Verifying Integration of ##\int_0^1 x^m \ln x \, \mathrm{d}x##

[murshid_islam]

TL;DR

Is my Integration ok?

I'm trying to compute $\int_0^1 x^m \ln x \, \mathrm{d}x$. I'm wondering if the bit about the application of L'Hopital's rule was ok. Can anyone check?

Letting $u = \ln x$ and $\mathrm{d}v = x^m$, we have $\mathrm{d}u = \frac{1}{x}\mathrm{d}x$ and $v = \frac{x^{m+1}}{m+1}$

$\int_0^1 x^m \ln(x) \, \mathrm{d}x$

$= \left. \frac{1}{m+1} x^{m+1} \ln(x) \right|_{0}^{1} - \int_{0}^{1} \frac{x^m}{m+1} \, \mathrm{d}x$

$= \frac{1}{m+1} \left(0 - \lim_{x \to 0} x^{m+1} \ln(x) \right) - \left. \frac{x^{m+1}}{(m+1)^2} \right|_{0}^{1}$

$= \frac{1}{m+1} \left(\lim_{x \to 0} x^{m+1} \ln\left(\frac{1}{x}\right) \right) - \frac{1}{(m+1)^2}$

$= \frac{1}{m+1} \left(\lim_{x \to 0}\frac{\ln(1/x)}{1/x^{m+1}} \right) - \frac{1}{(m+1)^2}$

$= \frac{1}{m+1} \left(\lim_{x \to 0}\frac{x (-1/x^2)}{-(m+1)x^{-m-2}} \right) - \frac{1}{(m+1)^2}$

$= \frac{1}{m+1} \left(\lim_{x \to 0}\frac{x^{m+1}}{m+1} \right) - \frac{1}{(m+1)^2}$

$= 0 - \frac{1}{(m+1)^2}$

$= \frac{-1}{(m+1)^2}$

 

[anuttarasammyak]

It seems OK. For verification, say m=0 the graphs of y=log x and y=e^x are symmetric wrt line y=x, the integration
\int_{-\infty}^0 e^x dx = 1
equals with your result putting minus sign.