Computing four infinite series through the Fourier series of cos(at)

        Today´s post is a bit long but I consider it one of the most important because  we will evaluate four very important  infinite series. To achieve this goal, I´ll start by computing the fourier series of the function \cos (a t) in the interval -\pi<t<\pi.  In addition to the four infinite series calculated here, the famous reflection formula for the gamma function can also be proved from this fourier series, but this will be shown in the next post.

The exponential fourier series of a function is given by the following equation:


f(t)=\sum_{k=-\infty}^{\infty} c_{k} e^{i k t},(1)

where

c_{k}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos (a t) e^{-i k t} d t

Recall from Euler´s formula that

\boxed{\cos(a t)=\frac{e^{i a t}+e^{-i a t}}{2}}

We start by computing c_{k}

c_{k}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-i k t}\left[\frac{e^{i a t}+e^{-i a t}}{2}\right] d t

split in two integrals

c_{k}=\frac{1}{4 \pi}\left[\int_{-\pi}^{\pi} e^{-i k t} e^{i a t} d t+\int_{-\pi}^{\pi} e^{-i k t} e^{-i a t} d t\right]


c_{k}=\frac{1}{4 \pi}\left[\int_{-\pi}^{\pi} e^{i(a-k) t} d t+\int_{-\pi}^{\pi} e^{-i(a+k) t} d t\right]


c_{k}=\frac{1}{4 \pi}\left[\left.\frac{e^{i(a-k) t}}{i(a-k)}\right|_{-\pi} ^{\pi}+\left.\frac{e^{-i(t+k) t}}{-i(a+k)}\right|_{-\pi} ^{\pi}\right]


c_{k}=\frac{1}{4 \pi i}\Big[ \frac{e^{i(a-k)\pi}-e^{-i(a-k)\pi}}{a-k}+\frac{e^{i(a+k)\pi}-e^{-i(a+k)\pi}}{a+k} \Big]


c_{k}=\frac{1}{2 \pi}\Big[\frac{1}{a-k} \cdot \frac{e^{i(a-k)\pi}-e^{-i(a-k)\pi}}{2i}+\frac{1}{a+k} \cdot\frac{e^{i(a+k)\pi}-e^{-i(a+k)\pi}}{2i} \Big]


c_{k}=\frac{1}{2 \pi}\Big[\frac{\sin[(a-k)\pi]}{a-k} +\frac{\sin[(a+k)\pi]}{a+k} \Big]


c_{k}=\frac{1}{2 \pi}\bigg[\frac{(a+k)\sin[(a-k)\pi]+(a-k)\sin[(a+k)\pi]}{(a^2-k^2)} \bigg]


c_{k}=\frac{1}{2 \pi}\bigg[\frac{(a+k) \Big[ \sin(a \pi)\overbrace{\cos(k \pi)}^{=(-1)^k}-\overbrace{\sin(k \pi)}^{=0}\cos(a \pi)\Big]}{(a^2-k^2)}+\frac{(a-k)\Big[ \sin(a \pi)\overbrace{\cos(k \pi)}^{=(-1)^k}+\overbrace{\sin(k \pi)}^{=0}\cos(a \pi) \Big]}{(a^2-k^2)} \bigg]


c_{k}=\frac{1}{2 \pi}\Big[\frac{(a+k)(-1)^k \sin(a \pi)}{(a^2-k^2)}+\frac{(a-k)(-1)^k \sin(a \pi)}{(a^2-k^2)} \Big]


c_{k}=\frac{1}{2 \pi}\Big[\frac{a(-1)^k \sin(a \pi)+a(-1)^k \sin(a \pi)+k(-1)^k \sin(a \pi)-k(-1)^k \sin(a \pi)}{(a^2-k^2)} \Big]


c_{k}=\frac{1}{2 \pi}\Big[\frac{2a(-1)^k \sin(a \pi)}{(a^2-k^2)} \Big]


\boxed{c_{k}=\Big[\frac{a(-1)^k \sin(a \pi)}{\pi(a^2-k^2)} \Big]}


Therefore plugging this result in (1) we arrive at the the fourier series of  \cos(at)  as


\boxed{\cos(at)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k \,a \, \sin(a \pi)}{\pi(a^2-k^2)}e^{ikt} }(2)



This last equation can be broken in three pieces


\cos(at)=\frac{a \sin(a \pi)}{\pi}\sum_{k=-\infty}^{-1}\frac{(-1)^k }{a^2-k^2}e^{ikt}+\frac{1}{a^2} \cdot\frac{a \sin(a\pi)}{\pi} +\frac{a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k }{a^2-k^2}e^{ikt}

By changing k\longmapsto-k in the first sum and noting that (-1)^k=(-1)^{-k}, we get the following


\cos(at)=\frac{ \sin(a\pi)}{a\pi}+\frac{a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{-k} }{a^2-k^2}e^{-ikt}+ \frac{a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k }{a^2-k^2}e^{ikt}

=\frac{ \sin(a\pi)}{a\pi}+\frac{a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\Big\{\frac{(-1)^{-k} }{a^2-k^2}e^{-ikt}+ \frac{(-1)^k }{a^2-k^2}e^{ikt} \Big \}

=\frac{ \sin(a\pi)}{a\pi}+\frac{a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k}(e^{-ikt}+e^{ikt}) }{a^2-k^2}

=\frac{ \sin(a\pi)}{a\pi}+\frac{2a \sin(a \pi)}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k} \cos(kt) }{a^2-k^2}


(3)

\boxed{\pi \frac{\cos(at)}{\sin(a\pi)}=\frac{ 1}{a}+2a\sum_{k=1}^{\infty}\frac{(-1)^{k} \cos(kt) }{a^2-k^2}}



Now, if we let t\longmapsto\pi in (3) we get

\pi \cot(a \pi)=\frac{ 1}{a}+2a\sum_{k=1}^{\infty}\frac{(-1)^{-k} \cos(k\pi) }{a^2-k^2}

since k is an integer  \cos(k\pi)=(-1)^{-k}  and


\pi \cot(a \pi)=\frac{ 1}{a}+2a\sum_{k=1}^{\infty}\frac{(-1)^{-k}(-1)^{-k} }{a^2-k^2}

Finally!


\boxed{\pi \cot(a \pi)-\frac{ 1}{a}=\sum_{k=1}^{\infty}\frac{2a }{a^2-k^2}}(4)



This is an amazing result to add in our repertoire!

If we let a\longmapsto ix in (4) we find the value of another important infinite serie

\pi \cot(i \pi x)=\frac{ 1}{ix}+\sum_{k=1}^{\infty}\frac{2ix }{(ix)^2-k^2}

\pi \frac{\Big(\frac{e^{ii \pi x}+e^{-ii \pi x}}{2}\Big)}{\Big(\frac{e^{ii \pi x}-e^{-ii \pi x}}{2i}\Big)}=-\frac{ i}{x}+\sum_{k=1}^{\infty}\frac{2ix }{(ix)^2-k^2}

-i \pi \coth(\pi x)= -\frac{i}{x}+2ix\sum_{k=1}^{\infty}\frac{1 }{-x^2-k^2}

-i \pi \coth(\pi x)= -\frac{i}{x}-2ix\sum_{k=1}^{\infty}\frac{1 }{x^2+k^2}


And finally!


                                      \boxed{\pi \coth(\pi x)= \frac{1}{x}+\sum_{k=1}^{\infty}\frac{2x }{x^2+k^2}}                                    (5)

Another very important corollary that we obtain is by letting t\longmapsto 0 in (3)


\boxed{ \frac{\pi}{\sin(a\pi)}=\frac{ 1}{a}+2a\sum_{k=1}^{\infty}\frac{(-1)^{k} }{a^2-k^2}}(6)



Equation (6) above can be used to derive the famous reflection formula for the Gamma function which will be proved in the next post!


The last result that I want to show in this post is by letting a\longmapsto ix in (6)

\frac{\pi}{\sin(i\pi x)}=\frac{ 1}{ix}+2ix\sum_{k=1}^{\infty}\frac{(-1)^{k} }{(ix)^2-k^2}

-\frac{2\pi i}{e^{\pi x}-e^{-\pi x}}=-\frac{ i}{x}-2ix\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}

\frac{\pi }{\sinh(\pi x)}=\frac{ 1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}

And finally!


\boxed{\pi \mathrm{csch}(\pi x)=\frac{ 1}{x}+2x\sum_{k=1}^{\infty}\frac{(-1)^{k} }{x^2+k^2}}(7)

Ricardo Albahari


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