test the integral $\int_{0}^{\infty} \frac {x}{3x^4 + 5x^2 +1}dx$ for convergence.

My thought

Can I compare it with 1/(3x^4)?

Any hints for the solution are appreciated!

  • Is $3x^4+5x^2+1>x^4$? – John Wayland Bales Dec 7 at 6:16
  • 2
    you can try $$\frac{x}{3x^4+5x^2+1}\leq \frac{1}{x^3},\,\,\,x\geq 0$$ – Lau Dec 7 at 6:19
  • @Lau but I have to prove this fact first ....... is it proved here on math stack ? – hopefully Dec 7 at 6:40
  • 1
    This can be easily seen:$\forall x\geq0, \frac x{3x^4+5x^2+1}\leq\frac x{3x^4}\leq\frac x{x^4} \because$ the terms we ignore in the denominator are all positive. By ignoring them, we make the denominator smaller and thus the value of the ratio larger. – Shubham Johri Dec 7 at 6:44
  • I am not speaking about this @Lau I am speaking about proving the convergence of an integral that is different from the required one (the initial point differes) – hopefully Dec 7 at 6:55
up vote 2 down vote accepted

$I=\int_0^\infty \frac x{3x^4 + 5x^2 +1}dx$

$\forall x>0, 0\leq\frac x{3x^4 + 5x^2 +1}\leq\frac x{x^4}=\frac1{x^3}$

$\implies 0\leq I=\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac x{3x^4 + 5x^2 +1}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac 1{x^3}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\int_1^\infty \frac 1{x^3}dx\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_0^1 \frac x{3x^4 + 5x^2 +1}dx+\frac12$

Since $\frac x{3x^4 + 5x^2 +1}$ is continuous on $[0,1]$ (the denominator has no real zeroes), it is bounded there and so is the proper integral $\int_0^1 \frac x{3x^4 + 5x^2 +1}dx$.

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