Approximation for $\sum_{i=1}^\infty \frac{\sqrt i}{\sqrt{2\pi \epsilon}} e^{\frac{-i}{2\epsilon}}$?

Question

I am trying to get some nice number (not the polylogarithmic) with the series on the title. I know the exact result is $\dfrac{1}{\sqrt{2\pi \epsilon}}Li_{-\frac{1}{2}}(\dfrac{1}{^{\epsilon}\sqrt{e}})$. However, putting some values into Wolfram Alpha (I'm sorry I'm an engineering student, it's stronger than me) I came to notice that the result seems to be pretty close to $\epsilon$. In particular, it gets closer to $\epsilon$ as $\epsilon$ grows. Therefore, is there a way to approximate the remainder between the real result and $\epsilon$?

If it's of any use, I know that $\int_0^{\infty} \dfrac{\sqrt x}{\sqrt{2\pi \epsilon}} e^{\dfrac{-x}{2\epsilon}} dx = \epsilon $. It's kinda important because I'm trying to find a discrete-analogue of the gamma distribution $\dfrac{1}{\sqrt{2\pi \epsilon x}} e^{\dfrac{-x}{2\epsilon}} $ and this would be the mean. If anyone has another discrete gamma distribution similar to this one it would also be appreciated.

Edit: well, it seems that it was not as easy as my other questions. For the sake of not going crazy, it would be more than enough to get some result like: $\epsilon + C > \dfrac{1}{\sqrt{2\pi \epsilon}}Li_{-\frac{1}{2}}(\dfrac{1}{^{\epsilon}\sqrt{e}})$, where there is C is a function of the other variables or a constant

Answer 1

Denoting $\alpha=\frac1{2\epsilon}$ and considering the case $\alpha\to 0$ $$S(\epsilon)=\sum_{i=1}^\infty \frac{\sqrt i}{\sqrt{2\pi \epsilon}} e^{-\frac{i}{2\epsilon}}=\frac{\sqrt\alpha}{\sqrt\pi}\sum_{k=1}^\infty\sqrt k\,e^{-\alpha k}=-\sqrt{\frac\alpha\pi}\frac{\partial}{\partial \alpha}\sum_{k=1}^\infty \frac1{\sqrt k}e^{-\alpha k}$$ $$=-\frac{\sqrt\alpha}\pi\frac{\partial}{\partial \alpha}\sum_{k=1}^\infty\,e^{-\alpha k}\int_0^\infty t^{-\frac12}e^{-kt}dt=-\frac{\sqrt\alpha}\pi\frac{\partial}{\partial \alpha}\int_0^\infty\frac1{e^{t+\alpha}-1}\frac{dt}{\sqrt t}$$ $$=\frac{\sqrt\alpha}\pi\int_0^\infty\frac{e^{t+\alpha}}{(e^{t+\alpha}-1)^2}\frac{dt}{\sqrt t}$$ Integrating by part $$=-\frac{\sqrt\alpha}\pi\frac1{\sqrt t}\left(\frac1{e^{t+\alpha}-1}-\frac1{e^\alpha-1}\right)\,\bigg|_0^\infty-\frac{\sqrt\alpha}{2\pi}\int_0^\infty\left(\frac1{e^{t+\alpha}-1}-\frac1{e^\alpha-1}\right)\frac{dt}{t^\frac32}$$ $$=\frac{\sqrt\alpha}{2\pi}e^{-\alpha}\int_0^\infty\left(\frac1{1-e^{-\alpha}}-\frac1{e^t-e^{-\alpha}}\right)\frac{dt}{t^\frac32}$$ Denoting for a while $\beta=1-e^{-\alpha}=\alpha-\frac{\alpha^2}2+ ...\,(\,\ll1\,)$ $$S=\frac{\sqrt\alpha}{2\pi}(1-\beta)\int_0^\infty\left(\frac1\beta-\frac1{e^t-1+\beta}\right)t^{-\frac32}dt$$ $$=\frac{\sqrt\alpha}{2\pi}(1-\beta)\int_0^\infty\left(\frac1\beta-\frac1{t+\beta}+\frac1{t+\beta}-\frac1{e^t-1+\beta}\right)t^{-\frac32}dt$$ $$S=\frac{\sqrt\alpha}{2\pi}(1-\beta)\left(\int_0^\infty\frac1{\beta(t+\beta)}\frac{dt}{\sqrt t}+\int_0^\infty\frac{e^t-1-t}{(t+\beta)(e^t-1+\beta)}\frac{dt}{t^\frac32}\right)=I_1+I_2\tag{1}$$ $$I_1=\frac{\sqrt\alpha}{2\pi}(1-\beta)\int_0^\infty\frac1{\beta(t+\beta)}\frac{dt}{\sqrt t}=\frac{\sqrt\alpha}2\frac{1-\beta}{\beta^\frac32}\tag{1a}$$ $$I_2=\frac{\sqrt\alpha}{2\pi}(1-\beta)\int_0^\infty\left(\frac{e^t-1-t-\frac{t^2}2}{(t+\beta)(e^t-1+\beta)}+\frac{\frac{t^2}2}{(t+\beta)(e^t-1+\beta)}\right)\frac{dt}{t^\frac32}$$ The first integral in the parentheses converges at $\beta=0$ and, therefore, is $\sim\sqrt\alpha$

Hence, $$I_2=\frac{\sqrt\alpha}{4\pi}(1-\beta)\int_0^\infty\frac{t^2}{(t+\beta)(e^t-1+\beta)}\frac{dt}{t^\frac32}+O(\sqrt\alpha)$$ $$=\frac{\sqrt\alpha}{4\pi}(1-\beta)\int_0^\infty\frac{\sqrt t}{(t+\beta)(e^t-1+\beta)}dt+O(\sqrt\alpha)$$ $$=\frac{\sqrt\alpha}{4\pi}(1-\beta)\int_0^\infty\frac{\sqrt t}{t+\beta}\left(\frac1{e^t-1+\beta}-\frac1{t+\beta}+\frac1{t+\beta}\right)dt+O(\sqrt\alpha)$$ In the same way, two first terms give the convergent integral $\sim\sqrt \alpha$ at $\beta\to 0$ $$I_2=\frac{\sqrt\alpha}{4\pi}(1-\beta)\int_0^\infty\frac{\sqrt t}{(t+\beta)^2}dt+O(\sqrt\alpha)=\frac{\sqrt\alpha}8\frac{1-\beta}{\sqrt\beta}+O(\sqrt\alpha)\tag{1b}$$ Using (1a), (1b) and $\beta=1-e^{-\alpha}=\alpha-\frac{\alpha^2}2+ ...$ $$\boxed{\,\,S(\epsilon)=I_1+I_2=\frac{\sqrt\alpha}2\frac{1-\beta}{\beta^\frac32}+\frac{\sqrt\alpha}8\frac{1-\beta}{\sqrt\beta}+O(\sqrt\alpha)=\frac1{2\alpha}+O(\sqrt\alpha)=\epsilon+O\Big(\frac1{\sqrt{2\epsilon}}\Big)\,\,}$$

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$\bf{Addendum}$

One can show that $$\boxed{S(\epsilon)=\sum_{i=1}^\infty \frac{\sqrt i}{\sqrt{2\pi \epsilon}} e^{-\frac{i}{2\epsilon}}=\epsilon+\frac{\zeta\big(-\frac12\big)}{\sqrt{2\pi\epsilon}}+O\Big(\frac1\epsilon\Big);\quad\zeta\Big(-\frac12\Big)=-0.20788...\,\,}$$ To get the result we just have to evaluate $\,\sim\sqrt\alpha\,$ term in $I_2$. $$I_3=\frac{\sqrt\alpha}{2\pi}\int_0^\infty\frac{e^t-1-t-\frac{t^2}2}{e^t-1}\frac{dt}{t^\frac52}+\frac{\sqrt\alpha}{4\pi}\int_0^\infty\frac{1+t-e^t}{e^t-1}\frac{dt}{t^\frac32}$$ Making straightforward transformations $$=\frac{\sqrt\alpha}{4\pi}\int_0^\infty\left(2-t\frac{e^t+1}{e^t-1}\right)\frac{dt}{t^\frac52}$$ Integrating by part a couple of times one can get $$I_3=\frac{2\sqrt\alpha}{3\pi}\int_0^\infty\left(\frac1{e^t-1}+\frac1{(e^t-1)^2}+\frac{e^t}{(e^t-1)^2}-\frac{te^t}{(e^t-1)^2}-\frac{2te^t}{(e^t-1)^3}\right)\frac{dt}{\sqrt t}$$ To evaluate the integral we can use analytical continuation and consider $$I(\lambda)=\int_0^\infty\left(\frac1{e^t-1}+\frac1{(e^t-1)^2}+\frac{e^t}{(e^t-1)^2}-\frac{te^t}{(e^t-1)^2}-\frac{2te^t}{(e^t-1)^3}\right)t^\lambda dt$$ choosing $\lambda>1$ to evaluate every separate integral, and finally putting $\lambda=-\frac12$.

Integrating by part whenever needed and using $\int_0^\infty\frac{t^{s-1}}{e^t-1}dt=\Gamma(s)\zeta(s)$, $$I(\lambda)=(1-\lambda)\Gamma(1+\lambda)\zeta(\lambda)$$ Choosing $\lambda=-\frac12$, we finally get $$I_3=\frac{2\sqrt\alpha}{3\pi}\,\frac{3}2\Gamma\Big(\frac12\Big)\zeta\Big(-\frac12\Big)=\sqrt{\frac\alpha\pi}\zeta\Big(-\frac12\Big)=\frac{\zeta\big(-\frac12\big)}{\sqrt{2\pi\epsilon}}$$ Numeric evaluations confirm the answer.