Chapter 8.4, Problem 37E

To determine

To evaluate: Calculate the value of the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}$ or state its divergence.

Answer to Problem 37E

The value of the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}$ is $\pi$ .

Explanation of Solution

Given information: The integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}$ is given.

Concept used:

• If $f\left(x\right)$ is continuous on the interval $\left(a,b\right]$ , then ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\underset{c\to a^{+}}{\lim }{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx$ .
• If $f\left(x\right)$ is continuous on the interval $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ .

Calculation:

Rewrite the given integral.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}={\displaystyle \underset{0}{\overset{1}{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}+{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}$

Now, simplify the right side of the obtained integral using the fact that the integrand in the first term has an infinite discontinuity at interior point $s=0$ and as a result, continuous on the interval $\left(0,1\right]$ .

Therefore,

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}=\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{c}{\overset{1}{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}+\underset{c\to \infty }{\lim }{\displaystyle \underset{1}{\overset{c}{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}$

Consider $s=u^2$ such that $ds=2udu$ and the limit varies from $\sqrt{c}$ to $1$ for the first term and from 1 to $\sqrt{c}$ for the second term of the right side of the obtained equation. Simplify the right side of the obtained integral by using the mentioned substitutions and limits to get the required result.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}=\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{\sqrt{c}}{\overset{1}{\int }}\frac{2udu}{\left(1+u^2\right)\sqrt{u^2}}}+\underset{c\to \infty }{\lim }{\displaystyle \underset{1}{\overset{\sqrt{c}}{\int }}\frac{2udu}{\left(1+u^2\right)\sqrt{u^2}}}$

  $\begin{array}{c}=\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{\sqrt{c}}{\overset{1}{\int }}\frac{2udu}{\left(1+u^2\right)u}}+\underset{c\to \infty }{\lim }{\displaystyle \underset{1}{\overset{\sqrt{c}}{\int }}\frac{2udu}{\left(1+u^2\right)u}}\\ =\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{\sqrt{c}}{\overset{1}{\int }}\frac{2du}{\left(1+u^2\right)}}+\underset{c\to \infty }{\lim }{\displaystyle \underset{1}{\overset{\sqrt{c}}{\int }}\frac{2du}{\left(1+u^2\right)}}\end{array}$

  $\begin{array}{c}=\underset{c\to 0^{+}}{\lim }2{\left[ta{n}^{-1}\left(u\right)\right]}_{\sqrt{c}}^1+\underset{c\to \infty }{\lim }2{\left[ta{n}^{-1}\left(u\right)\right]}_1^{\sqrt{c}}\\ =\underset{c\to 0^{+}}{\lim }2\left[ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(\sqrt{c}\right)\right]+\underset{c\to \infty }{\lim }2\left[ta{n}^{-1}\left(\sqrt{c}\right)-ta{n}^{-1}\left(1\right)\right]\end{array}$

  $\begin{array}{c}=2\left[\frac{\pi }{4}-ta{n}^{-1}\left(0\right)\right]+2\left[ta{n}^{-1}\left(\infty \right)-\frac{\pi }{4}\right]\\ =2\left[\frac{\pi }{4}-0\right]+2\left[\frac{\pi }{2}-\frac{\pi }{4}\right]\\ =\frac{\pi }{2}+2\left[\frac{\pi }{4}\right]\\ =\frac{\pi }{2}+\frac{\pi }{2}\\ =\pi \end{array}$

As a result, ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{ds}{\left(1+s\right)\sqrt{s}}}=\pi$