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.
Best Answer
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 $0$