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Is there any standard way to evaluate the following summation?

$$ S(x):=\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!} $$

where $(k+1/2)! = (k+1/2)(k-1/2)\ldots (1/2)$.

EDIT : As some of you asked for motivation for this question, the only thing that I can come up with is that I was trying to think of an analogous formula for half integral arguments $s$,

$$ \Gamma(s,x) = (s-1)! e^{-x} \sum_{m=0}^{s-1} \frac{x^m}{m!} $$

Krishnarjun
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  • I guess this can only be calculated numerically. What does Wolfram Alpha reply ? – Peter Oct 13 '21 at 11:26
  • I haven't checked Wolfram Alpha yet... Didn't think of that... – Krishnarjun Oct 13 '21 at 11:27
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    Your Question would be improved by adding context, such as what motivated your interest in this problem, etc. – hardmath Oct 13 '21 at 14:08
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    Note that $y=x^{1/2}S(x)$ solves $$y'=y+\frac{1}{4\sqrt{x}}.$$ Solving this differential equation (for example, by using the method of integrating factor) and using $S(0)=1$, you can determine $S(x)$. – Sangchul Lee Oct 13 '21 at 19:26
  • @hardmath The expression popped up in one of my research works. I just wanted to make sure that there are no standard summation formulas before I went and tried to solve it from scratch... – Krishnarjun Oct 14 '21 at 07:06

1 Answers1

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$$S(x)=\sum_{k=0}^\infty \frac{x^k}{(k+1/2)!}=\sum_{k=0}^\infty \frac{(2x)^k (2k)!!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k k!}{(2k+1)!}=\sum_{k=0}^\infty \frac{(4x)^k }{k!}\frac{(k!)^2}{(2k+1)!}$$ Given that $$\frac{(k!)^2}{(2k+1)!}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=B(k+1;k+1)$$ $$S(x)=\sum_{k=0}^\infty \frac{(4x)^k }{k!}\int_0^1t^k(1-t)^kdt$$ Changing the order of summation and integration and summing the series $$S(x)=\int_0^1e^{4xt(1-t)}dt\,=\,\Big(t=z+\frac{1}{2}\Big)\,\,\int_{-\frac{1}{2}}^{\frac{1}{2}}e^{4x(\frac{1}{4}-z^2)}dz=2e^x\int_0^\frac{1}{2}e^{-4xz^2}dz$$ $$S(x)=\frac{e^x}{\sqrt x}\int_0^\sqrt xe^{-t^2}dt=\frac{\sqrt\pi}{2}\frac{e^x}{\sqrt x}\text{erf}(\sqrt x)$$

Svyatoslav
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  • I had thought of another method a few minutes after posting the question. It involved mutiplying $S(x)$ by $\sqrt(x)$ and then differentiating and solving the resulting linear differential equation using some online computers. I got the same result. I am happy to see another 'hands on' proof. – Krishnarjun Oct 13 '21 at 13:13
  • @Krishnarjun this is nice that our answers coincides - it means the answer is correct. Good luck :) – Svyatoslav Oct 13 '21 at 13:18