<s>[INST] Continued fraction for $\int_{0}^{\infty}(e^{-xt}/\cosh t)\,dt$ 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 $\pi$.
 [/INST] 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.
</s>