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Convergence of Improper Integral $$\int^{\infty}_{0}\frac{2x}{(x^2+1)^3}dx$$ using comparison test

What i try

Put $x^2+1=t$ and $2xdx=dt$ and changing limits

So integration is $$\int^{\infty}_{1}\frac{1}{t^3}dt=-\frac{1}{2t^2}\bigg|^{\infty}_{1}=\frac{1}{2}.$$

So the integration is converges

But i did not understand How Do i find convèrgence of that Using Comparasion Test.

Help me please. Thanks

jacky
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    "Comparison test" means: show that the integrand is bounded by an integrand known to converge. For example, show that $0\leq 2x/(x^2+1)^3\leq 2x/x^5$ (and the integral over the latter converges on $[1, \infty]$ – retzler May 26 '20 at 01:54
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    hit: $1/x\geq1/x^3$ for $x\geq1$, try to use this to your problem – Simple May 26 '20 at 01:56
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    For $0 ≤ x ≤ 1$, you can use the fact that $\frac{2x}{(x^2+1)^3} ≤ \frac{2x}{1^3}$ since the denominator is smaller on the RHS. – Toby Mak May 26 '20 at 01:57

2 Answers2

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Let $I$ be the improper integral in question, then $I \le \displaystyle \int_{0}^\infty\dfrac{dx}{(1+x^2)^2} \le C + \displaystyle \int_{1}^\infty\dfrac{dx}{x^2}=C+1$ where $C$ is a constant.

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The idea of the solution below is to start the integral at some value of $x$ for which the choice of comparison is easy.

A lower limit of $x=1$ works nicely . . . \begin{align*} &\int_1^\infty \!\frac{2x}{(x^2+1)^3}\,dx\\[4pt] < \;& \int_1^\infty \!\frac{2x}{(x^2)^3}\,dx\\[4pt] =\;& \int_1^\infty \!\frac{2x}{x^6}\,dx\\[4pt] =\;& 2\int_1^\infty \!\frac{1}{x^5}\,dx\\[4pt] =\;& \frac{1}{2}\\[4pt] \end{align*} Thus ${\displaystyle{\int_1^\infty \!\frac{2x}{(x^2+1)^3}\,dx}}$ converges, and hence $$ \int_0^\infty \!\frac{2x}{(x^2+1)^3}\,dx = \int_0^1 \!\frac{2x}{(x^2+1)^3}\,dx + \int_1^\infty \!\frac{2x}{(x^2+1)^3}\,dx $$ also converges.

quasi
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