Question

\(5 - 32\)Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

6. \(\int_0^\infty {\frac{1}{{\sqrt(4){{1 + x}}}}} dx\)

Short answer

Integral is divergent.

Step-by-step solution

Definition.

Consider the integral\(\int_{\bf{a}}^{\bf{t}} {\bf{f}} {\bf{(x)dx}}\)is for the limit with the finite number\(t \ge a\). Then, the integral is written as:

\(\int_a^\infty f (x)dx = \mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)

Change limit for the integration.

In order to determine if integral converges or diverges we evaluate both integrals i.e; \(\int_a^\infty f (x)dx\,\& \,\mathop {\lim }\limits_{t \to \infty } \int_a^t f (x)dx\)& see if both integrals are equal or not.

Let \(u = 1 + x\)then \( \Rightarrow du = dx\)

The new limits of integrations are,

Lower limit,

\(\begin{aligned}{l}u &= x + 1\\u &= 0 + 1\\u &= 1\end{aligned}\)

Upper limit,

\(\begin{aligned}{l}u &= 1 + x\\u &= 1 + \infty \\u &= \infty \end{aligned}\)

Hence, the new limits are 1 to\(\infty \).

Evaluate\(\int_a^\infty  f (x)dx\).

Substituting\(u = 1 + x\)& change limit to evaluate the integral.

\(\begin{aligned}{c}\int_0^\infty {\frac{1}{{\sqrt(4){{1 + x}}}}} dx &= \int_1^\infty {\frac{1}{{\sqrt(4){u}}}} du\\ &= \mathop {\lim }\limits_{t \to \infty } \int_1^t {\frac{1}{{\sqrt(4){u}}}} du\\ &= \left( {\mathop {\lim }\limits_{t \to \infty } \int_1^t {{u^{ - \frac{1}{4}}}} du} \right)_1^t\\ &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{4}{3}{u^{\frac{3}{4}}}} \right)_1^t\end{aligned}\)

Solve further as:

\(\begin{aligned}{c}\int_0^\infty {\frac{1}{{\sqrt(4){{1 + x}}}}} dx &= \mathop {\lim }\limits_{t \to \infty } \left( {\frac{4}{3}{t^{\frac{3}{4}}} - \frac{4}{3}{1^{\frac{3}{4}}}} \right)\\ &= \infty - \frac{4}{3}\end{aligned}\)

Hence, the limit does not exist as a finite number and so the improper integral is divergent.