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

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 '18 at 6:16
• you can try $$\frac{x}{3x^4+5x^2+1}\leq \frac{1}{x^3},\,\,\,x\geq 0$$ – Lau Dec 7 '18 at 6:19
• @Lau but I have to prove this fact first ....... is it proved here on math stack ? – hopefully Dec 7 '18 at 6:40
• 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 '18 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 '18 at 6:55

## 1 Answer

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

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

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

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 $$\int_0^1 \frac x{3x^4 + 5x^2 +1}dx$$.