Asymptotic expression around $x=\infty$ of Infinite sum of the exponential integral: $\sum_{i=1,3,5..}\text{Ei}\left(\frac{-i^2 \pi^3}{4 x^2}\right)$

Question

I ran across the following conjecture, which I checked numerically and seemed to check out. For $x\to \infty$ the sum $$\Theta(x)=\sum_{i=1,3,5..} \text{Ei}{\left(-\frac{i^2\pi^3}{4 x^2}\right)},$$ has the asymptotic expression $$\Theta(x)_{x\to \infty}=\ln{2}-\frac{x}{\pi},$$ where $\text{Ei}(x)$ is defined as $$\text{Ei}(x)=\int_{-\infty}^x\frac{e^t}{t}\text{d}t.$$

Showing that this is actually true proves to be a significantly hard task that once again, I'm asking for your help. How would one go about proving this?

Answer 1

Assume that $x>0$. Using the alternative notation $E_1$, we find \begin{align*} \Theta (x) & = - \sum\limits_{n = 0}^\infty {E_1 \!\left( {\frac{{\pi ^3 }}{{4x^2 }}(2n + 1)^2 } \right)} = - \sum\limits_{n = 0}^\infty {\int_{(\pi /x)^2 }^{ + \infty } {\frac{{e^{ - (\pi /4)(2n + 1)^2 t} }}{t}dt} } \\ & = - \frac{1}{2}\int_{(\pi /x)^2 }^{ + \infty } {\frac{{\theta _2 (0|it)}}{t}dt} = - \int_{\pi /x}^{ + \infty } {\frac{{\theta _2 (0|is^2 )}}{s}ds} \\ & = - \int_{\pi /x}^{ + \infty } {\frac{{\theta _4 (0|is^{ - 2} )}}{{s^2 }}ds} = - \int_0^{x/\pi } {\theta _4 (0|it^2 )dt} \\ & = \int_0^{x/\pi } {(1 - \theta _4 (0|it^2 ))dt} - \frac{x}{\pi } \end{align*} where $\theta_i$ are the theta functions and we used the known transformation formula. Hence, $$ \mathop {\lim }\limits_{x \to + \infty } \left( {\Theta (x) + \frac{x}{\pi }} \right) = \int_0^{ + \infty } {(1 - \theta _4 (0|it^2 ))dt} $$ provided the improper integral on the right-hand side exists. Your claim is that the integral exists and it is $\log 2$. It is known that (see $(1.14.29)$ in F. Oberhettinger's Tables of Mellin Transforms) $$ \int_0^{ + \infty } {t^{s - 1} (1 - \theta _4 (0|it^2 ))dt} = \frac{{\Gamma (s/2)}}{{\pi ^{s/2} }}\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }}{{n^s }}} $$ provided $\Re s>0$. This can be proved for $\Re s>1$ via term-by-term integration and can be extended to $\Re s>0$ via analytic continuation. Thus, with $s=1$, $$ \int_0^{ + \infty } {(1 - \theta _4 (0|it^2 ))dt} = \frac{{\Gamma (1/2)}}{{\pi ^{1/2} }}\sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }}{n}} = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^{n + 1} }}{n}} = \log 2 $$ which is the desired result.

Note that we have an exact expansion for $x>0$ in terms of the complementary error function: $$ \Theta (x) = - \frac{x}{\pi } + \log 2 + \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n }}{n}\operatorname{erfc}\left( {\frac{x}{{\sqrt \pi }}n} \right)} . $$

Answer 2

To the the professional solution by @Gary we can add the amateur's one :)

Basically, to solve the problem we will need only two equations: $$\theta(s)=\sum_{n=-\infty}^\infty e^{-\pi n^2s}=1+2\sum_{n=1}^\infty e^{-\pi n^2s}$$ $$\theta(s)=\frac{1}{\sqrt s}\theta\Big(\frac{1}{s}\Big)$$ The second equation is the functional equation for theta-function; the easy proof can be found, for instance, here.

Let's denote $S=\sum_{i=1,3,5..}\operatorname{Ei}\big(-\frac{i^2\pi^3}{4x^2}\big)$. Using the definition $\operatorname{Ei}(z)=\int_{-\infty}^z\frac{e^t}{t}dt$, we can present the sum in the form $$S=-\sum_{i=1,3,5..}\int_1^\infty e^{-\frac{i^2\pi^3}{4x^2}s}\frac{ds}{s}$$ Changing the order of summation and integration
$$\sum_{i=1,3,5..}e^{-\frac{i^2\pi^3}{4x^2}s}=\sum_{i=1,2,3..}e^{-\frac{i^2\pi^3}{4x^2}s}-\sum_{i=2,4,..}e^{-\frac{i^2\pi^3}{4x^2}s}=\sum_{i=1,2,3..}e^{-\frac{i^2\pi^3}{4x^2}s}-\sum_{i=1,2,3,..}e^{-\frac{4i^2\pi^3}{4x^2}s}$$ $$=\frac{1}{2}\bigg(\sum_{i=-\infty}^\infty e^{-\frac{i^2\pi^3}{4x^2}s}-1\bigg)-\frac{1}{2}\bigg(\sum_{i=-\infty}^\infty e^{-\frac{i^2\pi^3}{x^2}s}-1\bigg)$$ $$=\frac{1}{2}\bigg(\theta\Big(\frac{\pi^2s}{4x^2}\Big)-\theta\Big(\frac{\pi^2s}{x^2}\Big)\bigg)=\frac{1}{2}\bigg(\frac{2x}{\pi\sqrt s}\theta\Big(\frac{4x^2}{\pi^2s}\Big)-\frac{x}{\pi\sqrt s}\theta\Big(\frac{x^2}{\pi^2s}\Big)\bigg)$$ In the last step the functional equation for theta-function has been used. Using the first formula for theta-function and putting all in the initial integral $$S=-\frac{1}{2}\int_1^\infty\frac{ds}{s\sqrt s}\bigg(\frac{2x}{\pi}\Big(2\sum_{i=1}^\infty e^{-\frac{4x^2i^2}{\pi s}}+1\Big)-\frac{x}{\pi}\Big(2\sum_{i=1}^\infty e^{-\frac{x^2i^2}{\pi s}}+1\Big)\bigg)$$ $$=-\frac{x}{2\pi}\int_1^\infty\frac{ds}{s\sqrt s}+\frac{x}{\pi}\sum_{i=1}^\infty\int_1^\infty\frac{ds}{s\sqrt s}\Big(e^{-\frac{x^2i^2}{\pi s}}-2e^{-\frac{4x^2i^2}{\pi s}}\Big)$$ $$=-\frac{x}{\pi}+\frac{x}{\pi}\sum_{i=1}^\infty\int_1^\infty\frac{ds}{s\sqrt s}(-1)^{i-1}e^{-\frac{x^2i^2}{\pi s}}$$ Making the substitution $t=\frac{1}{\sqrt s}$ $$S=-\frac{x}{\pi}+\frac{2x}{\pi}\sum_{i=1}^\infty(-1)^{i-1}\int_0^1e^{-\frac{x^2i^2}{\pi}t^2}dt$$ But $$\int_0^1e^{-\frac{x^2i^2}{\pi}t^2}dt=\int_0^\infty e^{-\frac{x^2i^2}{\pi}t^2}dt-\int_1^\infty e^{-\frac{x^2i^2}{\pi}t^2}dt=\frac{\sqrt\pi}{2}\frac{\sqrt\pi }{i x}-\frac{\sqrt\pi}{2}\frac{2}{\sqrt\pi}\frac{\sqrt\pi}{xi}\int_\frac{xi}{\sqrt\pi}^\infty e^{-s^2}ds$$ and the desired sum is $$S=-\frac{x}{\pi}+\sum_{i=1}^\infty\frac{(-1)^{i-1}}{i}-\sum_{i=1}^\infty\frac{(-1)^{i-1}}{i}\frac{2}{\sqrt\pi}\int_\frac{xi}{\sqrt\pi}^\infty e^{-s^2}ds$$ $$S=-\frac{x}{\pi}+\ln 2+\sum_{i=1}^\infty\frac{(-1)^{i}}{i}\operatorname{erfc}\Big(\frac{xi}{\sqrt\pi}\Big)$$