Chapter 7.6, Problem 65P

To determine

To evaluate the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sqrt{x}}{e^x-1}dx}$ .

Explanation of Solution

Given:

The given integral is ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sqrt{x}}{e^x-1}dx}$ .

Formula used:

Write the value of standard integral.

  ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sqrt{x}}{e^x-1}}=2.3152$

Write the integration.

  $I={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sqrt{x}}{e^x-1}}$ …… (1)

Calculation:

Substitute2.3152 for ${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sqrt{x}}{e^x-1}}$ in equation (1).

  $I=2.3152$

Conclusion:

Thus, the value of integral is $2.3152$ .