How to evaluate $\sum\limits_{x=0}^\infty \text{erfc}(x)= 1.1619990479471263635323…?$

Question

This will be the $5$th in a series of an infinite series of a single function. Here are 2 related sums:

A Kelvin-Bessel Sum: $$\mathrm{\sum\limits_{\Bbb N} ker(x)+i\ kei(x)= \sum\limits_1^\infty K_0\left(\sqrt ix\right)= 0.133691752… - 0.7256312077… i}$$

and

On $$\mathrm{\sum\limits_{n=0}^\infty \left(C(n)-\frac{\sqrt\pi}{2\sqrt2}\right)+ \sum\limits_{n=0}^\infty \left(S(n)-\frac{\sqrt\pi}{2\sqrt2}\right)}$$

Our goal sum uses the Complementary Error function integrating term by term. Note the Fresnel Integrals. There are also Gamma type functions:

$$\sum_{\Bbb N^0}\text{erfc}(x)=\sum_{x=0}^\infty \text{erfc}(x) =\lim_{n\to\infty} \left(n-\sum_0^n \text{erf}(x)\right)= \frac{2}{\sqrt \pi}\sum_{x\in \Bbb N^0}\int_x^\infty e^{-t^2} dt= \sum_{x\in \Bbb N^0}\left(1-\frac 2\pi \int_0^\infty \frac{\sin(2tx)}{te^{t^2}}dt\right)=\sum_{x\in\Bbb N^0}\left(1 - (1 + i) \left(\text C\left(\frac{((1 - i) z)}{\sqrt\pi}\right) - i\,\text S\left(\frac{((1 - i) z)}{\sqrt\pi}\right)\right)\right)=\sum_{\Bbb N^0}\text{erf}(x,\infty)= \frac{1}{\sqrt \pi}\sum_{\Bbb N^0}Γ\left(\frac12,x^2\right)= \sum_{\Bbb N^0}Q\left(\frac12,x^2\right) =\frac{1}{\sqrt \pi} \sum_{\Bbb N^0}x \text E_\frac12 \left(x^2\right)= \frac{1}{\sqrt \pi}\int_1^\infty t^{-\frac12}\sum_{x\in\Bbb N^0}xe^{-tx^2}dt =1.16199904794712636353230832245579717…$$

As @JohnBarber found:

$$\sum_{\Bbb N^0}\text{erfc}(x) =1+\frac{2}{\sqrt\pi}\int_1^\infty \lfloor x\rfloor e^{-x^2} dx$$ The final integral-sum representation reminds me of a Differentiated Jacobi Theta function of the Third Kind. Here is a question about it although there are others.

How can I evaluate the constant? Please correct me and give me feedback!

Answer 1

How about this: $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \sum_{n=0}^{\infty}\frac{2}{\sqrt{\pi}}\int_n^{\infty}e^{-t^2}\, dt $$ In this sum of integrals, the interval $[0,1)$ will be counted only once, in the $n = 0$ term. The interval $[1,2)$ will be counted twice, in the $n = 0$ and $n = 1$ terms. And so on. This means we can write: $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{2}{\sqrt{\pi}}\int_0^{\infty}\, \lfloor t+1\rfloor \,e^{-t^2}\, dt $$ I don't know if this is the sort of alternative integral representation you were looking for.

Edited to add two other ways to write this expression:

First way: The quantity $\lfloor t+1\rfloor$ can be written as $$ \lfloor t+1\rfloor \;=\; (t+1) \;-\; S(t)\, . $$ where $S(t)$ is a sawtooth wave of period $1$ with minimum value $0$ and a maximum value $1$. One way we could write this sawtooth is as $S(t) \,=\, t\;\mathrm{mod}\;1$. Substituting this expression for $\lfloor t+1\rfloor$ into the integral above and using the fact that $\int_0^{\infty} dt \, (t+1)\,e^{-t^2} = (1+\sqrt{\pi})/2$ yields $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{2}{\sqrt{\pi}}\left[\frac{1}{2}(1+\sqrt{\pi}) \;-\; \int_0^{\infty}dt\, S(t)\, e^{-t^2}\right]. $$ Evaluating this numerically in Mathematica with 20 digits of precision yields $1.16200283409802758182$, which is greater than Mathematica's estimate for the original sum by roughly $3.8\times {10}^{-6}$. Close enough for the vagaries of numerically evaluating weird sums and integrals.

Second way: Poisson's summation formula states that if $f(x)$ is a function and $$ \hat{f}(q) \;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, f(x) $$ is its Fourier transform, then $$ \sum_{n = -\infty}^{+\infty} f(n) \;=\; \sum_{n=-\infty}^{+\infty} \hat{f}(2\pi n)\, . $$ Define the even function $f(x) = \mathrm{erfc}(|x|)$. Since $f$ is even, we can write the original sum as $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{1}{2}f(0) \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} f(n) \;=\; \frac{1}{2} \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} \hat{f}(2\pi n)\, .\hspace{0.5in}\text{(1)} $$ The Fourier transform of $f$ is: \begin{align} \hat{f}(q) &\;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, \mathrm{erfc}(|x|)\\[0.1in] &\;=\; 2\int_0^{+\infty} dx\, \cos(q x)\, \mathrm{erfc}(x)\hspace{0.5in}\text{Since $\mathrm{erfc}(|x|)$ is even}\\[0.1in] &\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dx\, \cos(q x)\,\int_x^{\infty}dt\, e^{-t^2} \hspace{0.5in}\text{Definition of $\mathrm{erfc}$}\\[0.1in] &\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\int_0^t dx\, \cos(q x) \hspace{0.5in}\text{Reverse order of integration}\\[0.1in] &\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\frac{\sin(q t)}{q}\\[0.1in] &\;=\; \frac{4}{\sqrt{\pi}}\frac{D(q/2)}{q} \end{align} In the last line above, "$D$" is the Dawson function according to the third definition here. This expression for $\hat{f}(q)$ is valid everywhere except at $q = 0$, where $\hat{f}(0) = 2/\sqrt{\pi}$. Plugging all of this into (1) results in: $$ \sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{1}{2}\left[1 + \frac{2}{\sqrt{\pi}} \,+\,\frac{8}{\sqrt{\pi}}\sum_{n = 1}^{\infty} \frac{D(\pi n)}{2\pi n}\right]\, . $$ Evaluating this numerically in Mathematica yields $1.16199904795$.

Answer 2

Here's a general approach that I believe is valid when $f(0)$ is finite and the sum over $f(x)$ converges. This approach uses convolution with the Dirac comb $\operatorname{\text{Ш}}(x)$ where Mellin convolution, defined in formula (1) below, is denoted as $*_\mathcal{M}$ and Fourier convolution, defined in formula (4) below, is denoted as $*_\mathcal{F}$.

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$$\left[g(x)\,*_\mathcal{M}\,f(x)\right](y)=\int_0^\infty g(x)\,\frac{f\left(\frac{y}{x}\right)}{x} \, dx\tag{1}$$

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$$\sum\limits_{n=1}^\infty f(n)=-\frac{1}{2}\,f(0)+\left[(x\,\operatorname{\text{Ш}}(x))\,*_\mathcal{M}\,f\left(\frac{1}{x}\right)\right](1)=-\frac{1}{2}\,f(0)+\int_0^\infty\operatorname{\text{Ш}}(x)\,f(x)\,dx\tag{2}$$

$$\sum\limits_{n=0}^\infty f(n)=\frac{1}{2}\,f(0)+\left[(x\,\operatorname{\text{Ш}}(x))\,*_\mathcal{M}\,f\left(\frac{1}{x}\right)\right](1)=\frac{1}{2}\,f(0)+\int_0^\infty\operatorname{\text{Ш}}(x)\,f(x)\,dx\tag{3}$$

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$$\left[g(x)\,*_\mathcal{F}\,f(x)\right](y)=\int_{-\infty}^\infty g(x)\,f(y-x)\,dx\tag{4}$$

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$$\sum\limits_{n=-\infty}^\infty f(n)=\left[f(x)\,*_\mathcal{F}\,\operatorname{\text{Ш}}(x)\right](0)=\int_{-\infty}^\infty f(x)\,\operatorname{\text{Ш}}(-x)\,dx=\int_{-\infty}^\infty\operatorname{\text{Ш}}(x)\,f(x)\,dx\tag{5}$$

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Using the following analytic representation of the Dirac comb where the evaluation frequency $f$ is assumed to be a positive integer

$$\operatorname{\text{Ш}}(x)=\underset{f\to\infty}{\text{lim}}\left(1+2\sum\limits_{k=1}^f\cos(2 \pi k x)\right)\tag{6}$$

and the following related integral results

$$\int_0^\infty\text{erfc}(x)\,dx=\frac{1}{\sqrt{\pi}}\tag{7}$$

$$\int_0^\infty 2\cos(2 \pi k x)\,\text{erfc}(x)\,dx=\frac{2\,F(\pi k)}{\pi ^{3/2}\,k}\tag{8}$$

and noting that $\frac{1}{2}\text{erfc}(0)=\frac{1}{2}$, formula (3) above leads to the following result for $f(x)=\text{erfc}(x)$ where $F(x)$ is Dawson's integral:

$$\sum\limits_{n=0}^\infty\text{erfc}(n)=\frac{1}{2}+\frac{1}{\sqrt{\pi}}+\frac{2}{\pi^{3/2}}\sum\limits_{k=1}^\infty\frac{F(\pi k)}{k}\tag{9}$$

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Mathematica gives the numerical approximation of formula (9) above as $\sum_\limits{n=0}^\infty\text{erfc}(n)\approx 1.1619990479420759$.

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Note formula (9) above appears to be consistent with the answer posted by John Barber which uses the floor function $\lfloor x\rfloor =x-\text{SawtoothWave}[x]$. Also note the floor function is the integral of the Dirac comb and has the following analytic representation where the evaluation frequency $f$ is assumed to be a positive integer.

$$\lfloor x\rfloor=\underset{f\to\infty}{\text{lim}}\left(x-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^f\frac{\sin(2 \pi k x)}{k}\right)\right)\tag{10}$$