To Prove: \[\sum _{ k=1 }^{ \infty }{ \frac { \cos { \left( kx \right) } }{ { k }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 } -\frac { \pi x }{ 2 } +\frac { x^{ 2 } }{ 4 } \]

\(0\leq x\leq 2\pi\)

**Proof:**

Write \(\cos(kx)\) as: \(\displaystyle \cos { \left( kx \right) } =\frac { { e }^{ ikx }+{ e }^{ -ikx } }{ 2 } \)

Then we get \[S=\sum _{ k=1 }^{ \infty }{ \frac { \frac { { e }^{ ikx }+{ e }^{ -ikx } }{ 2 } }{ { k }^{ 2 } } } \]

Then on using the definition of polylogarithm I'll write it as: \[S=\frac { { Li }_{ 2 }\left( { e }^{ ikx } \right) +{ Li }_{ 2 }\left( { e }^{ -ikx } \right) }{ 2 } \]

Now, I'll use the following relation of polylogarithm and bernoulli numbers: \[{ Li }_{ m }(z)\quad =\quad { (-1) }^{ m-1 }{ Li }_{ m }\left( \frac { 1 }{ z } \right) -\frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { ln(-z) }{ 2\pi i } +\frac { 1 }{ 2 } \right) /z\notin (0,1)\]

Therefore, I'll re-write \(S\) as: \[S=\frac { 1 }{ 2 } \frac { { \left( 2\pi \right) }^{ 2 } }{ m! } { B }_{ 2 }\left( \frac { x }{ 2\pi } \right) =\frac { { \pi }^{ 2 } }{ 6 } -\frac { \pi x }{ 2 } +\frac { x^{ 2 } }{ 4 } \]

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## Comments

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TopNewestAlternatively, show that the LHS is the Fourier Series of the RHS for \(0\leq x\leq 2\pi\). It suffices to show that \(\displaystyle \int_{0}^{2\pi}\left(\frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4}\right)\cos(kx)dx=\frac{\pi}{k^2}\), a simple exercise in calculus.

Note that the equation holds for \(0\leq x\leq 2\pi\) only, a fact that should be stated in the problem.

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Wow. Thanks.

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Sir why did you remove your problem?

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I had no idea the same problem was posed before... It can happen to all of us

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this your book? I seriously liked it.

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There is a sign error: We have \(\cos(kx)=\frac{e^{ikx}+e^{-ikx}}{2}\)

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Thanks for mentioning it sir. It was a typo. While doing it on paper I hadn't made that mistake.

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