$$f(x)= \sum_{n=-\infty}^{\infty} \frac{\mathrm{e}^{i nx}}{n^2+1}$$

on $[-\pi, \pi]$. Prove that $f(x)>0$ for any $x \in [-\pi, \pi]$.

How to use Fourier analysis to show that function is positive? I try to differentiate it, but it seems that the derivative is not convergent. Right?



Since $$ \frac1{2\pi}\int_{-\infty}^\infty e^{-|x|}e^{-inx}\,\mathrm{d}x=\frac1\pi\frac1{n^2+1} $$ we have $$ \frac1{2\pi}\int_{-\pi}^\pi\pi\sum_{k\in\mathbb{Z}}e^{-|x+2k\pi|}e^{-inx}\,\mathrm{d}x=\frac1{n^2+1} $$ Therefore, inverting the transform, we get $$ \sum_{n\in\mathbb{Z}}\frac{e^{inx}}{n^2+1}=\pi\sum_{k\in\mathbb{Z}}e^{-|x+2k\pi|} $$

Convergence of the Derivative

The derivative converges everywhere except for $x\in2\pi\mathbb{Z}$. However, the convergence is only conditional. We can apply the Generalized Dirichlet Convergence Test to $$ \sum_{k\in\mathbb{Z}}\frac{ine^{inx}}{n^2+1} $$ since $\frac{in}{n^2+1}$ has bounded variation, tends to $0$ as $|n|\to\infty$, and the partial sums of $\sum\limits_{n\in\mathbb{Z}}e^{inx}$ are bounded except where $e^{ix}=1$.


What prompted me to look at $e^{-|x|}$ was to note that the second derivative of $$ f(x)=\sum_{n\in\mathbb{Z}}\frac{e^{inx}}{n^2+1} $$ is $$ f''(x)=-\sum_{n\in\mathbb{Z}}\frac{n^2e^{inx}}{n^2+1} $$ which does not converge in the standard sense, but in the sense of distributions $$ \begin{align} f(x)-f''(x) &=\sum_{n\in\mathbb{Z}}e^{inx}\\ &=\sum_{k\in\mathbb{Z}}\delta(x-2k\pi) \end{align} $$ where $\delta(x)$ is the Dirac delta distribution, which is $0$ away from $2\pi\mathbb{Z}$. Now, two solutions to $f(x)-f''(x)=0$ are $e^x$ and $e^{-x}$, so this brought to mind $e^{-|x|}$, whose first derivative has a jump discontinuity, which would lead to a Dirac delta distribution at $x=0$.

Adding $$ \int_0^\infty e^{-x}e^{-inx}\,\mathrm{d}x=\frac1{1+in} $$ and $$ \int_{-\infty}^0e^xe^{-inx}\,\mathrm{d}x=\frac1{1-in} $$ gave $$ \int_0^\infty e^{-|x|}e^{-inx}\,\mathrm{d}x=\frac2{1+n^2} $$ which is where I started the first section.

  • $\begingroup$ +1 Just for reference: this is also called Poisson summation formula. $\endgroup$ – Fabian Nov 21 '14 at 10:59
  • $\begingroup$ @Fabian: Indeed. This is another way of proving that $$\sum_{n\in\mathbb{Z}}\frac1{n^2+1}=\pi\coth(\pi)$$ $\endgroup$ – robjohn Nov 21 '14 at 11:08

Hint: $$ f(x)= \sum_{n=-\infty}^{\infty} \frac{\mathrm{e}^{i nx}}{n^2+1}= \sum_{n=-\infty}^{\infty} \frac{\cos nx + i\sin nx}{n^2+1}=\cdots $$ What happens with the imaginary part ($\sin$ is odd)?

  • $\begingroup$ But how does this help to show $f(x)>0$? $\cos(nx)$ can be both positive and negative. $\endgroup$ – Winther Nov 21 '14 at 8:39
  • $\begingroup$ @Winther, bounding the tail(s) of the series. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 21 '14 at 8:42
  • $\begingroup$ Yes that is the usual approach. However, I tried a simple take out $n=0,1$ and bound $n>2$ using $1/(n^2+1) \leq 1/n^2$ does not seem to be enough. $\endgroup$ – Winther Nov 21 '14 at 8:45
  • $\begingroup$ @Winther, I will think in it. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 21 '14 at 8:46
  • $\begingroup$ @Winther, T've found a rather convoluted solution. Isn't the same problem, but the recycling is esay. See books.google.es/… $\endgroup$ – Martín-Blas Pérez Pinilla Nov 21 '14 at 9:29

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