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Find the value of the constant C for which the following integral $$\int_{0}^{\infty} \bigg(\frac{x}{x^2+1} - \frac{C}{3x+1}\bigg)dx$$

converges. Evaluate the integral for that value of C. Make sure to fully justify your answer. \

My solution:

$$\lim_{A\to\infty}\int_{0}^{A} \frac{x}{x^2+1} dx - \lim_{A\to\infty} \int_{0}^{A} \frac{C}{3x+1} dx$$

$$= \lim_{A\to\infty} \frac{1}{2} ln(x^2+1) \bigg|_{0}^{A} - \lim_{A\to\infty} \frac{C}{3} ln|3x+1| \bigg|_{0}^{A}$$

$$= \frac{1}{2}(\infty + 0) - \frac{C}{3} (\infty + 0)$$

Idk how to make it converge.

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    You shouldn't manipulate the symbol $\infty$ as if it were a number. You should rather compute $\int_0^A$ and deduce how $C$ must be chosen in order to get a finite limit as $A\to+\infty$. – 2017-02-13
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    Bad idea splitting it as the sum of two limits: the sum of two limits makes sense only if both limits exist; and if both limits exist, the exercise is pointless. You should have kept it as $$\lim_{A\to\infty}\left(\frac12\ln(A^2+1)-\frac C3\ln(3A+1)\right)$$ – 2017-02-13

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The integral is $$I=\int_0^\infty \frac{(3-C)x^2+3x-C}{(x^2+1)(3x+1)}\,dx\ .$$ The integrand is continuous for $x\ge0$, so the only possible problem is the infinite interval. If $C=3$ then the integrand is of the order of $x^{-2}$ and the integral converges. If $C\ne3$ then it is of order $x^{-1}$ and the integral diverges.

Evaluating the integral by standard methods gives $$I=\lim_{A\to\infty}\ln\frac{\sqrt{A^2+1}}{3A+1} =\lim_{A\to\infty}\ln\frac{1+A^{-2}}{3+A^{-1}}=-\ln3\ .$$

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    Okay I got it. Thank you ! – 2017-02-13