# Conjectured identity involving an elliptic integral and a Bessel function

Let $$K$$ be the complete elliptic integral of the first kind and $$K_\nu$$ a modified Bessel function of the second kind. I'm wondering whether the following identity is true and if so, how to prove it: $$\int_{1}^\infty e^{-2 a u^2} u^{1/2}K\left(\frac{u-1}{2u}\right) du \overset{?}{=}\sqrt{\frac{\pi}{32a}} e^{-a}K_{1/4}(a)\tag{1}$$ for $$a$$ real and strictly positive. I've searched several integral tables but could not find anything that looks related.

The reason why I suspect this might be true is the following. Start from the integral $$Z(g)=\int_{-\infty}^{\infty} \frac{dx}{\sqrt{2\pi}}\, \exp{\left(-\frac{1}{2} x^2 -\frac{g}{4!}x^4\right)} \tag{2}$$ This can be expressed in terms of a Bessel function, for example as in this question, $$Z(g)=\sqrt{\frac{3}{2\pi g}} \exp{\left(\frac{3}{4g}\right)} K_{1/4}\left(\frac{3}{4 g}\right)$$ However, going back to $$(2)$$, expanding the second exponential and exchanging the order of integration and summation, one gets the asymptotic series $$Z(g)\sim \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!n!}\left(-\frac{g}{24}\right)^n$$ Its Borel transform sums to a complete elliptic integral of the first kind, as in this question: $$\mathcal{B}Z(g)= \sum_{n=0}^\infty \frac{(4n)!}{2^{2n}(2n)!(n!)^2}\left(-\frac{g}{24}\right)^n = \frac{2}{\pi} \frac{1}{\left(1+2g/3\right)^{1/4}}K\left(\frac{-1+\sqrt{1+2g/3}}{2\sqrt{1+2g/3}}\right)$$ The Borel transform converges for $$|g| < 3/2$$, however the right hand side makes sense for all $$g > -3/2$$. Now the statement $$(1)$$ is equivalent to the assertion that $$Z$$ is equal to its Borel sum, that is $$Z(g) \overset{?}{=} \int_0^\infty e^{-t} \mathcal{B}Z(gt)$$ after replacing $$u^2 = 1+2gt/3$$ and $$a=3/(4g)$$ in the integral for cleanliness.

• Numerically, it seems to be true. Oct 4, 2021 at 13:55
• Try writing the elliptical integral as an integral and switch them if a theorem allows. Oct 4, 2021 at 14:01

With the change of integration variables $$u = \sqrt {1 + t}$$, your original identity reads $$K_{1/4} (a) = \sqrt \pi e^{ - a} \sqrt {2a} \int_0^{ + \infty } {e^{ - 2at} \frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t} - 1}}{{2\sqrt {1 + t} }}} \right)dt} .$$ By entry 19.2 on page 195 of F. Oberhettinger, L. Badii, Tables of Laplace Transforms, we have $$K_\nu (a) = \sqrt \pi e^{ - a} (2a)^{\lambda - 1/2} \int_0^{ + \infty } {e^{ - 2at} t^{\lambda - 1} \frac{1}{{\Gamma (\lambda )}}{}_2F_1 \! \left( {\nu + \tfrac{1}{2}, - \nu + \tfrac{1}{2};\lambda ; - t} \right)dt}$$ for any $$\nu \in \mathbb C$$, $$\Re a>0$$ and $$\Re \lambda >0$$. Here $${}_2 F_1$$ stands for the Gauss hypergeometric function. Thus it is enough to show that $$\frac{2}{{\pi (1 + t)^{1/4} }}K\!\left( {\frac{{\sqrt {1 + t} - 1}}{{2\sqrt {1 + t} }}} \right) = {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right),\;\;t>0.$$ This is shown in https://math.stackexchange.com/q/4267825.
Proof using the asymptotics of the Bessel function. We start with the expansion $$F(a): = \sqrt {\frac{2}{\pi a}} e^a K_{1/4} (a) = \sum\limits_{n = 0}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!(2n)!}}\frac{1}{{a^{n + 1} }}} + R_{N} (a),$$ where $$\tag{1} \left| {R_{N} (a)} \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{1}{{\left| a \right|^{N + 2} }}$$ provided that $$|\arg a|\leq\frac{\pi}{2}$$ and $$N\geq 0$$, see for instance https://doi.org/10.1098/rspa.1990.0058. Now with an arbitrary $$c>0$$ $$f(t): = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} F(a)da} = \sum\limits_{n = 1}^N {( - 1)^n \frac{{(4n)!}}{{128^n n!^2 (2n)!}}t^n } + \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} R_{N} (a)da}$$ for any $$t>0$$. Differentiating $$N$$ times gives $$\tag{2} f^{(N)} (t) = ( - 1)^N \frac{{(4N)!}}{{128^N N!(2N)!}} + \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} a^N R_{N} (a)da}$$ where, by $$(1)$$, $$\tag{3} \left| {\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{at} a^N R_{N} (a)da} } \right| \le \frac{{(4N + 4)!}}{{128^{N + 1} (N + 1)!(2N + 2)!}}\frac{{e^{ct} }}{{2c}}.$$ Stirling's formula then implies that $$\frac{{\left| {f^{(N)} (t)} \right|}}{{N!}} \le C\frac{N}{{2^N }}\frac{{e^{ct} }}{c}$$ with an absolute $$C>0$$. Thus, by Taylor's formula and the Cauchy–Hadamard theorem, $$f(t)$$ is analytic in the tubular neighbourhood $$U:=\left\{ {t:\text{dist}(t,\mathbb R^ + ) < 2} \right\}$$ of the positive axis. Employing $$(2)$$ and $$(3)$$ with $$t=0$$ and letting $$c\to +\infty$$ shows that $$\frac{{f^{(N)} (0)}}{{N!}} = ( - 1)^N \frac{{(4N)!}}{{128^N N!^2 (2N)!}},$$ i.e., the Maclaurin series of $$f(t)$$ and $${}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$$ are the same. By analytic continuation, $$f(t)={}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \frac{t}{2}} \right)$$ for any $$t\in U$$. Accordingly, $$F(a) = \int_0^{ + \infty } {e^{ - at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - \tfrac{t}{2}} \right)dt} = 2\int_0^{ + \infty } {e^{ - 2at} {}_2F_1\! \left( {\tfrac{3}{4}, \tfrac{1}{4};1; - t} \right)dt}$$ provided $$\Re a >0$$.