Question

Explain why each of the following integrals is improper.

(a) \(\int_1^2 {\frac{x}{{x - 1}}} dx\)

(b) \(\int_0^\infty {\frac{1}{{1 + {x^3}}}} dx\)

(c) \(\int_{ - \infty }^\infty {{x^2}} {e^{ - {x^2}}}dx\)

(d) \(\int_0^{\pi /4} {\cot } xdx\)

Short answer

(a) Improper integral

(b) Improper integral

(c) Improper integral

(d) Improper integral

Step-by-step solution

Definition.

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration.

For option (a).

Consider the integral\(\int_1^2 {\frac{x}{{x - 1}}} dx\)

Assume\(f(x) = \frac{x}{{x - 1}}\).

Clearly \(f(x)\) is not defined at \(x = 1\) i.e; function is discontinuous at\(x = 1\), as \(\mathop {\lim }\limits_{x \to 1} \frac{x}{{x - 1}}\) does not exist.

Also the function is continuous in\((1,2)\).

Therefore, the integral \(\int_1^2 {\frac{x}{{x - 1}}} dx\) is improper integral.

For option (b):

Consider the integral\(\int_0^\infty {\frac{1}{{1 + {x^3}}}} dx\)

Assume\(f(x) = \frac{1}{{1 + {x^3}}}\)

Clearly here the interval of integral is\((0,\infty )\).

So, the integral is improper integral by definition of improper integral.

For option (c):

Consider integral\(\int_{ - \infty }^\infty {{x^2}} {e^{ - {x^2}}}dx\).

Here the limit of integration is\(( - \infty ,\infty )\) so the integral is improper by definition of improper integral.

For option (d):

Consider the integral\(\int_0^{\pi /4} {\cot } xdx\).

This integral can be rewritten as:

\(\int_0^{\pi /4} {\cot } xdx = \int\limits_0^{\frac{\pi }{4}} {\frac{{\cos x}}{{\sin x}}} dx\)

Clearly when\(\sin x = 0\) this function is discontinuous at\(x = 0\), because

\(\begin{array}{c}\mathop {\lim }\limits_{x \to 0} \cot x = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\sin x}}\\ = \infty \end{array}\)

Also the function is continuous in\(\left( {0,\frac{\pi }{4}} \right)\).

So, the integral is improper integral.