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Today I discussed the following integral in the chat room

$$\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}$$ where $0\leq a, b\leq \pi$ and $k>0$.

Some users suggested me that I can use Frullani's theorem:

$$\int_0^\infty \frac{f(ax) - f(bx)}{x} = \big[f(0) - f(\infty)\big]\ln \left(\frac ab\right)$$ So I tried to work with that way. \begin{align} I&=\int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}\\ &=\int_0^\infty \frac{\ln \left( x^2+2kx\cos b+k^2\right)-\ln \left( x^2+2kx\cos a+k^2\right)}{x}\mathrm dx\tag{1}\\ &=\int_0^\infty \frac{\ln \left( 1+\dfrac{2k\cos b}{x}+\dfrac{k^2}{x^2}\right)-\ln \left( 1+\dfrac{2k\cos a}{x}+\dfrac{k^2}{x^2}\right)}{x}\mathrm dx\tag{2}\\ \end{align} The issue arose from $(1)$ because $f(\infty)$ diverges and the same issue arose from $(2)$ because $f(0)$ diverges. I then tried to use D.U.I.S. by differentiating w.r.t. $k$, but it seemed no hope because WolframAlpha gave me this horrible form. Any idea? Any help would be appreciated. Thanks in advance.

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6 Answers 6

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First note that by substituting $x\mapsto kx$, we get $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x\cos(b)+1}\right)\frac{\mathrm{d}x}{x} $$ Let $u=\frac{x+\cos(a)}{\sin(a)}$. Then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} &=-2\int_0^\infty\frac{\sin(a)}{x^2+2x\cos(a)+1}\,\mathrm{d}x\\ &=-2\int_0^\infty\frac{\sin(a)}{(x+\cos(a))^2+\sin^2(a)}\,\mathrm{d}x\\ &=-\frac2{\sin(a)}\int_0^\infty\frac1{\frac{(x+\cos(a))^2}{\sin^2(a)}+1}\,\mathrm{d}x\\ &=-2\int_{\cot(a)}^\infty\frac1{u^2+1}\,\mathrm{d}u\\[9pt] &=-2a \end{align} $$ Integrating in $a$ gives $$ \int_0^\infty\log\left(\frac{x^2+2x\cos(a)+1}{x^2+2x+1}\right)\frac{\mathrm{d}x}{x} =-a^2 $$ Therefore, by subtraction, $$ \int_0^\infty\log\left(\frac{x^2+2kx\cos(a)+k^2}{x^2+2kx\cos(b)+k^2}\right)\frac{\mathrm{d}x}{x} =b^2-a^2 $$

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    $\begingroup$ Hhmmm, your answer gives me a new idea to evaluate this integral. (+1) $\endgroup$
    – Venus
    Dec 15, 2014 at 17:14
  • $\begingroup$ Brilliant ! :-$)$ $\endgroup$
    – Lucian
    Dec 15, 2014 at 21:14
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\begin{align} \int_0^\infty \ln \left( \frac{x^2+2kx\cos b+k^2}{x^2+2kx\cos a+k^2}\right) \;\frac{\mathrm dx}{x}&=\int_0^\infty\frac{1}{x}\int_a^b \frac{\mathrm d}{\mathrm dy}\ln \left(x^2+2kx\cos y+k^2\right) \;\mathrm dy\,\mathrm dx\\[10pt] &=-\int_0^\infty\frac{1}{x}\int_a^b \frac{2kx\sin y}{x^2+2kx\cos y+k^2} \;\mathrm dy\,\mathrm dx\\[10pt] &=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{x^2+2kx\cos y+k^2} \;\mathrm dy\\[10pt] &=2k\int_b^a\sin y\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2-k^2\cos^2y} \;\mathrm dy\\[10pt] &=2k\int_b^a\sin y\underbrace{\int_0^\infty \frac{\mathrm dx}{\left(x+k\cos y\right)^2+k^2\sin^2y}}_{\large\color{blue}{(k\sin y)\, u\,=\,x+k\cos y}} \;\mathrm dy\\[10pt] &=2\int_b^a\int_{\cot y}^\infty \frac{\mathrm du}{u^2+1} \;\mathrm dy\\[10pt] &=2\int_b^a\left(\frac{\pi}{2}-\arctan\left(\cot y\right)\right)\;\mathrm dy\\[10pt] &=2\int_b^ay\;\mathrm dy\\[10pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large a^2-b^2}} \end{align}

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  • $\begingroup$ I see that you've encapsulated the differentiation in a double integral instead of integrating twice. $\endgroup$
    – robjohn
    Dec 15, 2014 at 17:49
  • $\begingroup$ @robjohn Yes. So, I didn't need to work twice. But my answer is inspired by your answer. Thanks. ^^ $\endgroup$
    – Venus
    Dec 15, 2014 at 17:51
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Here is a complex-analytic method: Notice that

$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = 2 \int_{0}^{\infty} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx. \tag{1} $$

Let $R$ be a positive large number. Then the function $z \mapsto \log(1+z)/z$ is analytic on $\Bbb{C} \setminus (-\infty, 1]$ with the standard branch cut and we get

\begin{align*} \int_{0}^{R} \frac{\log(1 + e^{ib}x)}{x} \, dx &=\int_{0}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \qquad (z = e^{ib}x) \\ &= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + \int_{R}^{Re^{ib}} \frac{\log(1 + z)}{z} \, dz \\ &= \int_{0}^{R} \frac{\log(1 + z)}{z} \, dz + i \int_{0}^{b} \log(1 + Re^{i\theta}) \, d\theta, \quad (z=Re^{i\theta}) \end{align*}

and likewise for the integral with $b$ replaced by $a$. This shows that

$$ \int_{0}^{R} \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx = i \int_{a}^{b} \log(1 + Re^{i\theta}) \, d\theta. $$

Multiplying by 2 and taking real part, we get

$$ 2 \int_{0}^{R} \Re \frac{\log(1 + e^{ib}x) - \log(1 + e^{ia}x)}{x} \, dx = - 2 \Im \int_{a}^{b} \log(R^{-1} + e^{i\theta}) \, d\theta. $$

(Here we exploited the fact that $\log R$ is real.) Taking $R \to \infty$, in view of the identity $\text{(1)}$ it follows that

$$ \int_{0}^{\infty} \log\left( \frac{x^{2} + 2x\cos b + 1}{x^{2} + 2x\cos a + 1} \right) \frac{dx}{x} = - \int_{a}^{b} 2\theta \, d\theta = a^{2} - b^{2}. $$

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  • $\begingroup$ Thank you very much for helping me again, (+1). Anyway, Hatsune Miku looks very kawaii with that hat ^^ $\endgroup$
    – Venus
    Dec 16, 2014 at 5:26
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    $\begingroup$ @Venus, thank you! I tried many ways until the hat fits her perfectly. Your compliment rewards my effort :) $\endgroup$ Dec 16, 2014 at 6:41
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Another approach is to use the Fourier series $$\sum_{k=1}^{\infty}\frac{x^{k} \cos(ka)}{k} = - \frac{1}{2} \log \left(x^{2} - 2 x \cos(a) +1 \right) \ , \ |x| <1 $$ which can be derived from the Maclaurin series of $\log(1-z)$ by replacing $z$ with $xe^{ia}$ and equating the real parts on both sides.

$$ \begin{align} \int_0^\infty\log\left(\frac{x^2+2kx\cos(b)+k^2}{x^2+2kx\cos(a)+k^2}\right)\frac{dx}{x} &=\int_0^\infty\log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \tag{1} \\ &= 2 \int_{0}^{1} \tag{2} \log\left(\frac{u^2+2u\cos(b)+1}{u^2+2u\cos(a)+1}\right)\frac{du}{u} \\ &= 4 \int_{0}^{1} \frac{1}{u} \sum_{k=1}^{\infty} (-u)^{k} \frac{\cos(ka) - \cos(kb)}{k} \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) -\cos(kb)}{k} \int_{0}^{1} u^{k-1} \ du \\ &= 4 \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka) - \cos(kb)}{k^{2}} \\ &= 4 \left(\frac{a^{2}}{4} - \frac{b^{2}}{4} \right) \tag{3} \\ &= a^{2}- b^{2} \end{align} $$

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$(1)$ Let $ \displaystyle u= \frac{x}{k}.$

$(2)$ Separate the integral into two integrals, namely one over the interval $(0,1)$ and one over the interval $(1, \infty)$, and replace $u$ with $\frac{1}{u}$ in the second integral.

$(3)$ By integrating the Fourier series $ \displaystyle \sum_{k=1}^{\infty} (-1)^{k} \frac{\sin(ka)}{k} = - \frac{a}{2} \ , \ |a| < \pi$, we get $$ \sum_{k=1}^{\infty} (-1)^{k} \frac{\cos(ka)}{k^{2}} = \frac{a^{2}}{4} - \frac{\pi^{2}}{12} \ , \ |a| < \pi .$$

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  • $\begingroup$ Very nice approach, (+1). Thanks for giving the nice Fouries series. $\endgroup$
    – Venus
    Dec 16, 2014 at 5:29
  • $\begingroup$ @Venus Thanks. I had accidentally put the parameter $a$ in the numerator and the parameter $b$ in the denominator. $\endgroup$ Dec 16, 2014 at 8:39
  • $\begingroup$ How do you prove that Fourier series? $\endgroup$
    – clathratus
    Feb 13, 2019 at 1:06
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}\ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}:\ {\large ?} \,,\qquad 0 <\ a\,,\ b\ <\pi\,,\quad k > 0.}$

\begin{align}&\overbrace{\color{#66f}{\large\int_{0}^{\infty} \ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}}} ^{\dsc{x \over k}\ \ds{\mapsto}\ \dsc{x}} \\[5mm]&=\int_{0}^{\infty} \ln\pars{x^{2} + 2x\cos\pars{b} + 1\over x^{2} + 2x\cos\pars{a} + 1} \,{\dd x \over x} =\lim_{R\ \to\ \infty}\bracks{\fermi\pars{a,R} - \fermi\pars{b,R}} \\[5mm]&\mbox{where}\ \begin{array}{|c|}\hline\\ \ \fermi\pars{\mu,R}\equiv \int_{0}^{R}\ln\pars{x} {2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x\,,\quad 0 < \mu < \pi\ \\ \\ \hline \end{array}\qquad\qquad\quad\pars{1} \end{align}

Note that $\ds{r_{-} \equiv \exp\pars{\bracks{\pi - \mu}\ic}}$ and $\ds{r_{+} \equiv \exp\pars{\bracks{\pi + \mu}\ic}}$ are the roots of $\ds{x^{2} + 2x\cos\pars{\mu} + 1 = 0}$ such that $\ds{\pars{~\mbox{with}\ R > 1~}}$:

\begin{align}&\dsc{\int_{0}^{R}\ln^{2}\pars{x} {x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x} \\[5mm]&=2\pi\ic\bracks{\half\,\ln^{2}\pars{r_{-}} + \half\,\ln^{2}\pars{r_{+}}} -\int_{R}^{0}\bracks{\ln\pars{x} + 2\pi\ic}^{2} {x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x \\[5mm]&=\pi\ic\bracks{-\pars{\pi - \mu}^{2} - \pars{\pi + \mu}^{2}} +\dsc{\int_{0}^{R}\ln^{2}\pars{x} {x + \cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x} \\[5mm]&+2\pi\ic\ \overbrace{\int_{0}^{R}\ln\pars{x} {2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x} ^{\dsc{\fermi\pars{\mu,R}}}\ +\ \pars{2\pi\ic}^{2}\int_{0}^{R} {x + \cos\pars{\mu} \over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x \\[5mm]&-{\mathfrak C}\pars{R,\mu} \end{align} where $\ds{\left.{\mathfrak C}\pars{R,\mu} \equiv\oint\ln^{2}\pars{z} {z + \cos\pars{\mu}\over z^{2} + 2z\cos\pars{\mu} + 1}\,\dd z\,\right\vert _{z\ \equiv\ R\expo{\ic\theta}\,,\ 0\ <\ \theta\ <\ 2\pi}}$

This expression leads to: \begin{align} 0&=-2\pi\ic\pars{\pi^{2} + \mu^{2}} + 2\pi\ic\int_{0}^{R}\ln\pars{x} {2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x \\[5mm]&+\pars{2\pi\ic}^{2}\int_{\cos\pars{\mu}}^{R + \cos\pars{\mu}} {x \over x^{2} + \sin^{2}\pars{\mu}}\,\dd x - {\mathfrak C}\pars{R,\mu} \end{align}

and \begin{align} \fermi\pars{\mu,R}&=\int_{0}^{R}\ln\pars{x} {2x + 2\cos\pars{\mu}\over x^{2} + 2x\cos\pars{\mu} + 1}\,\dd x \\[5mm]&=\pi^{2} + \mu^{2} -\pi\ic\ln\pars{\bracks{R + \cos\pars{\mu}}^{2} + \sin^{2}\pars{\mu}} + {{\mathfrak C}\pars{R,\mu} \over 2\pi\ic} \end{align}

In the $\ds{R \to \infty}$ we'll have: \begin{align} \lim_{R \to\ \infty}\bracks{\fermi\pars{a,R} - \fermi\pars{b,R}} &=a^{2} - b^{2} \end{align}

such that expression $\pars{1}$ is reduced to: \begin{align}&\color{#66f}{\large\int_{0}^{\infty} \ln\pars{x^{2} + 2kx\cos\pars{b} + k^{2}\over x^{2} + 2kx\cos\pars{a} + k^2}\,{\dd x \over x}} =\color{#66f}{\large a^{2} - b^{2}} \end{align}

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  • $\begingroup$ Thanks for your answer Felix, (+1). It's good to have many different ways to evaluate this integral. $\endgroup$
    – Venus
    Dec 16, 2014 at 6:46
  • $\begingroup$ @Venus You're welcome. I left some details that you can recover. I did it because it was growing out of my energy. $\endgroup$ Dec 16, 2014 at 7:10
  • $\begingroup$ Thanks for the details Felix, you're so kind. I hope I can learn some complex analysis techniques from your answer. Anyway, where's your hat? You should wear it. :-) $\endgroup$
    – Venus
    Dec 16, 2014 at 8:19
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1st solution

We note that: $$\displaystyle{\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx\cos a+k^2} \right )=\ln \left ( x^2+2kx \cos \alpha +k^2 \right ) \bigg|_{\alpha =a}^{\alpha =b}}$$

Then we have \begin{aligned} \int^{\infty}_{0} \frac{1}{x}\ln\left(\frac{x^2+2kx\cdot \cos b+k^2}{x^2+2k x\cdot \cos a+k^2}\right)\,dx &= \int_{0}^{\infty}\frac{1}{x}\ln \left ( x^2+2k x \cos \alpha +k^2 \right )\bigg|_{\alpha =a}^{\alpha =b}\,dx \\ &= \int_{0}^{\infty}\frac{1}{x}\int_{a}^{b}\frac{\mathrm{d} }{\mathrm{d} \alpha }\ln \left ( x^2+2k x \cos \alpha +k^2\right )\,d\alpha \;dx\\ &= -\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{x^2+2k\cos a x +k^2}\,dx\;d\alpha \\ &=-\int_{a}^{b}\int_{0}^{\infty}\frac{2k\sin \alpha }{\left ( x+k\cos a \right )^2+k^2\sin^2 \alpha }\,dx\;d\alpha \\ &= -2\int_{a}^{b}\tan^{-1}\left ( \frac{x}{k\sin \alpha } +\frac{1}{\tan \alpha }\right )\bigg|_{0}^{\infty}d\alpha \\ &=-2\int_{a}^{b}\left [ \frac{\pi}{2}-\tan^{-1}\left ( \frac{1}{\tan \alpha } \right ) \right ]\,d\alpha \\ &=-2\int_{a}^{b}\tan^{-1}\left ( \tan \alpha \right )\, d\alpha \\ &=-2\int_{a}^{b}\alpha \,d\alpha =a^2-b^2 \end{aligned}

whence the request.

2nd solution

We will use the following rows ${\rm Fourier}$ :

$$\bullet\displaystyle{\sum_{n=1}^{\infty}\frac{x^n \cos (na)}{n}=-\frac{1}{2}\ln \left ( x^2-2x \cos a+1 \right ), \; \left | x \right |<1 \;\;\;\; (*)}$$ which ${\rm MacLaurin}$ comes out if in its $\ln (1-z)$ development we put $z=xe^{ia}$ and equalize the real parts.

$$\bullet \displaystyle{\sum_{n=1}^{\infty}(-1)^n \frac{\sin n x}{n}=-\frac{x}{2}, \; \left | x \right |<\pi}$$

We conclude the second with respect to x and then by setting where x=0 (see note) the order arises $$\displaystyle{\sum_{n=1}^{\infty}(-1)^n \frac{\cos nx }{n^2}=\frac{x^2}{4}- \frac{\pi^2}{12}, \; \left | x \right |<\pi}\;\;\;\; (**)$$

Then for the requested integral we have \begin{aligned} \int_{0}^{\infty}\frac{1}{x}\ln \left ( \frac{x^2+2kx \cos b+k^2}{x^2+2kx \cos a+k^2} \right )\,dx &\overset{u=x/k}{=\! =\! =\!}\int_{0}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\ &= \int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du + \int_{1}^{\infty}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\ &\overset{u \mapsto 1/u}{=\! =\! =\! =\!}2\int_{0}^{1}\frac{1}{u}\ln \left ( \frac{u^2+2u\cos b+1}{u^2+2u \cos a+1} \right )\,du \\ &\overset{(*)}{=}4\int_{0}^{1}\frac{1}{u}\sum_{n=1}^{\infty}(-u)^n \frac{\cos na-\cos nb}{n}\,du \\ &= 4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n}\int_{0}^{1}u^{n-1}\,du \\ &=4\sum_{n=1}^{\infty}(-1)^n \frac{\cos na-\cos nb}{n^2} \\ &\overset{(**)}{=\! =\!}4\left ( \frac{a^2}{4}-\frac{b^2}{4} \right ) \\ &=a^2-b^2 \end{aligned}

as we wanted.

Note: It is true that:

$$\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} &=-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} \\ &= -\eta (2)\\ &= -\left ( 1-2^{1-2} \right )\zeta(2)\\ &= -\frac{\zeta(2)}{2}=-\frac{\pi^2}{12} \end{aligned}}$$

where $\eta$ the function His ${\rm Dirichlet}$ and $\zeta$ the well-known zeta function of ${\rm Riemann}$ . We also used the well-known $$\displaystyle{\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}}$$

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