2)}dx=\frac{1}{2}\int{x\sin(x^2)}dx=-\frac{1}{2}\cos(x^2)+C$$

Using the definition of improper integral gives

$$\int_{0}^{\infty}x\sin(x^2)dx=\lim_{R\to \infty}\int_{0}^{R}x\sin(x^2)dx$$

$$=\lim_{R\to \infty}[-\frac{1}{2}\cos(x^2)]_{0}^{R}$$

$$=-\frac{1}{2}\lim_{R\to \infty}[\cos(R^2)]+\frac{1}{2}$$

This limit does not exist since \(cos(x)\) oscillates between \(−1\) and \(+1\). In particular, \(cos(x)\) does not approach any particular value as x gets larger and larger. Thus, \(\int_{0}^{\infty}x\sin(x^2)dx\) diverges, and hence, the integral \(\int_{-\infty}^{\infty}x\sin(x^2)dx\) diverges.

**Discontinuities**

When there is a discontinuity in \([a,b]\) or at an endpoint, then the improper integral is as follows.

If \(f(x)\) is continuous on \((a,b]\), then the improper integral of \(f\) over \((a,b]\) is

$$\int_{a}^{b}f(x)dx=\lim_{R\to a^+}\int_{R}^{b}f(x)dx.$$

If \(f(x)\) is continuous on \([a,b)\), then the improper integral of \(f\) over \([a,b)\) is

$$\int_{a}^{b}f(x)dx=\lim_{R\to b^+}\int_{a}^{R}f(x)dx.$$

**Convergence and divergence of an improper integral apply here as well: If the limit above exists and is a finite number, the improper integral converges. If not, we say the improper integral diverges.**

When there is a discontinuity in the interior of \([a,b]\), we use the following definition.

**Discontinuities Within Integration Interval**

If f has a discontinuity at \(x=c\) where \(c\in{[a,b]}\),and bothandare convergent, then \(f\) over \([a,b]\) is \(\int_{a}^{c}f(x)dx\) and \(\int_{c}^{b}f(x)dx\) are convergent, then \(f\) over \([a,b]\) is

$$\int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$$

**FAQs**

**What is improper integral with example?**

An improper integral is one that has either or both infinite limits or an integrand that approaches infinity at one or more points in the range of integration. Riemann integrals cannot be used to compute improper integrals.

**What makes an integral improper?**

Integration is improper if either the lower limit of integration is infinite, the upper limit of integration is infinite, or both the upper and lower limits of integration are infinite.

**What is proper and improper integral?**

An integral with neither infinite limit nor integrand that approaches infinity at any point in its range.

**What are the two types of improper integrals?**

Improper Integrals exist in two types: Type 1 – when the limits of integration are infinite, and Type 2 – when the limits of integration are finite. An improper integral of type 2 occurs when the integrand becomes infinite within the integration interval.