Ramanujan's transformation formula connected with $r_{2}(n)$

We have the identity $$\sum_{n\geq0}\frac{r_{2}\left(n\right)}{\sqrt{n+a}}e^{-2\pi\sqrt{\left(n+a\right)b}}=\frac{1}{\sqrt{\pi}}\sum_{n\geq0}r_{2}\left(n\right)\int_{0}^{\infty}e^{-\left(n+a\right)x-\pi^{2}b/x}\frac{dx}{\sqrt{x}}. \tag{1}$$ In fact $$\int_{0}^{\infty}e^{-\left(n+a\right)x-\pi^{2}b/x}\frac{dx}{\sqrt{x}}=e^{-2\pi\sqrt{\left(n+a\right)b}}\int_{0}^{\infty}e^{-\left(\sqrt{\left(n+a\right)x}-\pi\sqrt{b/x}\right)^{2}}\frac{dx}{\sqrt{x}} $$ $$\overset{x=\frac{u}{n+a}}{=}\frac{e^{-2\pi\sqrt{\left(n+a\right)b}}}{\sqrt{n+a}}\int_{0}^{\infty}e^{-\left(\sqrt{u}-\pi\sqrt{b\left(n+a\right)/u}\right)^{2}}\frac{du}{\sqrt{u}}\overset{v=\sqrt{u}}{=}\frac{e^{-2\pi\sqrt{\left(n+a\right)b}}}{\sqrt{n+a}}\int_{-\infty}^{\infty}e^{-\left(v-\pi\sqrt{b\left(n+a\right)}/v\right)^{2}}dv $$ and we recall that if $f $ is an integrable function and $k>0 $, holds (see proof) $$\int_{-\infty}^{\infty}f\left(x-\frac{k}{x}\right)dx=\int_{-\infty}^{\infty}f\left(x\right)dx $$ hence $$=\frac{e^{-2\pi\sqrt{\left(n+a\right)b}}}{\sqrt{n+a}}\int_{-\infty}^{\infty}e^{-v^{2}}dv=\frac{e^{-2\pi\sqrt{\left(n+a\right)b}}}{\sqrt{n+a}}\cdot\sqrt{\pi} $$ so we can write $(1)$ as $$=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-ax-\pi^{2}b/x}\left(\sum_{n\geq0}r_{2}\left(n\right)e^{-nx}\right)\frac{dx}{\sqrt{x}}=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-ax-\pi^{2}b/x}\theta^{2}\left(x\right)\frac{dx}{\sqrt{x}} $$ where $\theta\left(x\right)=1+2e^{-x}+2e^{-4\pi x}+\dots $ and, recalling the functional identity $$\theta\left(x\right)=\sqrt{\frac{\pi}{x}}\theta\left(\frac{\pi^{2}}{x}\right) $$ we have $$=\sqrt{\pi}\int_{0}^{\infty}e^{-ax-\pi^{2}b/x}\theta^{2}\left(\frac{\pi^{2}}{x}\right)\frac{dx}{x\sqrt{x}}=\sqrt{\pi}\sum_{n\geq0}r_{2}\left(n\right)\int_{0}^{\infty}e^{-ax-\pi^{2}\left(b+n\right)/x}\frac{dx}{x\sqrt{x}}= $$ $$=\sqrt{\pi}\sum_{n\geq0}r_{2}\left(n\right)e^{-2\pi\sqrt{\left(n+b\right)a}}\int_{0}^{\infty}e^{-\left(\sqrt{ax}-\pi\sqrt{\left(n+b\right)/x}\right)^{2}}\frac{dx}{x\sqrt{x}} $$ $$\overset{w=\frac{n+b}{x}}{=}\sqrt{\pi}\sum_{n\geq0}r_{2}\left(n\right)\frac{e^{-2\pi\sqrt{\left(n+b\right)a}}}{\sqrt{n+b}}\int_{0}^{\infty}e^{-\left(\sqrt{a\left(n+b\right)/w}-\pi\sqrt{w}\right)^{2}}\frac{dw}{\sqrt{w}}$$ $$=\sum_{n\geq0}r_{2}\left(n\right)\frac{e^{-2\pi\sqrt{\left(n+b\right)a}}}{\sqrt{n+b}}. $$