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$\newcommand{LogI}{\operatorname{Li}}$ I know that $\LogI_{2}(-1)=-\frac{\pi^2}{12}$, but I have never seen a proof of this result without using a functional identity of the Dilogarithm or a series expansion. I would like to know how calculate $\LogI_{2}(-1)$ using the integral representation of the Dilogarithm function: $\LogI_{2}(z)=\int_{z}^{0}\frac{\ln{(1-t)}}{t}dt$ or $\LogI_{2}(z)=-\int_{0}^{1}\frac{\ln{(1-zt)}}{t}dt$. This amounts to calculating either of the following integrals: $$\LogI_{2}(-1)=\int_{-1}^{0}\frac{\ln{(1-t)}}{t}dt$$ $$\LogI_{2}(-1)=-\int_{0}^{1}\frac{\ln{(1+t)}}{t}dt$$ I am interested in seeing how to evaluate these integrals without using series expansion or functional identities of the Dilogarithm. Methods involving the use of Complex Analysis are welcome with some explanation.

Simply Beautiful Art
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Biggs
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    Apply the residue theorem to $\lim_{N \to \infty} \int_{|z|=N+1/2}\frac{z^{-2}}{e^{2i \pi z}-1}dz= 0$ – reuns Oct 27 '17 at 23:52
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    https://math.stackexchange.com/questions/890872/the-other-ways-to-calculate-int-01-frac-ln1-x2xdx?rq=1 1st answer – DanielC Oct 27 '17 at 23:58
  • @reuns I like your approach, but my knowledge of complex variables is extremely lacking. I understand that the residue theorem will yield $2i\pi$ times the residue at the pole $z=0$ which is $\frac{i\pi}{6}$, but that value is off by a factor of $\frac{1}{4}$. If you could explain your choice of contour and how it relates to one of the given integral representations I would be very grateful. – Biggs Oct 28 '17 at 01:37
  • @reuns You have now utterly confused me. Aren't the odd Bernoulli numbers equal to $0$? WA gives this for the residue: https://www.wolframalpha.com/input/?i=residue+of+(z%5E-2)%2F(e%5E(2i(pi)z)-1)+at+z+%3D+0 Am I missing something here? – Biggs Oct 28 '17 at 03:34
  • The residue at $ z=0$ is $2i \pi \frac{B_2}{2}$ (the Bernouilli number). Note how $\frac{z}{e^z−1}$ appears when integrating by parts followed by $t=e^{-z}$ in your integral representation. – reuns Oct 28 '17 at 03:39
  • This value can be calculate if you know the value of $\text{Li}2(1)$ see below. Therefore your question is to evaluate $\displaystyle \sum{n=1}^{\infty}\frac{1}{n^2}$ read https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2 – FDP Oct 28 '17 at 06:46

1 Answers1

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Enforce the substitution $x\to x^2$. Then,

$$\begin{align} \int_0^1\frac{\log(1-x)}{x}\,dx&=2\int_0^1\frac{\log(1-x^2)}{x}\,dx\\\\ &=2\int_0^1 \frac{\log(1-x)}{x}\,dx+2\int_0^1\frac{\log(1+x)}{x}\,dx\tag1 \end{align}$$

Hence, we see from $(1)$ that

$$\begin{align} -\int_0^1\frac{\log(1+x)}{x}\,dx&=\frac12\int_0^1\frac{\log(1-x)}{x}\,dx\tag2\\\\ &=-\frac12 \int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy\tag3 \end{align}$$

Next, in THIS ANSWER, I evaluated the integral in $(3)$ by using the transformation of coordinates $x=s+t$ and $y=s-t$ and using elementary integral analysis. The result as given in that answer was

$$\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy=\frac{\pi^2}{6}\tag 4$$

Using $(4)$ in $(3)$ reveals

$$-\int_0^1\frac{\log(1+x)}{x}\,dx=-\frac{\pi^2}{12}$$

as expected! And we are done!

metamorphy
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Mark Viola
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  • @reuns I presume we know the value of the dilogarithm at $1$. This can be calculated without appealing to series analysis. Rather, one can transform the single integral to a double integral and then making a coordinate transformation in which the double integral can be evaluated (see Tom Apostle's Method). – Mark Viola Oct 28 '17 at 01:15
  • @reuns I've edited and provided a link to an answer I posted HERE in which I evaluated the double integral $\int_0^1\int_0^1 \frac{1}{1-xy},dx,dy$ by transforming coordinates with $x=s+t$ and $y=s-t$ then applying straightforward elementary integral analysis to carry out the double integral. – Mark Viola Oct 28 '17 at 16:01