# Finding convergence of integral $\int_0^1 \frac{x^n}{1+x}dx$

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 $$n> -1$$.

## 3 Answers

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 $$n>-1$$.

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.

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.