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Test the convergence $$\int_0^1 \frac{x^n}{1+x}dx$$

I have used comparison test for improper integrals..by comparing with $1/(1+x)$... so I found it convergent .. But the solution set says that it is convergent if $n> -1$.

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We have that for $n\ge 0$ the integral is a proper integral, then consider $n<0$ and by $m=-n>0$ we have

$$\int_0^1 \frac{1}{x^m+x^{m+1}}dx$$

and since as $x\to 0^+$

$$\frac{1}{x^m+x^{m+1}} \sim \frac{1}{x^m}$$

the integral converges for $m<1$ that is $n>-1$.

As an alternative by $y=\frac1x$ we have

$$\int_0^1 \frac{x^n}{1+x}dx=\int_1^\infty \frac{1}{y^{n+1}+y^{n+2}}dx$$

and since as $x\to \infty$

$$ \frac{1}{y^{n+1}+y^{n+2}}\sim \frac{1}{y^{n+2}}$$

the integral converges for $n+2>1$ that is $n>-1$.

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Let $a_n = \displaystyle \int_{0}^1 \dfrac{x^n}{1+x}dx$. Since $\dfrac{x^n}{1+x} = x^{n-1}\left(1-\dfrac{1}{1+x}\right)=x^{n-1}-\dfrac{x^{n-1}}{1+x}$, taking the integral $\displaystyle \int_{0}^1$ both sides give: $a_n = \dfrac{1}{n}- a_{n-1}=\dfrac{1}{n}-\dfrac{1}{n-1}+a_{n-2}= \dfrac{1}{n}-\dfrac{1}{n-1}+\dfrac{1}{n-2}-a_{n-3}=...=-\ln 2+ 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\cdots + \dfrac{1}{n-2}-\dfrac{1}{n-1}+\dfrac{1}{n}$. $a_n$ converges when considered as a partial sum of an alternating harmonic series.

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Since $x^n$ is continuous on $[0,1]$ for $n\ge 0,$ the integral converges for $n\ge 0.$

For $n<0,$ $x^n$ blows up at $0.$ So we need to consider the integral over $[a,1]$ for small $a>0.$ Notice that for $x\in (0,1],$

$$\frac{x^n}{2}\le \frac{x^n}{1+x} \le x^n.$$

It follows that the integral of interest converges iff $\int_0^1 x^n\,dx $ converges. Things are easy now, so I'll stop here. Ask questions if you like.

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