For which $p$ does $\int_0^1 x^p(x+1)\log x \ dx $ converge?

Question

My question is this: for which value of $p$ does this improper integral converge?

$$\int_0^1 x^p(x+1)\log x \ dx $$

I know from a previous exercise that $$\int_0^1 x^p\log x \ dx $$ converges for $p > -1$.

So for this integral

$$\int_0^1 x^p(x+1)\log x \ dx $$

I found $$x^p(x+1)\log x = x^{p+1}\log x \ + x^p\log x$$

and $$x^{p+1}\log x \ + x^p\log x \le 2x^{p+1}\log x$$

and $$2x^{p+1}\log x \asymp x^{p+1}\log x$$

and I assume from the previous exercise that the integral must converge for $p > -2$ and since the polynomial beats the log with $x$ from $0$ to $1$ for values of $p > -2$.

Is my procedure right?

Answer 1

Note that for any $x\in(0,1)$, $x^{p}\ln x<0$ and $1<x+1<2$, so $x^{p}\ln x>x^{p}(x+1)\ln x>2x^{p}\ln x$. Therefore, for any $\delta\in(0,1)$, we have that $\int_{\delta}^{1}x^{p}\ln x\,dx\geq\int_{\delta}^{1}x^{p}(x+1)\ln x\,dx\geq2\int_{\delta}^{1}x^{p}\ln x\,dx$. (Actually we have strict inequality. However, we do not need this fact.)

Now, we can show that $\lim_{\delta\rightarrow0+}\int_{\delta}^{1}x^{p}\ln x\,dx$ exists iff $\lim_{\delta\rightarrow0+}\int_{\delta}^{1}x^{p}(x+1)\ln x\,dx$ exists.

Answer 2

Using Feynman's Trick \begin{align}I(p)&=\int _0^1x^p(x+1)\,\mathrm dx\\&=\int_0^1(x^{p+1}+x^p)\,\mathrm dx\\&=\dfrac{x^{p+2}}{p+2}+\dfrac{x^{p+1}}{p+1}\Bigg|_0^1\\&=\dfrac1{p+2}+\dfrac1{p+1}\\I'(p)&=\int_0^1x^p(x+1)\ln x\,\mathrm dx\\&=\dfrac{\mathrm d}{\mathrm dp}\left(\dfrac1{p+2}+\dfrac1{p+1}\right)\\&=-\dfrac1{(p+2)^2}-\dfrac1{(p+1)^2}\end{align}

$I'(p)$ is only valid iff $p>-1$