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Use Parseval's equation and the table of Fourier series to evaluate $$\sum\frac{1}{(1+n^2)^2}$$

So I have used this method before to show $$\sum\frac{1}{n^4} = \frac{\pi^4}{32}$$ however, for this question I am struggling.

Basically, off the sheet I have from university I have used $$F(x) = \left|\sin\left(\pi\times\frac xL\right)\right| , -L<x<L, \to \frac 2\pi - \frac 4\pi \sum \frac{1}{n^2-1}\times\cos\left(\frac{n\pi x}{L}\right)$$

I find $a_0$ and $a_n$ then

I apply Parseval's theorem then it becomes clear I have gone wrong...

Jack D'Aurizio
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Ben
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2 Answers2

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Instead of Parseval's identity, we may just use differentiation under the integral sign.

By integration by parts we have, for any $a>0$ and $n\in\mathbb{N}^+$: $$ \frac{1}{a^2+n^2} = \int_{0}^{+\infty}\frac{\sin(nx)}{n} e^{-ax}\,dx \tag{1}$$ hence by differentiating both sides with respect to $a$ we get: $$ \frac{a}{(a^2+n^2)^2} = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin(nx)}{n} x e^{-ax}\,dx \tag{2}$$ and by evaluating at $a=1$, then summing over $n\geq 1$, we get: $$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2} = \frac{1}{2}\int_{0}^{+\infty}W(x)\,x\, e^{-x}\,dx \tag{3} $$ where $W(x)$ is a sawtooth wave, a $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ on the interval $(0,2\pi)$.
By exploiting such periodicity we get that: $$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2}=\frac{1}{2}\int_{0}^{2\pi}\frac{\pi-x}{2}\sum_{m\geq 0}(x+2m\pi)e^{-x-2m\pi}\,dx \tag{4}$$ and by simplifying the RHS we get: $$ \sum_{n\geq 1}\frac{1}{(n^2+1)^2}=\frac{1}{4\left(e^\pi-e^{-\pi}\right)^2}\int_{0}^{2\pi}(\pi -x)e^{-x}(2\pi-x+ x e^{2\pi})\,dx \tag{5} $$ with the original problem boiling down to the evaluation of an elementary integral.
The final outcome is: $$ \boxed{\sum_{n\geq 1}\frac{1}{(n^2+1)^2} = \color{red}{-\frac{1}{2}+\frac{\pi\cosh(\pi)}{4\sinh(\pi)}+\frac{\pi^2}{4\sinh^2\pi}}}\tag{6}$$

Jack D'Aurizio
  • 353,855
3

It is easy to show that the complex Fourier Series for $e^{bx}$, $x\in[0,2\pi]$ is given by

$$e^{bx}=\frac{e^{2\pi b}-1}{2\pi}\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}$$

whereupon isolating the series becomes

$$\sum_{n=-\infty}^\infty \frac{e^{inx}}{b-in}=\frac{2\pi e^{bx}}{e^{2\pi b}-1} \tag1$$

Differentiating $(1)$ with respect to $b$ and multiplying by $-1$ reveals

$$\begin{align} \sum_{n=-\infty}^\infty \frac{e^{inx}}{(b-in)^2}&=-\frac{d}{db}\left(\frac{2\pi e^{bx}}{e^{2\pi b}-1} \right)\\\\ &=\left(\frac{2\pi e^{bx}}{(e^{2\pi b}-1)^2}\right)\left(2\pi e^{2\pi b}-(e^{2\pi b}-1)x \right)\tag 2 \end{align}$$

Next, setting $b=1$ in $(2)$ yields

$$\begin{align} \sum_{n=-\infty}^\infty \frac{e^{inx}}{(1-in)^2}&=\left(\frac{2\pi e^{x}}{(e^{2\pi }-1)^2}\right)\left(2\pi e^{2\pi }-(e^{2\pi }-1)x \right)\tag 3 \end{align}$$

Finally, applying Parseval's Theorem to $(3)$ we find that

$$\begin{align} \sum_{n=-\infty}^\infty \frac{1}{(n^2+1)^2}&=\left(\frac{2\pi}{(e^{2\pi}-1)^2}\right)^2\,\frac{1}{2\pi}\int_0^{2\pi }e^{2x}(2\pi e^{2\pi}-(e^{2\pi}-1)x)^2\,dx\\\\ &=\frac{\pi}2 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right) \end{align}$$

whereupon solving for the series of interest yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{(n^2+1)^2}=-\frac12 +\frac{\pi}4 \left(\pi \text{csch}^2(\pi)+\text{coth}(\pi)\right)}$$

Mark Viola
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