Improper Integrals

The concept of Riemann integral as we have developed, requires that the range of integration if finite and the integrand remains bounded in that domain. If either (or both) of these assumptions is not satisfied it is necessary to attach a new interpretation to the integral.
In the case the integrand f becomes ifinite in the interval a ≤ x ≤ b, i.e. f has points of infinite discontinuity (singular points) in [a,b], or the limits of integration (or infinite

An improper integral is a definite integral that possesses one or more of the following properties:

1. the interval of integration is either semi-infinite or infinite in lenght i.e. of the form [a; ∞]

2. the integrand becomes infinite at an interior point of the interval of integration;

3. the integrand becomes infinite at an ened point of the interval of integration.

Example. Verify that the following is not an improper integral

$${\int }_0^1\frac{\sin x}{x}\text{d}x$$

is an example of proper integral since

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

Definition 6.1.1. Let f : [a,b) ⟶ ℝ, a continuous function, such that

$$\underset{x\to b^{-}}{lim}f(x)=+\mathrm{\infty }$$

We define the improper integral, ∫^b_{a f(x) dx, as}

$${\int }_a^{b}f(x)\text{d}x=\underset{\epsilon \to 0}{lim}{\int }_a^{b-\epsilon }f(x)\text{d}x$$

If there limit exists we say the improper integral converges. If the limit is ±∞ we say that the integral diverges. If, however, no such limit exists, we say that the improper integral does not exists. An analogous definition can be given for f : (a,b] ⟶ ℝ.  □

An analogous definition is given if f: (a, b] ⟶ ℝ, is continuos and

$$\underset{x\to a^{+}}{lim}f(x)=\pm \mathrm{\infty }$$

We let

$${\int }_a^b=f(x)\text{d}x=\underset{\epsilon \to 0^{+}}{lim}{\int }_{a+\epsilon }f(x)\text{d}x$$

Example 6.1.2. Calculate the following integral

$${\int }_a^b\frac{\text{d}x}{(b-x{)}^{\alpha }}(\alpha >0)$$

Case α = 1. We have

$${\int }_a^{b-\epsilon }\frac{\text{d}x}{(b-x{)}^{\alpha }}=[-\log (b-x){]}_a^{b-\epsilon }=-\log \epsilon +\log (b-a)$$

lim_{ε ⟶0⁺} [log ε + log(b − a)] = ∞

Thus the integral is divergent.

Case α ≠ 1. We have

$${\int }_a^{b-\epsilon }\frac{\text{d}x}{(b-x{)}^{\alpha }}=\frac{1}{1-\alpha }{[-(b-x{)}^{1-\alpha }]}_a^{b-\epsilon }=\frac{1}{1-\alpha }[-{\epsilon }^{1-\alpha }+(b-a{)}^{1-\alpha }]$$

Several convergence (or divergence) criteria are now discussed.

Theorem 6.1.3 (Comparison Test with Non-negative functions). Let f and g to continuous functions f,g : [a,b) ⟶ ℝ, such that

if 0 ≤ f(x) ≤ g(x) in [a,b), then

Indeed by the monocity integral property

$$0\le {\int }_a^{b-\epsilon }f(x)\text{d}x\le {\int }_a^{b-\epsilon }g(x)\text{d}x$$

and passing to the limit as ε ⟶ 0⁺, the claim is proved.  □

Theorem 6.1.4 (Asymptotic Comparison Test with Non-negative functions). Let f and g to continuous functions f,g : [a,b) ⟶ ℝ, such that

then

The same criteria hold for f,g ⟶ ∞ for for x ⟶ a⁺ or if f,g ⟶ −∞. In this last case the inequalities of the comparison test are among the absolute values of f and g.

Example 6.1.5. Consider the following integral

$${\int }_0^1\frac{\text{d}x}{\sqrt[3]{1-x^2}}$$

The integrand can be written as

$$f(x)=\frac{\text{d}x}{\sqrt[3]{1-x^2}}=\frac{1}{\sqrt[3]{1-x}\sqrt[3]{1-x}}$$

is a continuos and positive function in [a,b] and tends to +∞ fro x ⟶ 1⁻. Further

$$f(x)\sim \frac{1}{\sqrt[3]{2}}\frac{1}{\sqrt[3]{1-x}}\text{for}x\to 1^{-}$$

We knwon that the function g(x) = 1/(2(1 − x)^1/3 is positive and integrable (example 6.1.2 case α = 1/3), then f as well is integrable. The integral is hence convergent.  ■

Example 6.1.5. Consider the integral

$${\int }_1^3\frac{\text{d}x}{x^2-5x+4}$$

The integrand can be written as

it is a negative and continuous function in [1,3] and tends to −∞ for x ⟶ 1⁺. Observe that −f(x) ~ 1/3(x − 1) and that g(x) = 1/3(x −1). We knwon that the function g(x) = 1/3(x −1) is positive and not integrable (example 6.1.2 case α = 1). Then it follows that also −f and f are not integrable. The integral is divergent to −∞ ■

The comparison test and asymptotic comparison test apply only to constant sign integrands in analogy to what we studied for alternating sign series. A function might be unbounded for x ⟶ a⁺ without being constant sign. For example

$$f(x)=\frac{1}{\sqrt{x}}\sin \frac{1}{x}$$

is unbounded for x ⟶ 0⁺, but it doesn't tend to ±∞ and its sign is not constant. For this kind of situations the followin holds true

$${\int }_a^b|f(x)|\text{d}x\text{convergent}\Rightarrow {\int }_a^b|f(x)|\text{d}x\text{convergent}$$

If f is integrable on [a,b] then f is said absolutely integrable on [a,b].

Example 6.1.6. The integral

$${\int }_0^1\frac{1}{\sqrt{x}}\sin \frac{1}{x}$$

converges since

$$|\frac{1}{\sqrt{x}}\sin \frac{1}{x}|\le \frac{1}{\sqrt{x}}$$

which has a convergent integral.  ■

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