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

In summary, the conversation discusses the computation of ##\int_0^1 x^m \ln x \, \mathrm{d}x## using u-substitution and integration by parts, and concludes that the result is ##\frac{-1}{(m+1)^2}##. The application of L'Hopital's rule to verify the result is also deemed satisfactory.
  • #1
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TL;DR Summary
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} ##
 
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  • #2
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
[tex]\int_{-\infty}^0 e^x dx = 1[/tex]
equals with your result putting minus sign.
 
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