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I want to prove the Basel Problem and I managed to turn it into an integral which I can't solve. I am interested to know how it can be evaluated without using Taylor series expansions, perhaps with the use of special functions?

$$\int_{0}^{1} \! \frac{-\ln(1-t)}{t} \, \mathrm{d}t.$$

I tried using substitutions that can relate it to the Gamma or Beta function but I always hit a seemingly insurmountable roadblock. I will highly appreciate a detailed answer. Thank you!

Sewer Keeper
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Math_Enthusiast
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    This is by definition $\left[\operatorname{Li}_2(x)\right]_0^1=\frac{\pi^2}6$. Is this the answer you are looking for? – user Jun 16 '20 at 06:15
  • This is the first time that I encountered that function. Is there an explicit formula for solving particular values of it? – Math_Enthusiast Jun 16 '20 at 06:23
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    https://en.wikipedia.org/wiki/Polylogarithm – user Jun 16 '20 at 06:26
  • Basel problem can be proved using elementary integration, but not a simple way like this. It seems that we can't avoid circular argument by using dilogarithm. – Infiniticism Jun 16 '20 at 06:56
  • Can you present how? – Math_Enthusiast Jun 16 '20 at 09:51
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    Have you seen these threads for example: https://math.stackexchange.com/q/3403819/321264, https://math.stackexchange.com/q/2493083/321264, https://math.stackexchange.com/q/2626072/321264, https://math.stackexchange.com/q/1638640/321264? – StubbornAtom Jun 16 '20 at 15:13

2 Answers2

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A similar problem and solution can be found here. Proposed by Khalef and solved by Sujee.

Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have

$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$

$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$

$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$

$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$

$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$

Remark:

This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$

Ali Shadhar
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{-\ln\pars{1 - t} \over t}\,\dd t & = -\int_{0}^{1}{\ln\pars{t} \over 1 - t}\,\dd t = \left.\partiald{}{\nu}\int_{0}^{1}{1 - t^{\nu} \over 1 - t}\,\dd t \,\right\vert_{\ \nu\ =\ 0} \\[5mm] &= \left.\partiald{\Psi\pars{\nu + 1}}{\nu}\,\right\vert_{\ \nu\ =\ 0} = \Psi\, '\pars{1} = \sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{2}} \\[5mm] & = \sum_{n = 1}^{\infty}{1 \over n^{2}} = \bbx{\pi^{2} \over 6\phantom{^{2}}} \\ & \end{align}

Felix Marin
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