Determine whether the improper integral $\int_{0}^{\infty}\frac{x^3}{1+x^4}\,dx$ exists

Question

While doing an exercise I need to prove that $\frac{x^3}{1+x^4}$ is integrable.

So I have to see if $\int_{0}^{\infty} |\frac{x^3}{1+x^4}| dx < \infty$. I tried to divide it in two integrals but I don't know how to continue...

$\int_{0}^{\infty} |\frac{x^3}{1+x^4}| dx = \int_{0}^{\infty} \frac{x^3}{1+x^4} dx =\int_{0}^{1} \frac{x^3}{1+x^4} dx + \int_{1}^{\infty} \frac{x^3}{1+x^4} dx \leq \frac{1}{4}\int_{0}^{1} \frac{4x^3}{1+x^4} dx + \int_{1}^{\infty} \frac{x^3}{1+x^4} dx$

So the first integral is the neperian logarithm and it is finite but the second one?

I don't know if that's the best way to do this...

Could anyone help me please?

Answer 1

Note that $\frac{x^3}{1+x^4}=\frac1x-\frac1{x(1+x^4)}$. Now $\int_1^\infty\frac1xdx$ is divergent, while $\int_1^\infty\frac1{x(1+x^4)}dx<\int_1^\infty\frac1{x^5}dx$ is convergent, so $\int_1^\infty\frac{x^3}{1+x^4}dx$ cannot be convergent.

Answer 2

We have $$\frac d{dx} \left(\frac{1}{4}\ln(1+x^4)\right) = \frac{x^3}{1+x^4}$$ and therefore $$\int_0^\infty \frac{x^3}{1+x^4}\,dx = \left.\frac{1}{4}\ln(1+x^4)\right|_0^\infty = \frac 14\lim_{x\to\infty}\ln(1+x^4) = \infty$$ so your integral doesn't converge.

Answer 3

We have $$\frac{x^3}{1+x^4} \ge \frac{1}{2x}$$ for $x\ge1$ so the integral on $[1,\infty)$ diverges.