$\mathbf{1.}$ R. P. Boas and H. Pollard $$ \sum_{n=-\infty}^\infty \frac{\sin (n+c)\alpha}{n+c}e^{int}=\int_{-\infty}^\infty \frac{\sin (x+c)\alpha}{x+c}e^{ixt}dn,\quad 0 < \alpha < \pi, ~-\pi < t < 2\pi $$ $$ \sum_{n=-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}=\int_{-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}dn,\quad 0 < \alpha < \pi $$
Generalizations of sinc sums can be found in R. Baillie, D. Borwein, and J. M. Borwein.
$\mathbf{2.}$ D. E. Dominici, P. M. W. Gill, and T. Limpanuparb $$\sum_{n=-\infty}^\infty \frac{J_\mu (a|n|) J_\nu(b|n|)}{|n|^{\mu+\nu-2k}}= \int_{-\infty}^\infty \frac{J_\mu (a|t|) J_\nu(b|t|)}{|t|^{\mu+\nu-2k}}dt,\quad a,b\in[0,\pi],~\text{Re}(\mu+\nu-2k) > -1,k\in\mathbb{N}_0 $$
$\mathbf{3.}$ Gradsteyn and Ryzhik, 3.876.3-4 $$ \sum_{n=-\infty}^\infty\frac{\cos\pi\sqrt{a^2+\alpha^2 n^2}}{c^2+\alpha^2 n^2}=\int_{-\infty}^\infty\frac{\cos\pi\sqrt{a^2+\alpha^2 x^2}}{c^2+\alpha^2 x^2}dx,\quad 0 < \alpha < 2,~\sqrt{a^2-c^2}+\tfrac12\in \mathbb{N} $$ $$ \sum_{n=-\infty}^\infty\frac{\sin\pi\sqrt{a^2+\alpha^2 n^2}}{(c^2+\alpha^2 n^2)\sqrt{a^2+\alpha^2 n^2}}=\int_{-\infty}^\infty\frac{\sin\pi\sqrt{a^2+\alpha^2 x^2}}{(c^2+\alpha^2 x^2)\sqrt{a^2+\alpha^2 x^2}}dx,\quad 0 < \alpha < 2,~\sqrt{a^2-c^2}\in \mathbb{N} $$
$\mathbf{4.}$ This example is due to T. Osler $$ \sum_{n=-\infty}^\infty \frac{e^{i\theta n}}{\Gamma(b+\alpha n)\Gamma(1-a-\alpha n)}=\int\limits_{-\infty}^\infty\frac{e^{i\theta t}dt}{\Gamma(b+\alpha t)\Gamma(1-a-\alpha t)},\quad 0 < \alpha < 1, -\pi < \theta < \pi $$
The case $\alpha=1$ was known to Ramanujan.
$\mathbf{5.}$ Ramanujan's Notebooks, vol. II, Chapter 14, Entries 5(i),5(ii),16(i). $$ \sum_{n=0}^\infty\frac{\sin^{2k+1}\alpha n}{n}=\int_0^\infty\frac{\sin^{2k+1}\alpha x}{x}dx,\quad 0 < \alpha < \frac{2\pi}{2k+1},~k\in\mathbb{N} $$ $$ \sum_{n=0}^\infty\frac{\sin^{2k+2}\alpha n}{n^2}=\int_0^\infty\frac{\sin^{2k+2}\alpha x}{x^2}dx,\quad 0 < \alpha < \frac{\pi}{k+1},~k\in\mathbb{N} $$
$\mathbf{6.}$ K. S. Krishnan, "On the equivalence of certain infinite series and the corresponding integrals" J. Indian Math. Soc. 12, 79–88 (1948). see also Gradsteyn and Ryzhik, 3.876.1 $$ \sum_{n=-\infty}^\infty\frac{\sin\pi\sqrt{a^2+\alpha^2 n^2}}{\sqrt{a^2+\alpha^2 n^2}}=\int_{-\infty}^\infty\frac{\sin\pi\sqrt{a^2+\alpha^2 x^2}}{\sqrt{a^2+\alpha^2 x^2}}dx,\quad 0 < \alpha < 2 $$
$\mathbf{7.}$ $$ \sum_{n=-\infty}^\infty \frac{(bq^n,p/aq^n;p)_\infty}{(-zq^n,-q/zq^n;q)_\infty} =\int_{-\infty}^\infty\frac{\left(bq^x,p/aq^x;p\right)_\infty}{\left(-zq^x,-q/zq^x;q\right)_\infty}dx,\quad |p| < |q| $$
Proof of this formula can be found in this article. Example 4 is a special case of this one. Ramanujan knew the case $|p|=|q| < 1, ~|b/a| < |z| < 1$.
$\mathbf{8.}$ The special case $s=0,\alpha=\beta=1$ was conjectured by A. Rüdinger
For $m,n\in\mathbb{N},\alpha > 0, \beta > 0, 0<\alpha+\beta<2$ \begin{align} \sum _{k=-\infty}^{\infty } &\binom{m}{\alpha k+\alpha s} \binom{n}{\beta k+\beta s} \binom{m+n+\beta k+\beta s}{m+n}\\ &=\int_{-\infty}^\infty \binom{m}{\alpha x+\alpha s} \binom{n}{\beta x+\beta s} \binom{m+n+\beta x+\beta s}{m+n}dx. \end{align}
$\mathbf{9.}$ \begin{align} \sum_{n=-\infty}^\infty\ln\frac{1+e^{-\pi\alpha\sqrt{(2n+1)^2+\beta^2}}}{1+e^{-\pi\sqrt{(2n+1)^2/\alpha^2+\beta^2}}}=\int\limits_{-\infty}^\infty\ln\frac{1+e^{-\pi\alpha\sqrt{(2x+1)^2+\beta^2}}}{1+e^{-\pi\sqrt{(2x+1)^2/\alpha^2+\beta^2}}}\ dx \end{align}
There are many other examples similar to 9, see this preprint.
$\mathbf{10.}$ $$ \sum _{k=0}^\infty \frac{k^{4 n+1}}{e^{2 \pi k}-1}=\int_0^\infty\frac{x^{4n+1}}{e^{2\pi x}-1}dx,\quad n\in\mathbb{N}. $$
This formula is due to Glaisher.
$\mathbf{11.}$ If $\frac{n-1}{2}$ is odd, then $$ \sum _{x=-\infty}^\infty \frac{\cos \frac{\pi (2x+1)n}{2}-e^{-n \pi\left|x+\frac{1}{2}\right| }}{\left|x+\frac{1}{2}\right| \left(\cosh \frac{\pi(2x+1)}{2}+\cos \frac{\pi(2x+1)}{2}\right)}=\int\limits_{-\infty}^\infty \frac{\cos \frac{\pi (2x+1)n}{2}-e^{-n \pi\left|x+\frac{1}{2}\right| }}{\left|x+\frac{1}{2}\right| \left(\cosh \frac{\pi(2x+1)}{2}+\cos \frac{\pi(2x+1)}{2}\right)}\, dx $$
See Berndt, B. C. (2016). Integrals associated with Ramanujan and elliptic functions. The Ramanujan Journal, 41(1-3), 369–389..
$\mathbf{11.}$ Brede, Markus. "On Taylor-like integral representations of analytic functions." The Ramanujan Journal 19 (2009): 305-324. $$ \sum_{k=-\infty}^\infty\frac{2^k}{\left(2^k+1\right) \left(2^k+2\right)}=\int_{-\infty}^\infty\frac{2^x}{\left(2^x+1\right) \left(2^x+2\right)}\, dx=1. $$ This MSE post has been helpful in preparation of this list.
