Accurate identities related to $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n$ and $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^4}x^n$

Question

Some time ago while playing around with maths, I derived the following pair of formulae:

$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}e^{4x\cos^2\theta}\;d\theta\tag{1}$$

$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^4}x^n=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}I_{0}\left(4\sqrt{x}\cos\theta\right)\;d\theta\tag{2}$$

as well as the following generalizations:

$$\sum_{n=0}^{\infty}\frac{(2n)!^m}{(n!)^{2m+1}}x^n=\left(\frac{2}{\pi}\right)^m\int_0^{\frac{\pi}{2}}\ldots\int_0^{\frac{\pi}{2}}e^{4^{m}x\prod_{k=1}^{m}\cos^2\theta_k}\;d\theta_1\ldots\;d\theta_m\tag{3}$$

$$\sum_{n=0}^{\infty}\frac{(2n)!^m}{(n!)^{2m+2}}x^n=\left(\frac{2}{\pi}\right)^m\int_0^{\frac{\pi}{2}}\ldots\int_0^{\frac{\pi}{2}}I_0\left(2^{m+1}\sqrt{x}\prod_{k=1}^{m}\cos\theta_k\right)\;d\theta_1\ldots\;d\theta_m\tag{4}$$

(In what follows $J_0$ and $I_0$ and Bessel are modified Bessel functions respectively of the first kind). The method I used to find these was extremely formal (in the second sense of this answer) and I know that such derivations can give spurious results, so my question is: are these formulae correct?

(Note that I have asked them in the same question because my derivation below proved them together, requiring one to prove the other. These series could be written in terms of binomial coefficients but this page does not seem to contain any formulae like these involving powers of binomial coefficients in the numerator of the summand.)

[Edit: I have just noticed this question which appears to mention a special case of $(2)$ (modified by the formal identity $J_0(ix)=I_0(x)$).]

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Outline of formal derivation: I started with the Laplace transform formula $L\left[t^{\frac{n}{2}}\int_{0}^{\infty}u^{-\frac{n}{2}}J_n(2\sqrt{ut})f(u)\;du\right](s)=\frac{1}{s^{n+1}}L[f(t)]\left(\frac{1}{s}\right)$ from Borelli and Coleman's Differential Equations: a modelling perspective (the proof is not too difficult), which formally implies that if $f(t)=\sum\limits_{n=0}^{\infty}a_{n}t^n$ then:

$$\sum_{n=0}^{\infty}\frac{a_{n}t^n}{n!}=L^{-1}\left[\frac{f(\frac{1}{s})}{s}\right](t)=\int_0^\infty J_0(2\sqrt{ut})L^{-1}[f(s)](u)\;du\tag{5}$$

Taking $f(t)=\sqrt{\frac{1}{1-t}}=\sum\limits_{n=0}^{\infty}\frac{(2n)!t^n}{(n!)^{2}4^n}$ (a binomial series) in $(5)$, by convolution and shifting we get $\sum\limits_{n=0}^{\infty}\frac{(2n)!t^n}{(n!)^{3}4^n}=L^{-1}\left[\frac{1}{s}\sqrt{\frac{1}{1-\frac{1}{s}}}\right](t)=\frac{1}{\pi}\int_0^t \frac{e^u}{\sqrt{u(t-u)}}\;du$, and $\color{#ff0000}{(1)}$ follows by setting $u=\frac{t\left(\cos(2\theta)+1\right)}{2}$. Taking $f(t)=\sum\limits_{n=0}^{\infty}\frac{(-1)^n(2n)!}{(n!)^3}t^n$ in $(5)$ and using $(1)$ and elementary trigonometric identities gives $\sum\limits_{n=0}^{\infty}\frac{(-1)^n(2n)!}{(n!)^4}t^n=\frac{1}{\pi}\int_0^\infty J_0(2\sqrt{ut})L^{-1}\left[e^{-2s}\int_0^\pi e^{-2s\cos\theta}\;d\theta\right](u)\;du$. But making the transformation $\theta=\cos^{-1}\left(\frac{t-2}{2}\right)$, the argument of the inverse Laplace transform becomes $\int_0^4\frac{e^{-st}}{\sqrt{4t-t^2}}\;dt$ which is the Laplace transform of $\frac{1-H(t-4)}{\sqrt{4t-t^2}}$ ($H(t)$ is the Heaviside function) so our formula reduces to $\frac{1}{\pi}\int_0^\infty J_0(2\sqrt{ut})\frac{1-H(u-4)}{\sqrt{4u-u^2}}\;du$ which by taking $u=2(1+\cos2\theta)$ and formally setting $J_0(ix)=I_0(x)$ gives $\color{#ff0000}{(2)}$.

Using the resulting formula $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^4}(-1)^n t^{2n}=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}J_{0}\left(4t\cos\theta\right)\;d\theta$ and taking the Laplace transform of both sides and interchanging integral and Laplace transform gives $\sum\limits_{n=0}^{\infty}\frac{(2n)!^2}{(n!)^4}\frac{(-1)^n}{ s^{2n+1}}=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{s^2+16\cos^2\theta}}\;d\theta$, which gives:

$$\sum_{n=0}^{\infty}\frac{(2n)!^2}{(n!)^4}(-1)^n t^n=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1+16t\cos^2\theta}}\;d\theta\tag{6}$$

This is an elliptic integral I believe. Applying $(5)$ to this, interchanging integral and Laplace transform, applying convolution and shifting, and taking $u=\frac{t}{2}(1+\cos2\varphi)$ lets us arrive at:

$$\sum_{n=0}^\infty \frac{(2n)!^2}{(n!)^5}x^n=\frac{4}{\pi^2}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}e^{16x\cos^2\theta\cos^2\varphi}\;d\theta\;d\varphi\tag{7}$$

The methods of deriving $(2)$ from $(1)$ and of deriving $(7)$ from $(2)$ may be easily extended to an induction deriving $\color{#ff0000}{(3)}$ and $\color{#ff0000}{(4)}$.

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Accuracy: The error of $(1)$ for $x=0.02$ according to Wolfram|Alpha is on the order of $10^{-16}$ which seems extremely small although I am not sure about the sizes of errors in Wolfram|Alpha (and there are very accurate near identities; I have also derived elsewhere identities with errors of $10^{-16}$ which appear to be mere approximations). I am not sure about the accuracy of $(2)$, $(3)$, or $(4)$ but noting the relation of $(6)$ to elliptic integrals we get the following $AGM$ formula:

$$\sum_{n=0}^{\infty}\frac{(2n)!^2}{(n!)^4}x^n=\frac{1}{AGM(\sqrt{1-16x},1)}\tag{8}$$

whose error at $x=0.02$ appears to be also on the order of $10^{-16}$. This seems to imply that the derivation may still be on the right track when $(6)$ is shown.

So my question is: Are some or all of the above (highlighted) identities correct?

Answer 1

Herein we present solutions to the $(1)$ and $(3)$ in the OP based on exploiting the uniform convergence of the series for $e^x$ on compact intervals. To that end, we proceed.

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To show that $(1)$ in the OP is correct, we proceed as follows.

$$\begin{align} \frac2\pi\int_0^{\pi/2}e^{4x\cos^2(\theta)}\,d\theta&=\frac2\pi\int_0^{\pi/2}\sum_{n=0}^\infty \frac{(4x)^n}{n!}\cos^{2n}(\theta)\,d\theta\\\\ &=\frac2\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\int_0^{\pi/2}\cos^{2n}(\theta)\,d\theta\\\\ &=\frac2\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\frac12 B\left(\frac12,n+\frac12\right)\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\frac{\Gamma(1/2)\Gamma(n+1/2}{\Gamma(n+1)}\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{\sqrt\pi(4x)^n}{(n!)^2}\Gamma(n+1/2)\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{\sqrt\pi(4x)^n}{(n!)^2}\frac{\sqrt \pi(2n-1)!!}{2^n}\\\\ &=\sum_{n=0}^\infty \frac{x^n\,(2n)!}{(n!)^3} \end{align}$$

as was to be shown!

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We can use the previous result to show that $(3)$ in the OP is indeed correct. Note that we have

$$\begin{align} \overbrace{\int_0^{\pi/2}\cdots \int_0^{\pi/2}}^{m\,\text{ terms}} e^{4^mx\prod_{k=1}^m \cos^2{\theta_k}}\,d\theta_1\cdots\,d\theta_m&=\sum_{k=0}^\infty \frac{(4^{mk})x^k}{k!}\left(\int_0^{\pi/2}\cos^{2k}(\theta)\,d\theta\right)^m\\\\ &=\sum_{k=0}^\infty \frac{(4^{mk})x^k}{k!}\left(\frac{\pi}{2}\frac{(2k)!}{4^k(k!)^2}\right)^m\\\\ &=\left(\frac{\pi}{2}\right)^m\sum_{k=0}^\infty \frac{\left((2k)!\right)^m}{(k!)^{2m+1}}\,x^k \end{align}$$

After multiplying by $(2/\pi)^m$, we arrive at the result $(3)$!