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How do I prove that $\int^{+\infty}_0\frac{x}{(1+x^7)^{1/3}}dx$ converges?

I wanted to do a comparison test with the integral $\int^{+\infty}_0\frac{x}{(1+x)^{1/3}}dx$, but it diverges. I can't seem to find a function which I would compare this to. If I could get a hint that would be great.

Neil hawking
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1 Answers1

3

HINT

Notice that

\begin{align*} \int_{0}^{\infty}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x & = \int_{0}^{1}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x + \int_{1}^{\infty}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x\\\\ & \leq \int_{0}^{1}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x + \int_{1}^{\infty}\frac{x}{x^{7/3}}\mathrm{d}x\\\\ & = \int_{0}^{1}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x + \int_{1}^{\infty}\frac{\mathrm{d}x}{x^{4/3}} \end{align*}

Can you take it from here?

user0102
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  • Thank you! How do I show that $\int_{0}^{1}\frac{x}{(1+x^{7})^{1/3}}\mathrm{d}x$ converges? I know it's convergent but not sure how exactly –  Jul 11 '21 at 20:09
  • @use80085 the integrand is continuous over $[0,1]$, so the integral is an ordinary Riemann integral. No need to establish convergence. – Alann Rosas Jul 11 '21 at 20:12
  • oh okay in know there's a rule for this, forgot that –  Jul 11 '21 at 20:12