Find values of constants a and b such that the given improper integral converges

Question

Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get: $$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\frac{b}{3x-2}\right) dx=k$$ then $$\lim_{N \to \infty} \left[\frac{2}{3}ln(x)+(a-\frac{2}{3})ln (x+3)-\frac{b}{3}ln(3x-2)\right]_3^N =k$$ By logarithm properties: $$\lim_{N \to \infty} \left[ln\frac{(x)^\frac{2}{3}(x+3)^\left(a-\frac{2}{3}\right)}{(3x-2)\frac{b}{3}}\right]_3^N =k$$ Evaluating: $$\lim_{N \to \infty} \left[ln\frac{(N)^\frac{2}{3}(N+3)^\left(a-\frac{2}{3}\right)}{(3N-2)^\frac{b}{3}}\right]-\left[ln\frac{(3)^\frac{2}{3}(6)^\left(a-\frac{2}{3}\right)}{(7)^\frac{b}{3}}\right] =k$$ But, I do not know what to do after this step. Any recommendation will be appreciated

Answer 1

Hint: It converges when

$$ \lim _{ n\to \infty } \ln { \left[ \sqrt [ 3 ]{ \frac { { n }^{ 2 }{ \left( n+3 \right) }^{ 3a-2 } }{ { \left( 3n-2 \right) }^{ b } } } \right] } \\ \\ 2+3a-2=b\quad \Rightarrow b=3a $$

Answer 2

Hint. From your third line, one may write, as $x \to \infty$, $$ \begin{align} &\color{blue}{\frac{2}{3}}\ln(x)+\color{blue}{\left(a-\frac{2}{3}\right)}\ln (x+3)\color{blue}{-\frac{b}{3}}\ln(3x-2) \\=&\color{blue}{\left(\frac{2}{3}+a-\frac{2}{3}-\frac{b}{3}\right)}\ln x+\left(a-\frac{2}{3}\right)\ln \left(1+\frac 3x\right)-\frac{b}{3}\cdot\ln\left(3-\frac 2x\right) \\=&\color{blue}{\left(a-\frac{b}{3}\right)}\ln x+O(1) \end{align} $$ then all one needs is to annihilate the factor of $\ln x$, that is taking $$ \color{blue}{a=\frac b3} $$ in order to guarantee the convergence of the given integral.

Answer 3

Since $\frac{ax+2}{x^2+3x}=\frac{a}x+O\!\left(\frac1{x^2}\right)$ and $\frac{b}{3x-2}=\frac{b}{3x}+O\!\left(\frac1{x^2}\right)$, it should be that $a=\frac{b}3$. $$ \begin{align} &\int_3^\infty\left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right)\,\mathrm{d}x\\ &=\int_3^\infty\left(\frac{\frac23}{x}+\frac{a-\frac23}{x+3}-\frac{b}{3x-2}\right)\,\mathrm{d}x\\ &=\left[\frac23\log(x)+\left(a-\frac23\right)\log(x+3)-\frac{b}3\log(3x-2)\right]_3^\infty\\[6pt] &=\left[\frac23\log\left(\frac{x}{x+3}\right)+a\log\left(1+\frac3x\right)-\frac{b}3\log\left(3-\frac2{x}\right)+\left(a-\frac{b}3\right)\log(x)\right]_3^\infty \end{align} $$ So we see that indeed, the convergence is when $a=\frac{b}3$.