Find Values of p for which the integral converges Find values of p for which the integral
$\int_0^1{x^pln(x)dx}$
converges and calculate the value of the integral for these values of p.  
I got $\int{x^pln(x)dx} = \frac{x^{p+1}ln(x)}{p+1}-\frac{x^{p+1}}{(p+1)^{2}}$  already. Would like to know how to check if it converges.
 A: Using integration by parts
$$
\int{x^pln(x)dx}=\frac{x^{p+1}\ln{x}-\int{x^{p}dx}}{p+1}=\frac{(x^{1+p} (-1+(1+p) \ln{x}))}{(1+p)^2}
$$
The integral between $0$ and $1$ exists iff
$$
\lim_{x\rightarrow0}{(x^{1+p}\ln{x})}=\lim_{x\rightarrow0}{\frac{\ln{x}}{x^{-1-p}}}
$$
converges
If $p<-1$ then $\lim_{x\rightarrow0}{(x^{1+p})}=\infty$ and $\lim_{x\rightarrow0}{\ln{x}}=-\infty$, then its easy to see why their product cannot be finite.
Using L'Hôpital's rule, it is quite easy to show that if $p>-1$ then the limit converges to $0$
A: You can try as follows: for $\;p>-1\;$ 
$$\lim_{x\to0^+}x^{p+1}\left(\log x-\frac1{p+1}\right)=\lim_{x\to0^+}\frac{\log x-\frac1{p+1}}{\frac1{x^{p+1}}}\stackrel{\text{l'H}}=\lim_{x\to0^+}-\frac{x^{p+1}}{p+1}=0$$
For $\;p=-1\;$ :
$$\int_0^1\frac{\log x}xdx=\left.\frac12\log^2x\right|_0^1=\lim_{x\to0^+}-\frac12\log^2x=-\infty$$
For $\;p<-1\;$ :
$$\lim_{x\to0^+}\overbrace{x^{p+1}}^{\rightarrow\infty}\left(\overbrace{\log x}^{\rightarrow-\infty}-\frac1{p+1}\right)=-\infty$$
