Chapter 5.10, Problem 64E

To determine

To calculate the value of $C$ for which the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{x}{x^2+1}-\frac{C}{3x\text{+1 }}\right)dx}$ is convergent

Answer to Problem 64E

This integral is convergent at $C=3$ , and the value of the integral is $-\ln (3)$

Explanation of Solution

Given Information: The integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{x}{x^2+1}-\frac{C}{3x\text{+1 }}\right)dx}$ is convergent

Concept used: If the value of the integral is finite, it is convergent.

Formula used:

 i ${\displaystyle \int \frac{1}{x}dx=}\ln (x)$

 ii ${\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=}\ln \left[f(x)\right]$

The given integral becomes as:

  ${\left[\frac{1}{2}\ln \left(x^2+1\right)-\frac{C}{3}\ln \left(3x+1\right)\right]}_0^{\infty }$

  $={\left[\ln {\left(x^2+1\right)}^{1/2}-\ln {\left(3x+1\right)}^{C/3}\right]}_0^{\infty }$

  $={\left[\ln \left[\frac{{\left(x^2+1\right)}^{1/2}}{{\left(3x+1\right)}^{C/3}}\right]\right]}_0^{\infty }$

  $={\left[\ln \left[\frac{x{\left(1+(1/x^2\right)}^{1/2}}{{(3x)}^{C/3}{\left(1+1/3x\right)}^{C/3}}\right]\right]}_0^{\infty }$

  $={\left[\ln \left[\frac{{\left(1+(1/x^2\right)}^{1/2}}{{(3)}^{C/3}{(x)}^{[(C/3)-1]}{\left(1+1/3x\right)}^{C/3}}\right]\right]}_0^{\infty }$

  $={\left[\ln \left[\frac{{\left(1+(1/x^2\right)}^{1/2}}{{(3)}^{C/3}{(x)}^{[(C-3)/3)]}{\left(1+1/3x\right)}^{C/3}}\right]\right]}_0^{\infty }$

At $C=3$ , the integral has finite value.

The value of the integral is $\ln \left(\frac{1}{3}\right)$

This integral has value $\ln \left(\frac{1}{3}\right)$ at $C=3$ .