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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?

Dave L. Renfro
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User160
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    As $x^{3}, x^{4} \ge 0$ on $x \in [0, \infty)$, the integrand is strictly positive and hence $$\int_{0}^{\infty} \frac{x^{3}}{1+x^{4}} dx = \frac{1}{4} \int_{0}^{\infty} \frac{4x^{3}}{1+x^{4}} dx = \frac{1}{4} \int_{0}^{\infty} \frac{(1+x^{4})'}{1+x^{4}} dx = \frac{1}{4} \ln(1+x^{4}) \bigg\lvert_{0}^{\infty}$$ – Matthew Cassell Jan 02 '22 at 15:55
  • You might also want to see this – FShrike Jan 02 '22 at 15:56
  • so it is not integrable? @mattos – User160 Jan 02 '22 at 15:58
  • No it is not. In the link I provide, note also that you find the integral to feature $\csc(\pi)$ which is divergent. – FShrike Jan 02 '22 at 15:59
  • @User160 No, it isn't. Also, I meant to say 'the integrand is strictly non-negative' in my previous comment. – Matthew Cassell Jan 02 '22 at 16:01
  • Notice that $\frac{x^3}{1+x^4}\sim \frac{1}{x}$ when $x\to \infty $... so no chance to be integrable. – Surb Jan 02 '22 at 16:05
  • And is there any choice to bound $|\frac{x^3 \cos(xt)}{1+x^4}|$ by an integrable function $g(x)$ ? – User160 Jan 02 '22 at 16:07

3 Answers3

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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.

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

jjagmath
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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.

Ennar
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