Continued fraction for $\int_{0}^{\infty}(e^{-xt}/\cosh t)\,dt$

Question

In one of the comments to a question I posted on MSE, I got this wonderful continued fraction $$\int_{0}^{\infty}\frac{e^{-xt}}{\cosh t}\,dt = \frac{1}{x}\genfrac{}{}{0pt}{}{}{+}\frac{1^{2}}{x}\genfrac{}{}{0pt}{}{}{+}\frac{2^{2}}{x}\genfrac{}{}{0pt}{}{}{+}\frac{3^{2}}{x}\genfrac{}{}{0pt}{}{}{+\dots}\tag{1}$$ Another article here (search for "Conjecture 1") mentions that this is a conjecture for $x \geq 1$. I believe this formula $(1)$ should be true whenever both sides of the equation are defined.

Let me know if there is an elementary proof available for $(1)$.

Update: As mentioned in this comment the integral in $(1)$ can be easily shown to be equal to an infinite series as follows \begin{align} \int_{0}^{\infty}\frac{e^{-xt}}{\cosh t}\,dt &= 2\int_{0}^{\infty}\frac{e^{-xt}}{e^{t} + e^{-t}}\,dt\notag\\ &= 2\int_{0}^{1}\frac{v^{x}}{1 + v^{2}}\,dv \text{ (putting }v = e^{-t})\notag\\ &= 2\int_{0}^{1}\sum_{k = 0}^{\infty}(-1)^{k}v^{x + 2k}\,dv\notag\\ &= 2\sum_{k = 0}^{\infty}(-1)^{k}\int_{0}^{1}v^{x + 2k}\,dv\notag\\ &= 2\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{x + 2k + 1}\notag \end{align}

Also it is mentioned there that there is a general result by Rogers which converts the above sum into the continued fraction on the right of $(1)$. If there are any references to this result of Rogers, it would be of great help here.

Further Update: It turns out that the continued fraction expansion mentioned above is quite helpful in approximating $\pi$.

Answer 1

Looks like the "EGF of the convergents" way is the easiest here, at least for me (here and again).

The CF in $(1)$ is a particular case of the one I consider at the beginning of this answer, but this time the things are easier, since the ODE we get is of the first order. Fix "arbitrary" $z\in\mathbb{C}$; then $$\cfrac{1}{z+\cfrac{1^2}{z+\cfrac{2^2}{\ddots+\cfrac{n^2}{z}}}}=\frac{P_{n+1}}{Q_{n+1}},$$ where $R_n=(P_n,Q_n)$ meets $R_0=(0,1)$, $R_1=(1,z)$, and $R_{n+1}=zR_n+n^2 R_{n-1}$ for $n>0$.

Then the series $R(t)=\sum_{n=0}^\infty R_n t^n/n!$ converges for $|t|<1$ and satisfies $$(1-t^2)R'(t)-(z+t)R(t)=R_1-zR_0,$$ the solution of which, under the obvious $R(0)=R_0$, is $$R(t)=\frac{R_0+(R_1-zR_0)\int_0^t(1+\tau)^{-(1+z)/2}(1-\tau)^{-(1-z)/2}\,d\tau}{(1+t)^{(1-z)/2}(1-t)^{(1+z)/2}}.$$

The singularity analysis of $R(t)$ as $|t|\to1^-$ (as covered in Analytic Combinatorics by P. Flajolet and R. Sedgewick), for $\color{red}{\Re z>0}$, gives $\lim\limits_{n\to\infty}n^{(1-z)/2}R_n/n!=L/\Gamma\big((1+z)/2\big)$, where $$L=\lim_{t\to1^-}(1-t)^{(1+z)/2}R(t)=2^{(z-1)/2}\big(R_0+(R_1-zR_0)I\big),\\I=\int_0^1(1+\tau)^{-(1+z)/2}(1-\tau)^{-(1-z)/2}\,d\tau\underset{\big[\tau=\tanh t\big]}{=}\int_0^\infty\frac{e^{-zt}\,dt}{\cosh t}.$$ Since $R_n=(P_n,Q_n)$ with $R_0=(0,1)$ and $R_1-zR_0=(1,0)$, we get $\lim\limits_{n\to\infty}P_n/Q_n=I$.

A direct approach (for a more general class of continued fractions), using recurrences for the corresponding integrals, is taken in the book The Applications Of Continued Fractions And Their Generalizations To Problems In Approximation Theory by A. N. Khovanskii (in fact hard to read for me, but perhaps would be easier if simplified for the present case).

Answer 2

In addition to the nice answer by @metamorphy, one can also transform the series into the continued fraction by expressing it as an hypergeometric function \begin{align} S(x)&=2\sum_{k = 0}^{\infty}\frac{(-1)^{k}}{x + 2k + 1}\\ &=\frac{1}{x+1}\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right) \end{align} (see below).

Then the ratio \begin{align} \frac{1}{S(x)}&=(x+1)\frac{1}{\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)}\\ &=(x+1)\frac{\,_2F_1\left( 0,0;\frac{x+1}{2};\frac12 \right)}{\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right)}\\ &=2\frac{\mathbf F\left( 0,0;\frac{x+1}{2};\frac12 \right)}{\mathbf F\left( 1,1;\frac{x+3}{2};\frac12 \right)} \end{align} where the regularized HG functions are introduced. This ratio can be expressed as a continued fraction: \begin{equation} \frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a+1,b+1;c+1;z\right)}={x _{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_{2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots}}}} \end{equation} where \begin{align} x_{n}&=c+n-(a+b+2n+1)z\\ y_{n}&=(a+n)(b+n)z(1-z) \end{align} Here, $a=b=0,c=\frac{x+1}{2},z=1/2$. Then \begin{align} x_n&=\frac x2\\ y_n&=\frac{n^2}{4} \end{align} We obtain \begin{align} \frac{1}{2S(x)}&={\frac x2+\cfrac{1^2/4}{x/2+\cfrac{2^2/4}{x/2+\cfrac{3^2/4}{x/2+\cdots}}}}\\ &={\frac x2+\cfrac{1^2/2}{x+\cfrac{2^2}{x+\cfrac{3^2}{x+\cdots}}}} \end{align} and thus \begin{equation} S(x)=\cfrac1{ x+\cfrac{1^2}{x+\cfrac{2^2}{x+\cfrac{3^2}{x+\cdots}}}} \end{equation} which is the desired result.

---

The hypergeometric form of the series can be obtained by denoting the general term of the series \begin{equation} c_k=\frac{2(-1)^{k}}{x + 2k + 1} \end{equation} the successive ratio of the terms is \begin{align} \frac{c_{k+1}}{c_k}&=-\frac{x+2k+1}{x+2k+3}\\ &=\frac{(k+1)(k+\frac{x+1}{2})}{k+\frac{x+3}{2}}\frac{-1}{k+1} \end{align} with $c_0=\frac{2}{x+1}$, the series writes \begin{equation} S(x)=\frac{2}{x+1}\,_2F_1\left( 1,\frac{x+1}{2};\frac{x+3}{2};-1 \right) \end{equation} Now, the transformation \begin{equation} \,_2F_1(a,b;c;z)=(1-z)^{-a}\,_2F_1\left(a,c-b;c;\frac z{z-1}\right) \end{equation} gives the expression \begin{equation} S(x)=\frac{1}{x+1}\,_2F_1\left( 1,1;\frac{x+3}{2};\frac12 \right) \end{equation} which is also tabulated here.