# Test for convergence of improper integral

I need to test if, integrals below, either converge or diverge:

1) $\displaystyle\int_{0}^{1}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$

2) $\displaystyle\int_{1}^{\infty}\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}dx$

I tried comparing with $\displaystyle\int_{0}^{1}\frac{1}{(1+x)\ln^3(1+x)}dx$, $\displaystyle\int_{0}^{1}\frac{\sqrt{x}}{(1+x)}dx$ but ended up with nothing.

Do you have any suggestions? Thanks!

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Near $0$, $\log(1+x)=x(1+O(x))$ so $$\frac{\sqrt{x}}{(1+x)\log^3(1+x)}=x^{-5/2}(1+O(x))$$ and because $-5/2\le-1$, the integral in 1) does not converge.

$$\left(\frac{\sqrt{x}}{\log^3(1+x)}\right)^{1/3}=\frac{x^{1/6}}{\log(1+x)}$$ By L'Hospital, \begin{align} \lim_{x\to\infty}\frac{x^{1/6}}{\log(1+x)} &=\lim_{x\to\infty}\frac{\frac16x^{-5/6}}{1/(1+x)}\\ &=\lim_{x\to\infty}\frac16\left(x^{-5/6}+x^{1/6}\right)\\[12pt] &=\infty \end{align} Therefore, $$\lim_{x\to\infty}\frac{\sqrt{x}}{\log^3(1+x)}=\infty$$ Thus, the integral in 2) does not converge by comparison to $\int_1^\infty\frac1{1+x}\,\mathrm{d}x$

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Both integrals diverge. The first diverges because the integrand behaves as $x^{-5/2}$ as $x \to 0$, which is a non-integrable singularity. You can see that the second diverges upon substituting $x=e^{u}-1$ - the integrand behaves as $e^{u/2}/u^3$ as $u \to \infty$.

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Very nice substitution in that second integral. Great answer, +1 – Alex Wertheim Jun 3 '13 at 18:28
@AWertheim: thanks! – Ron Gordon Jun 3 '13 at 18:28

1) The integral diverges since as $x\sim 0$

$$\frac{\sqrt{x}}{(1+x)\ln^3(1+x)}\sim \frac{\sqrt{x}}{(1)(x^3) }\sim \frac{1}{x^{5/2}}.$$

Note:

$$\ln(1+x) = x - \frac{x^2}{2} + \dots.$$

2) For the second integral, just replace $x \leftrightarrow 1/x$, so the integrand will behave as $x\sim 0$ as

$$\frac{\sqrt{1/x}}{(1+1/x)\ln^3(1+1/x)}= \frac{\sqrt{x}}{(1+x)(\ln^3(1+1/x))}\sim \frac{\sqrt{x}}{(x)(\ln^3(1/x))} = -\frac{1}{\sqrt{x}\ln^3(x)}.$$

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