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# Calculus

Integration of Functions

# Improper Integrals

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Problem 1
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Problems 2-10

The definite integral $$\int\limits_a^b {f\left( x \right)dx}$$ is called an improper integral if one of two situations occurs:

• The limit $$a$$ or $$b$$ (or both the limits) are infinite;
• The function $${f\left( x \right)}$$ has one or more points of discontinuity in the interval $$\left[ {a,b} \right].$$

### Infinite Limits of Integration

Let $${f\left( x \right)}$$ be a continuous function on the interval $$\left[ {a,\infty} \right).$$ We define the improper integral as
${\int\limits_a^\infty {f\left( x \right)dx} }={ \lim\limits_{n \to \infty } \int\limits_a^n {f\left( x \right)dx} .}$ Consider the case when $${f\left( x \right)}$$ is a continuous function on the interval $$\left( {-\infty, b} \right].$$ Then we define the improper integral as
${\int\limits_{ – \infty }^b {f\left( x \right)dx} }={ \lim\limits_{n \to – \infty } \int\limits_n^b {f\left( x \right)dx} .}$ If these limits exist and are finite then we say that the improper integrals are convergent. Otherwise the integrals are divergent.

Let $${f\left( x \right)}$$ be a continuous function for all real numbers. We define:
${\int\limits_{ – \infty }^\infty {f\left( x \right)dx} } = {\int\limits_{ – \infty }^c {f\left( x \right)dx} }+{ \int\limits_c^\infty {f\left( x \right)dx} .}$ If, for some real number $$c,$$ both of the integrals in the right side are convergent, then we say that the integral $$\int\limits_{ – \infty }^\infty {f\left( x \right)dx}$$ is also convergent; otherwise it is divergent.

### Comparison Theorems

Let $${f\left( x \right)}$$ and $${g\left( x \right)}$$ be continuous functions on the interval $$\left[ {a,\infty } \right).$$ Suppose that $$0 \le g\left( x \right) \le f\left( x \right)$$ for all $$x$$ in the interval $$\left[ {a,\infty } \right).$$

1. If $$\int\limits_a^\infty {f\left( x \right)dx}$$ is convergent, then $$\int\limits_a^\infty {g\left( x \right)dx}$$ is also convergent;
2. If $$\int\limits_a^\infty {g\left( x \right)dx}$$ is divergent, then $$\int\limits_a^\infty {f\left( x \right)dx}$$ is also divergent;
3. If $$\int\limits_a^\infty {\left| {f\left( x \right)} \right|dx}$$ is convergent, then $$\int\limits_a^\infty {f\left( x \right)dx}$$ is also convergent. In this case, we say that the integral $$\int\limits_a^\infty {f\left( x \right)dx}$$ is absolutely convergent.

### Discontinuous Integrand

Let $${f\left( x \right)}$$ be a function which is continuous on the interval $$\left[ {a,b} \right),$$ but is discontinuous at $$x = b.$$ We define the improper integral as
${\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\tau \to 0 + } \int\limits_a^{b – \tau } {f\left( x \right)dx} .}$ Similarly we can consider the case when the function $${f\left( x \right)}$$ is continuous on the interval $$\left( {a,b} \right],$$ but is discontinuous at $$x = a.$$ Then
${\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\tau \to 0 + } \int\limits_{a + \tau }^b {f\left( x \right)dx} .}$ If these limits exist and are finite then we say that the integrals are convergent; otherwise the integrals are divergent.

Let $${f\left( x \right)}$$ be a continuous function for all real numbers $$x$$ in the interval $$\left[ {a,b} \right],$$ except for some point $$c \in \left( {a,b} \right).$$ We define:
${\int\limits_a^b {f\left( x \right)dx} } = {\int\limits_a^c {f\left( x \right)dx} }+{ \int\limits_c^b {f\left( x \right)dx} ,}$ and say that the integral $$\int\limits_a^b {f\left( x \right)dx}$$ is convergent if both of the integrals in the right side are also convergent. Otherwise the improper integral is divergent.

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Determine for what values of $$k$$ the integral $${\int\limits_1^\infty} {\large\frac{{dx}}{{{x^k}}}\normalsize}\;\left( {k > 0,k \ne 1} \right)$$ converges.

### ✓Example 2

Calculate the integral $$\int\limits_0^\infty {\large\frac{{dx}}{{{x^2} + 16}}\normalsize} .$$

### ✓Example 3

Determine whether the integral $${\int\limits_1^\infty} {\large\frac{{dx}}{{{x^2}{e^x}}}\normalsize}$$ converges or diverges?

### ✓Example 4

Calculate the integral $${\int\limits_{ – 2}^2} {\large\frac{{dx}}{{{x^3}}}\normalsize}.$$

### ✓Example 5

Determine whether the improper integral $${\int\limits_{ – \infty }^\infty} {\large\frac{{dx}}{{{x^2} + 2x + 8}}\normalsize}$$ converges or diverges?

### ✓Example 6

Determine whether the integral $${\int\limits_1^\infty} {{\large\frac{{\sin x}}{{\sqrt {{x^3}} }}\normalsize} dx}$$ converges or diverges?

### ✓Example 7

Determine whether the integral $${\int\limits_0^4} {\large\frac{{dx}}{{{{\left( {x – 2} \right)}^3}}}\normalsize}$$ converges or diverges?

### ✓Example 8

Determine for what values of $$k$$ the integral $${\int\limits_0^1} {\large\frac{{dx}}{{{x^k}}}\normalsize}$$ $$\left( {k \gt 0,k \ne 1} \right)$$ converges?

### ✓Example 9

Find the area above the curve $$y = \ln x$$ in the lower half-plane between $$x = 0$$ and $$x = 1.$$

### ✓Example 10

Find the circumference of the unit circle.

### Example 1.

Determine for what values of $$k$$ the integral $${\int\limits_1^\infty} {\large\frac{{dx}}{{{x^k}}}\normalsize}\;\left( {k > 0,k \ne 1} \right)$$ converges.

#### Solution.

By the definition of an improper integral, we have
${\int\limits_1^\infty {\frac{{dx}}{{{x^k}}}} } = {\lim\limits_{n \to \infty } \int\limits_1^n {\frac{{dx}}{{{x^k}}}} } = {\lim\limits_{n \to \infty } \int\limits_1^n {{x^{ – k}}dx} } = {\lim\limits_{n \to \infty } \left. {\left( {\frac{{{x^{ – k + 1}}}}{{ – k + 1}}} \right)} \right|_1^n } = {\frac{1}{{1 – k}} \cdot \lim\limits_{n \to \infty } \left. {\left( {{x^{ – k + 1}}} \right)} \right|_1^n } = {\frac{1}{{1 – k}} \cdot \lim\limits_{n \to \infty } \left( {{n^{ – k + 1}} – {1^{ – k + 1}}} \right) } = {\frac{1}{{k – 1}} \cdot \lim\limits_{n \to \infty } \left( {1 – {n^{1 – k}}} \right).}$ As seen from the expression, there are $$2$$ cases:

• If $$0 \lt k \lt 1,$$ then $${n^{1 – k}} \to \infty$$ as $$n \to \infty$$ and the integral diverges;
• If $$k \gt 1,$$ then $${n^{1 – k}}$$ $$= {\large\frac{1}{{{n^{k – 1}}}\normalsize}} \to 0$$ as $$n \to \infty$$ and the integral converges.
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