value of λ so that the improper integral exists

Question

For which value of $\lambda \in\Re $ the improper integral I exists $$I=\int_0^\infty (\frac{x+1}{3x^2+λ}-\frac{λ}{2x+1})dx. $$

I tried to calculate the integrals separately but no effect. Thanks in advance

Answer 1

Clearly if $\lambda=0$, $I$ diverges. Now suppose $\lambda\neq0$. Note $$I=\int_0^\infty (\frac{x+1}{3x^2+λ}-\frac{λ}{2x+1})dx=\int_0^\infty \frac{(2-3\lambda)x^2+2x^2+3x+(1-a^2)}{(3x^2+λ)(2x+1)}dx. $$ If $\lambda\neq\frac23$, then $$ \frac{(2-3\lambda)x^2+2x^2+3x+(1-a^2)}{(3x^2+λ)(2x+1)}\approx\frac{2-3\lambda}{6x}$$ when $x$ is large. Since $\int_1^\infty\frac1xdx$ diverges, hence $I$ diverges.

If $\lambda=\frac23$, then $$ \frac{(2-3\lambda)x^2+2x^2+3x+(1-a^2)}{(3x^2+λ)(2x+1)}\approx\frac{1}{3x^2} $$ when $x$ is large. Since $\int_1^\infty\frac1{x^2}dx$ converges, hence $I$ converges.

Answer 2

If $\lambda = 0$, then the integrand behaves like $\frac{1}{3x}$ as $x \to 0$, whose integral around zero diverges. So we must have $\lambda \ne 0$.

If $\lambda \ne 0$, then the integrand behaves like $\frac{1}{\lambda} - \lambda$ as $x \to 0$ which is fine. It remains to check the convergence of the upper tail.

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As $x \to \infty$ the integrand behaves like $(\frac{1}{3} - \frac{\lambda}{2}) \frac{1}{x}$, so if $\frac{1}{3} \ne \frac{\lambda}{2}$, there exists a constant $c > 0$ such that for all large $x$ one of the following happens. $$\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} \ge \frac{c}{x} \qquad \text{or} \qquad \frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} \le -\frac{c}{x}.$$ In either case, the integral diverges as $x \to \infty$. So we need $\lambda = 2/3$.