My Attempt
Consider the integral
\begin{align*}
I_1 &= \int_{0}^{1} \log\left(1-u^{\frac{1}{3}}\right) \, d u.
\end{align*}
Integrating by parts, letting $a=\log\left(1-u^{\frac{1}{3}}\right)$ and $\, d b = \, d u$, we obtain
\begin{align*}
d a = -\frac{1}{3u^{2/3}(1-u^{1/3})} \, d u \quad \text{and} \quad b = u,
\end{align*}
so that
\begin{align}
\nonumber I_1 &= \left[ u\log\left(1-u^{\frac{1}{3}}\right) \right]_{0}^{1} – \frac{1}{3} \int_{0}^{1} \frac{u^{1/3}}{1-u^{1/3}} \, d u\\
&= \left[ u\log\left(1-u^{\frac{1}{3}}\right) \right]_{0}^{1} – \frac{1}{3} I_2. \tag{1}
\end{align}
Note that
\begin{align*}
I_2 &= \int_{0}^{1} \frac{u^{1/3}}{1-u^{1/3}} \, d u.
\end{align*}
Letting $v=1-u^{\frac{1}{3}}$, we have
\begin{align*}
u=(1-v)^{3} \quad \text{and} \quad \, d u = -3(1-v)^2 \, d v.
\end{align*}
Hence,
\begin{align*}
I_2 &= \int_{1}^{0} \frac{1-v}{v} \left[ -3(1-v)^2 \right] \, d v\\
&= 3 \int_{0}^{1} \frac{(1-v)^3}{v} \, d v\\
&= 3 \int_{0}^{1} \frac{1-3v+3v^2-v^3}{v} \, d v\\
&= 3 \int_{0}^{1} \left( \frac{1}{v} – 3 + 3v – v^2 \right) \, d v\\
&= 3 \left[ \log(v) – 3v + \frac{3}{2}v^2 – \frac{1}{3}v^3 \right]_{0}^{1}.
\end{align*}
Substituting into equation (1), we have
\begin{align*}
I_1 &= \left[ u\log\left(1-u^{\frac{1}{3}}\right) \right]_{0}^{1} – \frac{1}{3} \left\{ 3 \left[ \log(v) – 3v + \frac{3}{2}v^2 – \frac{1}{3}v^3 \right]_{0}^{1} \right\}\\
&= \left[ u\log\left(1-u^{\frac{1}{3}}\right) \right]_{0}^{1} – \left[ \log(v) – 3v + \frac{3}{2}v^2 – \frac{1}{3}v^3 \right]_{0}^{1}\\
&= \left[ \lim_{u\to 1} u\log(1-u^{\frac{1}{3}}) – 0 \right] + \left[ -\frac{11}{6} – \left( \lim_{v\to 0} \log(v) + 0 \right) \right]\\
&= -\frac{11}{6} + \left[ \lim_{u\to 1} u\log(1-u^{\frac{1}{3}}) – \lim_{v\to 0} \log(v) \right]\\
&= -\frac{11}{6} + \lim_{u\to 1} \left[ u\log(1-u^{\frac{1}{3}}) – \log(1-u) \right]\\
&= -\frac{11}{6} + 0\\
&= -\frac{11}{6}.
\end{align*}
Question
I am not sure whether I am integrating correctly or not. I can't seem to prove (or further elaborate) the last limit, which is
\begin{align*}
\lim_{u\to 1} \left[ u\log(1-u^{\frac{1}{3}}) – \log(1-u) \right],
\end{align*}
to be zero. Can anyone help me with this problem? Any help/feedback is much appreciated. Thank you!
Best Answer
When you integrate by parts, the two terms separately diverge as $u \nearrow 1$, owing to the $\log$ terms, but the diverging parts turn out to cancel.
We can avoid the issues of divergence when integrating by parts by choosing an antiderivative of $3 (v - 1)^2$ for which the two terms cancel separately. We can see that it's the constant term of the antiderivative that gives rise to the diverging parts, so we choose the antiderivative with zero constant term, namely, $v^3 - 3 v^2 + 3 v$. Then, \begin{multline} \int_0^1 3 (v - 1)^2 \log v \,dv = \left.\left(v^3 - 3 v^2 + 3 v\right) \log v\right\vert_0^1 - \int_0^1 \left(v^3 - 3 v^2 + 3 v\right) \cdot \frac1v \,dv \\ = -{\int_0^1} \left(v^2 - 3 v + 3\right) dv = -\frac{11}6 . \end{multline}