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I was reading the book (Almost) Impossible Integrals, Sums and Series when I came across this problem (1.7) $$\int_0^1\dfrac{\log^2(1+x)}{x}dx=\dfrac{1}{4}\zeta(3)$$ Earlier in the book I had derived the results $$\int_0^1x^m\log^nxdx=(-1)^n\dfrac{n!}{(m+1)^{n+1}}\tag{1}$$ $$\int_0^ax^m\log^nxdx=a^{m+1}\sum_{k=0}^n(-1)^k\binom{n}{k}\dfrac{k!}{(m+1)^{k+1}}\log^{n-k}a\tag{2}$$ My Attempt-
Taking $u=x+1$, $$I=\int_0^1\dfrac{\log^2(1+x)}{x}dx=\int_1^2\dfrac{\log^2u}{u-1}du=-\int_1^2\sum_{n=1}^\infty u^{n-1}\log^2udu=-\sum_{n=1}^\infty\int_1^2u^{n-1}\log^2udu$$ Now, $$\int_1^2u^{n-1}\log^2udu=\int_0^2u^{n-1}\log^2udu-\int_0^1u^{n-1}\log^2udu$$ By $(1)$, $$\int_0^1u^{n-1}\log^2udu=\dfrac{2}{n^3}$$ By $(2)$, $$\int_0^2u^{n-1}\log^2udu=2^n\sum_{k=0}^2(-1)^k\binom{2}{k}\dfrac{k!}{n^{k+1}}\log^{2-k}2=2^n\left[\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right]$$ So, $$\int_1^2u^{n-1}\log^2udu=2^n\left[\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right]-\dfrac{2}{n^3}$$ Hence, $$I=-\sum_{n=1}^\infty\left[2^n\left(\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right)-\dfrac{2}{n^3}\right]=-\sum_{n=1}^\infty2^n\left(\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right)+2\zeta(3)$$ According to the final result, the first term should give $\dfrac{7}{4}\zeta(3)$, which I don't think is possible.

I don't need alternate methods, those are already there in the book. Can anyone spot the mistake or find out how the expression I found can be simplified to get the result?

Souparna
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3 Answers3

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The summation does not seem to converge.

Consider the partial sum $$S_p=\sum_{n=1}^p2^n\left(\dfrac{\log^2(2)}{n}-\dfrac{2\log(2)}{n^2}+\dfrac{2}{n^3}\right)$$ which has an explicit expression in terms of the Lerch transcendent function $$\sum_{n=1}^p \frac{2^n}{n^k}=\text{Li}_k(2)-2^{p+1}\, \Phi (2,k,p+1)$$ makes $S_p$ to be very close to an exponential function.

Plot $\log(S_p)$ as a function of $p$.

Claude Leibovici
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The mistake is

$$ \int_{1}^{2}\frac{\log^{2}u}{u-1}du=-\int_{1}^{2}\sum_{n=1}^{\infty}u^{n-1}\log^{2}udu $$

because the equality

$$ \frac{1}{u-1}=-\sum_{n=1}^{\infty}u^{n-1} $$

only converges when $|u|<1$. From the integral, $u$ goes from $1$ to $2$, which is outside the interval of convergence of the series.

There is also a divergence issue in the line

$$ I=-\sum_{n=1}^\infty\left[2^n\left(\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right)-\dfrac{2}{n^3}\right]=-\sum_{n=1}^\infty2^n\left(\dfrac{\log^22}{n}-\dfrac{2\log2}{n^2}+\dfrac{2}{n^3}\right)+2\zeta(3) $$

and one could confirm this by using the Divergence Test for series. Since $2^n$ grows faster than $n$, $n^2$, and $n^3$, the limit of the sequence terms equals $\infty$ which makes the series diverge.

Just for fun, I decided to numerically check the other lines, which you can see here in this Desmos link and fortunately, they seem correct. Cheers :)

Accelerator
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$$\sum_{k=0}^\infty\frac{H_kx^{k+1}}{k+1}=\frac{\log^2(1-x)}2\implies2\sum_{k=0}^\infty\frac{H_k(-1)^{k+1}x^k}{k+1}=\frac{\log^2(1+x)}x$$$$\int_0^1\frac{\log^2(1+x)}xdx=2\sum_{k=0}^\infty\frac{(-1)^{k+1}H_k}{(k+1)^2}$$The sum is $\frac{\zeta(3)}8$

Kamal Saleh
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