Evaluate $\int_{0}^{1}{x\ln x dx}$.

Question

Evaluate $\displaystyle\int_{0}^{1} x \ln x\, dx$.

I used integration by parts with $u=\ln x$ and $dv=x\,dx$.

Then I got this: $\displaystyle\frac{x^2 \ln x}{2}- \int_{0}^{1}\frac{x}{2}\,dx$.

But the problem is $\ln x$ is undefined when $x=0$. So how do I solve this?

Answer 1

First of all, you've miscalculated $v\,du.$ Also, since you're dealing with an improper integral, you'll need to find $$\lim_{a\searrow 0}\left[1^2\ln(1)-a^2\ln(a)-\int_a^1v\,du\right].$$ L'Hospitals rule should do the trick.