Theorem 1. The reciprocal of $\pi$ assumes the following family of integral representations
\begin{equation}
\frac{1}{\pi}= \frac{\left[\prod_{k=1}^{m+n}(2k-1)\right]^2 \lambda^{2m+1}}{(2m)!\, 2^{4m+2n}} \int_0^{\infty} \frac{x^{2m} J_{2m+1}(\lambda x)}{\Gamma\left(m+ n+\frac{1}{2} + x\right) \Gamma\left(m+n+\frac{1}{2}-x\right)}\,{\rm d}x
\end{equation}
\begin{equation}
\frac{1}{\pi}= \frac{\left[\prod_{k=1}^{m+n}(2k-1)\right]^2 \lambda^{2m+2}}{(2m+1)!\, 2^{4m+2n+1}} \int_0^{\infty} \frac{x^{2m+1} J_{2m+2}(\lambda x)}{\Gamma\left(m+n+ \frac{1}{2} + x\right) \Gamma\left(m+n+\frac{1}{2}-x\right)}\,{\rm d}x
\end{equation}
for $m=0, 1, 2, \dots$, $n=1, 2, 3,\dots$ and $\lambda\geq\pi$, where $J_n(x)$ is the Bessel function of the first kind and $\Gamma(x)$ is the Euler gamma function.
\begin{equation}
\frac{1}{\pi}= \frac{\left[\prod_{k=1}^{m+n}(2k-1)\right]^2 \lambda^{2m+1}}{(2m)!\, 2^{4m+2n}} \int_0^{\infty} \frac{x^{2m} J_{2m+1}(\lambda x)}{\Gamma\left(m+ n+\frac{1}{2} + x\right) \Gamma\left(m+n+\frac{1}{2}-x\right)}\,{\rm d}x
\end{equation}
\begin{equation}
\frac{1}{\pi}= \frac{\left[\prod_{k=1}^{m+n}(2k-1)\right]^2 \lambda^{2m+2}}{(2m+1)!\, 2^{4m+2n+1}} \int_0^{\infty} \frac{x^{2m+1} J_{2m+2}(\lambda x)}{\Gamma\left(m+n+ \frac{1}{2} + x\right) \Gamma\left(m+n+\frac{1}{2}-x\right)}\,{\rm d}x
\end{equation}
for $m=0, 1, 2, \dots$, $n=1, 2, 3,\dots$ and $\lambda\geq\pi$, where $J_n(x)$ is the Bessel function of the first kind and $\Gamma(x)$ is the Euler gamma function.