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Problem: Evaluate:

$$\dfrac{u^{2n+1}\ln(u)}{2n+1}\bigg|^{u=1}_{u=0}$$

This actually came up when I tried to solve the Integral $$\int_0^1 u^{2n}\ln(u) du$$ by using IBP. The problem is that I am unable to evaluate this at $u=0$.$$$$ Would somebody please be so kind as to help me evaluate $\dfrac{u^{2n+1}\ln(u)}{2n+1}\bigg|^{u=1}_{u=0}?$ I would be indeed very grateful for your assistance. Many thanks in advance!

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    $\begingroup$ What do you call this in "I am unable to evaluate this at $u=0$" ? The integral does not depend on $u$. $\endgroup$ – Yves Daoust Jun 23 '15 at 15:48
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The evaluation at $u=1$ is not a real issue, and since: $$ \lim_{u\to 0^+} u\log u = 0 $$ is easy to prove through a change of variable, for instance, neither the evaluation in the other endpoint is difficult.

In any case, $$ \int_{0}^{1} u^{2\alpha}\log(u)\,du = \frac{1}{2}\frac{d}{d\alpha}\int_{0}^{1}u^{2\alpha}\,du = \frac{1}{2}\frac{d}{d\alpha}\frac{1}{2\alpha+1}=-\frac{1}{(2\alpha+1)^2}.$$

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  • $\begingroup$ Sir, could you please show me how to evaluate the limit? Also, Sir, could you explain why we take a limit? I've never understood why we take a limit in such situations. $\endgroup$ – Ishan Jun 23 '15 at 15:49
  • $\begingroup$ @BetterWorld: We take a limit because the function $u\log u$ is not defined at $u=0$, but for the sake of integration that does not matter. In virtue of that limit, if we define $u\log u$ as zero in zero we have a continuous function, and everything works in the smooth way. At last, that limit is the same as: $$\lim_{t\to -\infty} t e^{t} = \lim_{r\to +\infty} -\frac{r}{e^r}$$ by substituting $u=e^t$ and $t=-r$. $\endgroup$ – Jack D'Aurizio Jun 23 '15 at 15:53
  • $\begingroup$ Sorry Sir, but I was unable to understand what you meant by :"but for the sake of integration that does not matter". Sir, nor could I understand the meaning of "as zero in zero ".$$$$Thanks a lot for showing me how to evaluate the Limit:) $\endgroup$ – Ishan Jun 23 '15 at 15:57
  • $\begingroup$ @BetterWorld: If $f$ is defined over $(a,b)$, $g$ is defined over $[a,b]$, they are both integrable and for every $c\in(a,b)$ we have $f(c)=g(c)$, then $\int_{a}^{b}f(x)\,dx = \int_{a}^{b}g(x)\,dx$. So we may re-define $f$ at the points $a$ and $b$, but the value of its integral is still the same. In our case, we have $f(u)=u\log u$ defined over $u\in(0,1]$. We re-define $f(0)$ as $\lim_{u\to 0^+}f(u)=0$ and the integral is still the same. $\endgroup$ – Jack D'Aurizio Jun 23 '15 at 16:11
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    $\begingroup$ You can also do $t \ln(t) = \frac{\ln(t)}{\frac{1}{t}}$, then L'Hopital gives $-\frac{\frac{1}{t}}{\frac{1}{t^2}}=-t$, which goes to zero. $\endgroup$ – Ian Jun 23 '15 at 16:11
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Using $\partial_{x} \, u^{ax} = \partial_{x} \, e^{ax \ln(u)} = a \ln(u) \, u^{ax}$ then \begin{align} I &= \int_{0}^{1} u^{a x} \, \ln(u) \, du = \frac{1}{a} \partial_{x} \, \int_{0}^{1} u^{ax} \, du \\ &= \frac{1}{a} \partial_{x} \left[ \frac{u^{ax+1}}{ax+1} \right]_{0}^{1} \\ &= \frac{1}{a} \partial_{x} \left( \frac{1}{ax+1} \right) \\ &= \frac{(-1)^{1}}{(ax+1)^{2}} \end{align} Now let $a=2$ to obtain \begin{align} \int_{0}^{1} u^{2n} \, \ln(u) \, du = - \frac{1}{(2n+1)^{2}}. \end{align} The process may be continued to obtain \begin{align} \int_{0}^{1} u^{2n} \, \ln^{m}(u) \, du = \frac{(-1)^{m} \, m!}{(2n+1)^{m+1}}. \end{align}

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