Improper Integrals: Videos & Practice Problems

Improper integrals involve integrating functions with infinite bounds, requiring limits to define the process. For an upper bound of positive infinity, rewrite the integral as a limit approaching infinity. Similarly, for a lower bound of negative infinity, use a limit approaching negative infinity. If both bounds are infinite, split the integral into two parts. Convergence occurs when the limit exists, while divergence indicates it does not. Understanding these concepts is crucial for evaluating the behavior of functions at extreme values.

Improper Integrals: Infinite Intervals

In calculus, improper integrals arise when one or both bounds of an integral extend to infinity. To evaluate these integrals, we must use limits to handle the infinite bounds appropriately. An improper integral can be defined as follows: if we want to integrate a function from a lower bound \( a \) to an upper bound of positive infinity, we rewrite the integral as:

\[\int_a^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_a^{t} f(x) \, dx\]

This means we first integrate the function from \( a \) to a variable \( t \), and then take the limit as \( t \) approaches infinity. This process allows us to find the area under the curve as we extend towards infinity.

Similarly, if the lower bound is negative infinity, we express the integral as:

\[\int_{-\infty}^{b} f(x) \, dx = \lim_{t \to -\infty} \int_{t}^{b} f(x) \, dx\]

Again, we integrate from \( t \) to \( b \) and take the limit as \( t \) approaches negative infinity. This approach helps us calculate the area under the curve extending towards negative infinity.

In cases where both bounds are infinite, we can split the integral into two parts. For example:

\[\int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{c} f(x) \, dx + \int_{c}^{\infty} f(x) \, dx\]

Here, \( c \) is a constant chosen between the two infinities. Each part can then be evaluated using the limits as described above.

To determine whether an improper integral is convergent or divergent, we check if the limit exists. If the limit yields a finite number, the integral is convergent; if it results in infinity or does not exist, the integral is divergent.

For example, consider the integral:

\[\int_{0}^{\infty} e^{-x} \, dx\]

We rewrite it as:

\[\lim_{t \to \infty} \int_{0}^{t} e^{-x} \, dx\]

Calculating the integral gives us:

\[\lim_{t \to \infty} \left[-e^{-x}\right]_{0}^{t} = \lim_{t \to \infty} \left(-e^{-t} + 1\right) = 1\]

Since the limit exists and equals 1, this integral is convergent.

In contrast, for the integral:

\[\int_{0}^{\infty} x \, dx\]

We rewrite it as:

\[\lim_{t \to \infty} \int_{0}^{t} x \, dx\]

Calculating this gives:

\[\lim_{t \to \infty} \left[\frac{x^2}{2}\right]_{0}^{t} = \lim_{t \to \infty} \frac{t^2}{2} = \infty\]

Since the limit does not exist (it approaches infinity), this integral is divergent.

Understanding these concepts and procedures is crucial for effectively working with improper integrals in calculus. Practice with various examples will help solidify these techniques and enhance your problem-solving skills.

Improper Integrals: Infinite Intervals Example 1

To evaluate the improper integral from negative infinity to zero of the function \( x \sin(x) \, dx \), we first recognize that the presence of negative infinity as a limit indicates that we need to approach this integral using limits. We rewrite the integral as:

\[\lim_{t \to -\infty} \int_{t}^{0} x \sin(x) \, dx\]

Since the integrand is a product of two functions, we cannot use a simple substitution. Instead, we apply integration by parts, which is based on the formula:

\[\int u \, dv = uv - \int v \, du\]

We choose \( u = x \) (thus \( du = dx \)) and \( dv = \sin(x) \, dx \) (which integrates to \( v = -\cos(x) \)). Applying integration by parts, we have:

\[\int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx\]

Integrating \( \cos(x) \) gives us \( \sin(x) \), so we can express the integral as:

\[-x \cos(x) + \sin(x)\]

Now, we evaluate this expression from \( t \) to \( 0 \):

\[\lim_{t \to -\infty} \left[ -x \cos(x) + \sin(x) \right]_{t}^{0}\]

Plugging in the upper limit \( 0 \), we find:

\[-0 \cdot \cos(0) + \sin(0) = 0\]

For the lower limit \( t \), we have:

\[-t \cos(t) + \sin(t)\]

Thus, the expression becomes:

\[0 - \left( -t \cos(t) + \sin(t) \right) = t \cos(t) - \sin(t)\]

As \( t \) approaches negative infinity, we analyze the behavior of \( t \cos(t) \) and \( -\sin(t) \). The term \( t \cos(t) \) oscillates between \( -\infty \) and \( +\infty \) because \( \cos(t) \) oscillates between -1 and 1, while \( t \) approaches negative infinity. Similarly, \( -\sin(t) \) oscillates between -1 and 1. Therefore, both terms do not converge to a single value.

Since both components diverge, we conclude that the original integral diverges. Thus, the integral from negative infinity to zero of \( x \sin(x) \, dx \) is divergent.

Improper Integrals: Infinite Intervals Example 2

To determine the convergence or divergence of an improper integral, we start by rewriting the integral with an infinite upper bound. For the integral in question, we express it as a limit:

$$\int_0^{\infty} \frac{2}{1 + x^2} \, dx = \lim_{t \to \infty} \int_0^t \frac{2}{1 + x^2} \, dx$$

Next, we can factor out the constant 2 from the integral:

$$= 2 \cdot \lim_{t \to \infty} \int_0^t \frac{1}{1 + x^2} \, dx$$

The integral $$\int \frac{1}{1 + x^2} \, dx$$ is a well-known form that results in the inverse tangent function:

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

Thus, we can rewrite our expression as:

$$= 2 \cdot \lim_{t \to \infty} \left[ \tan^{-1}(x) \right]_0^t$$

Evaluating this from 0 to t gives us:

$$= 2 \cdot \lim_{t \to \infty} \left( \tan^{-1}(t) - \tan^{-1}(0) \right)$$

Since $$\tan^{-1}(0) = 0$$, we simplify to:

$$= 2 \cdot \lim_{t \to \infty} \tan^{-1}(t)$$

As $$t$$ approaches infinity, the inverse tangent function approaches its horizontal asymptote at $$\frac{\pi}{2}$$:

$$= 2 \cdot \frac{\pi}{2} = \pi$$

Since we obtained a finite value of $$\pi$$, we conclude that the integral converges. Therefore, the result of the improper integral is:

$$\pi$$

Improper Integrals: Infinite Intervals Example 3

In evaluating improper integrals, it is essential to determine whether they converge or diverge, and to compute their values if they converge. For instance, consider the integral of the function \( x e^x \) from 0 to infinity. Since the upper limit is infinity, we rewrite the integral as a limit:

\[ \int_0^\infty x e^x \, dx = \lim_{t \to \infty} \int_0^t x e^x \, dx \]

To solve this integral, we apply integration by parts, which is based on the formula:

\[ \int u \, dv = uv - \int v \, du \]

Choosing \( u = x \) (thus \( du = dx \)) and \( dv = e^x \, dx \) (which gives \( v = e^x \)), we can express the integral as:

\[ \lim_{t \to \infty} \left( x e^x - \int e^x \, dx \right) \]

Evaluating the integral of \( e^x \) yields:

\[ \lim_{t \to \infty} \left( x e^x - e^x \right) \bigg|_0^t \]

Substituting the bounds, we find:

\[ \lim_{t \to \infty} \left( t e^t - e^t \right) - \left( 0 - 1 \right) \]

As \( t \) approaches infinity, both \( t e^t \) and \( e^t \) diverge to infinity, leading to the conclusion that the integral diverges. Thus, the result for this example is that the integral diverges.

In the second example, we evaluate the integral of \( x e^x \) from negative infinity to zero. We express this as:

\[ \int_{-\infty}^0 x e^x \, dx = \lim_{t \to -\infty} \int_t^0 x e^x \, dx \]

Using the same integration by parts approach, we find that the integral evaluates to:

\[ \lim_{t \to -\infty} \left( 0 - (t e^t - e^t) \right) \bigg|_t^0 \]

Substituting the bounds gives:

\[ 0 - (-1) - \lim_{t \to -\infty} (t e^t - e^t) \]

As \( t \) approaches negative infinity, \( e^t \) approaches zero, and thus the entire expression converges to -1. Therefore, this integral converges to -1.

Finally, for the integral from negative infinity to positive infinity of \( x e^x \), we can split it into two parts:

\[ \int_{-\infty}^\infty x e^x \, dx = \int_{-\infty}^0 x e^x \, dx + \int_0^\infty x e^x \, dx \]

From previous evaluations, we know the first integral converges to -1, while the second diverges. Consequently, since one part diverges, the entire integral diverges. Thus, the conclusion for this example is that the integral diverges.

In summary, understanding the behavior of the function as it approaches the limits is crucial in determining the convergence or divergence of improper integrals. Integration by parts is a valuable technique for evaluating these integrals, especially when dealing with products of functions.

Evaluate the integral or state that it diverges. 
${\int }_{-\infty }^{-1}\frac{2}{x^3}dx$

• A. The integral diverges.
• B. $-2$; converges.
• C. $-1$; converges.
• D. $1$; converges.

Answer: (C)

Evaluate the integral or state that it diverges. 
${\int }_2^{\infty }\frac{1}{x(lnx{)}^4}dx$

• A. The Integral diverges.
• B. $2$; converges.
• C. $\frac{1}{5{\left(\ln 2\right)}^5}$ ; converges.
• D. $\frac{1}{3{\left(\ln 2\right)}^3}$; converges.

Answer: (D)

Here’s what students ask on this topic:

What is an improper integral and how is it different from a regular definite integral?

An improper integral is a type of definite integral where one or both of the bounds are infinite, or the integrand becomes unbounded within the interval of integration. Unlike regular definite integrals, improper integrals require the use of limits to evaluate them. For example, if the upper bound is infinity, the integral is rewritten as a limit: $lim\left(t_{\to }\right)\int a_{t}f\left(x\right)dx$. This process ensures the integral is properly defined even when dealing with infinite or unbounded values.

How do you evaluate an improper integral with an infinite upper bound?

To evaluate an improper integral with an infinite upper bound, rewrite the integral using a limit. For example, if the integral is $\int a_{\infty }f\left(x\right)dx$, rewrite it as $lim\left(t_{\to }\right)\int a_{t}f\left(x\right)dx$. Then, compute the integral as usual and apply the limit. If the limit exists and is finite, the integral converges. Otherwise, it diverges.

What does it mean for an improper integral to converge or diverge?

An improper integral converges if the limit used to evaluate it exists and results in a finite value. For example, if $lim\left(t_{\to }\right)\int a_{t}f\left(x\right)dx$ equals a finite number, the integral converges. Conversely, the integral diverges if the limit does not exist or approaches infinity. Divergence indicates that the area under the curve is unbounded or undefined.

How do you handle improper integrals with both bounds being infinite?

When both bounds of an improper integral are infinite, split the integral into two parts at a finite constant $c$. For example, $\int {\mathrm{-\infty }}_{\infty }f\left(x\right)dx$ becomes $\int {\mathrm{-\infty }}_{c}f\left(x\right)dx+\int c_{\infty }f\left(x\right)dx$. Evaluate each part separately using limits. If both parts converge, the entire integral converges. If either part diverges, the integral diverges.

What are some common examples of improper integrals that diverge?

Common examples of improper integrals that diverge include integrals where the function grows too quickly as it approaches infinity. For instance, $\int 1_{\infty }{x}^{-1}dx$ diverges because the area under the curve becomes unbounded. Similarly, $\int 0_{\infty }{e}^{x}dx$ diverges because the exponential function grows infinitely large as $x$ approaches infinity.

How do you determine if an improper integral with a vertical asymptote converges?

To determine if an improper integral with a vertical asymptote converges, rewrite the integral using a limit that approaches the point of the asymptote. For example, for $\int 0_1{x}^{-1}dx$, rewrite it as $lim(t_{\to }+)\int t_1{x}^{-1}dx$. Evaluate the integral and apply the limit. If the limit exists and is finite, the integral converges; otherwise, it diverges.