mark
Write down one or two complete sentences explaining why
\int_1^{\infty} \frac{x}{x^4+1} \, dx
converges.
You may assume that
\int_1^{\infty} \frac{1}{x^p} dx
converges whenever p>1.
An archived instance of a Calc II forum
Prove that an improper integral converges
chowell1
\int^\infty_1\frac{x}{x^4}dx
converges to 0 because the denominator grows faster, therefore
\int^\infty_1\frac{x}{x^4+1}
also converges towards 0.
rstahles
We know the integral:
\int_1^{\infty} \frac{x}{x^p} \, dx
converges as p>1. This, in return, must mean that the integral:
\int_1^{\infty} \frac{1}{x^3} \, dx
converges. So:
\int_1^{\infty} \frac{x}{x^4+1} \, dx
must also converge.
mark
@chowell1 and @rstahles
I think you need to write down an inequality that compares the integrand x/(x^4+1) with 1/x^3. Of course, you know what
\int_1^{\infty}\frac{1}{x^3}dx
does.
knguyen3
\int_1^\infty\frac{x}{x^4+1}dx converges by comparison with \int_1^\infty\frac{1}{x^3}dx, since 0<=\frac{x}{x^4+1}<=\frac{x}{x^4}=\frac{1}{x^3} and \int_1^\infty\frac{1}{x^3}dx converges.
agooch
We know the integral:
\int_1^{\infty} \frac{x}{x^4+1} \, dx
converges by comparison with
\int_1^{\infty} \frac{x}{x^p} \, dx
since
0\leq \frac{x}{x^4+1} \leq \frac{1}{x^p}
and
\int_1^{\infty} \frac{x}{x^p} \, dx
converges when p>1.