Question

Find the value of the constantC for which the integral \(\int\limits_0^\infty {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} \)converges. Evaluate the integral for this value ofC.

Short answer

C = 3; - ln 3

Step-by-step solution

Finding the value of C

Given \(\int\limits_0^\infty {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} \)

\(\begin{aligned}{c}\int\limits_0^\infty {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} = \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} \\ \Rightarrow \mathop {\lim }\limits_{t \to \infty } \int_0^t {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} \end{aligned}\)

\(\int {\frac{x}{{{x^2} + 1}} = \frac{1}{2}\ln \left| {{x^2} + 1} \right|} + C\;{\rm{and}}\;\int {\frac{1}{{bx + a}}dx = \frac{1}{b}\ln (bx + a) + C} \)

\(\begin{aligned}{c} \Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{2}\ln \left| {{x^2} + 1} \right| - \frac{C}{3}\ln (3x + 1)} \right)_0^t\\ \Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln {{\left( {{x^2} + 1} \right)}^{\frac{1}{2}}} - \ln {{(3x + 1)}^{\frac{C}{3}}}} \right)_0^t\\ \Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln \frac{{{{\left( {{x^2} + 1} \right)}^{\frac{1}{2}}}}}{{{{(3x + 1)}^{\frac{C}{3}}}}}} \right)_0^t\end{aligned}\)

If here we put \(t = \infty \), then the integral will be divergent \(\int\limits_0^\infty {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} = \infty \) but we want to converge it.

So, for making it convergent, we will take x common and try to eliminate it

\( \Rightarrow \)\( \Rightarrow \mathop {\lim }\limits_{t \to \infty } \left( {\ln \frac{{x{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{1}{2}}}}}{{{x^{^{\frac{C}{3}}}}{{(3 + \frac{1}{x})}^{\frac{C}{3}}}}}} \right)_0^t\)

In above equation x should be equal to \({x^{^{\frac{C}{3}}}}\) for convergence of the given integral

\(\begin{aligned}{c} \Rightarrow x = {x^{^{\frac{C}{3}}}}\\ \Rightarrow C = 3\end{aligned}\)

Hence the value of C is 3

Evaluation of Integral \(\int\limits_0^\infty  {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} \)

From above equation \(\mathop {\lim }\limits_{t \to \infty } \left( {\ln \frac{{x{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{1}{2}}}}}{{{x^{^{\frac{C}{3}}}}{{(3 + \frac{1}{x})}^{\frac{C}{3}}}}}} \right)_0^t\)

After solving the given equation for that it converges, we get

\(C = 3\)and \(\int\limits_0^\infty {\left( {\frac{x}{{{x^2} + 1}} - \frac{C}{{3x + 1}}} \right)dx} = - \ln \;3\)

Hence the value of the integral is – ln 3