Question

For what value of \(c\) does the integral \(\int_{1}^{\infty}\left(\frac{c x}{x^{2}+2}-\frac{1}{3 x}\right) d x\) converge? Evaluate the integral for this value of \(c\)

Short answer

The integral will converge for \(c ≤ 1\). The value of the integral for the range of convergence is \(c / 2\) ln 2.

Step-by-step solution

Determine The Value of \(c\)

Apply the comparison test to determine the condition under which the integral will converge. In general, if \(0 ≤ g(x) ≤ f(x)\) for \(a ≤ x < ∞\) and \(\int_{a}^{\infty} g(x) dx\) converges, then \(\int_{a}^{\infty} f(x) dx\) also converges. Comparing the function \(f(x) = \frac{c x}{x^{2}+2}-\frac{1}{3 x}\) with \(g(x) = \frac{c}{x} + \frac{1}{3x} \), \(c > 0\).Hence, \(\int_{1}^{\infty} \frac{c}{x} dx\) + \(\int_{1}^{\infty} \frac{1}{3x} dx\) is going to be finite, obviously implies that \(c ≤ 1\). So, the integral will converge only for \(c ≤ 1\).

Perform Integration

Now that we've established the condition under which the integral will converge (when \(c ≤ 1\)), we find it relevant to evaluate the integral with this value of \(c\). To do that, we split the integral into two parts and solve each one separately.\(\int_{1}^{\infty} \frac{c x}{x^{2}+2} dx\) - \(\int_{1}^{\infty} \frac{1}{3 x} dx\)The first integral can simplify to \(c / 2\) ln\((x^2+2)\) from 1 to infinity which simplifies to \(c / 2\) ln 2.The second integral results in \(-1 / 3\) ln x from 1 to infinity which is 0.Combining the two values, we can write the solution of the integral as \(c / 2\) ln 2.