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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.

I posted before but I don't get it still some help

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

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

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    Instead of evaluating the integrals separately, bring them together. I'm saying, consider $\lim_{A \to \infty} \bigg( \frac12 \ln(x^2+1) - \frac C3 \ln|3x+1| \bigg)\bigg|_0^A$, and now try to find $C$ which works. This way, you can avoid the infinity. – 2017-02-18
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    Then what? Do I have to use ln laws because I don't know if I could? – 2017-02-18
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    Happily use $\ln$ laws. – 2017-02-18
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    See the above link. I knew I recognized this - the OP already asked the *exact same question* – 2017-02-18
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    I did say i posted before. I just didn't understand where he got the ln 3 explanation. I prefer this one – 2017-02-18

2 Answers 2

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Regardless of the value of $C$ we get a rational function for $$ \frac{x}{x^2+1} - \frac{C}{3x+1}, $$ namely $$ \frac{3x^2 + x - C x^2 - C}{3 x^3 + x^2 + 3x + 1} = \frac{(3 - C)x^2 + x - C}{3 x^3 + x^2 + 3x + 1} $$ If $C \neq 3$ the difference of degrees between denominator and numerator is $1$ and the integral diverges at $\infty.$ If $C=3$ the difference is $2$ and the integral converges; $$ \frac{ x - 3}{3 x^3 + x^2 + 3x + 1}. $$ This also converges at $0,$ the denominator is positive (at least $1$) for $x \geq 0.$

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    thx I get it now – 2017-02-18
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For large $x$ the first term behaves like $\sim \frac1x$, the second term as $\sim -\frac{C}{3x}$. So in order to cancel the divergence associated with integrating $1/x$ starting up to $x=\infty$, you need to set $C=3$.