Asymptotic expression around $x=\infty$ of Infinite sum of the exponential integral: $\sum_{i=1,3,5..}\text{Ei}\left(\frac{-i^2 \pi^3}{4 x^2}\right)$ 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?
 A: 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:
$$
\Theta (x) =  - \frac{x}{\pi } + \log 2 + \sum\limits_{n = 1}^\infty  {\frac{{( - 1)^n }}{n}\operatorname{erfc}\left( {\frac{x}{{\sqrt \pi  }}n} \right)} .
$$
A: 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
$$S=-\frac{x}{\pi}+\sum_{i=1}^\infty\frac{(-1)^{i-1}}{i}-\sum_{i=1}^\infty\frac{(-1)^{i-1}}{i}\frac{2}{\sqrt\pi}\int_\frac{xi}{\sqrt\pi}^\infty e^{-s^2}ds$$
$$S=-\frac{x}{\pi}+\ln 2+\sum_{i=1}^\infty\frac{(-1)^{i}}{i}\operatorname{erfc}\Big(\frac{xi}{\sqrt\pi}\Big)$$
