Summing up factorials (1)

We have $\begin{aligned} \sum_{r=0}^\infty \frac{(r!)^2}{(2r+1)!} & = \sum_{r=0}^\infty \frac{\Gamma(r+1)^2}{\Gamma(2r+2)} \; = \; \sum_{r=0}^\infty B(r+1,r+1) \\ & = \sum_{r=0}^\infty \int_0^1 x^r(1-x)^r\,dx \; = \; \int_0^1 \frac{1}{1-x(1-x)}\,dx \; = \; \int_0^1 \frac{1}{x^2 - x + 1}\,dx \\ & = \Big[\tfrac{2}{\sqrt{3}}\tan^{-1}\left(\tfrac{2x-1}{\sqrt{3}}\right)\Big]_0^1 \; = \; \tfrac{4}{\sqrt{3}}\tan^{-1}\tfrac{1}{\sqrt{3}} \\ & = \tfrac{4}{\sqrt{3}} \times \tfrac16\pi \; = \; \tfrac{2\pi}{3\sqrt{3}} \end{aligned}$ making the answer $2+3=\boxed{5}$ .

Notice that $r! = \int_{0}^{\infty} t^r e^{-t} \, \mathrm{d}t \stackrel{(t=x^2)}{=} \int_{0}^{\infty} 2x^{2r+1}e^{-x^2} \, \mathrm{d}x.$ Using this, $\begin{aligned} \sum_{r=0}^{\infty}\frac{r!^2}{(2r+1)!} &= \sum_{r=0}^{\infty}\frac{1}{(2r+1)!}\left( \int_{0}^{\infty} 2 x^{2r+1} e^{-x^2} \, \mathrm{d}x \right)\left( \int_{0}^{\infty} 2 y^{2r+1} e^{-y^2} \, \mathrm{d}y \right) \\ &= 4 \int_{0}^{\infty}\int_{0}^{\infty} \sum_{r=0}^{\infty} \frac{(xy)^{2r+1}}{(2r+1)!} e^{-x^2-y^2} \, \mathrm{d}x\mathrm{d}y \\ &= 4 \int_{0}^{\infty}\int_{0}^{\infty} \sinh(x y) e^{-x^2-y^2} \, \mathrm{d}x\mathrm{d}y \\ &= 2 \int_{0}^{\infty}\int_{0}^{\infty} \left( e^{-x^2+xy-y^2} - e^{-x^2-xy-y^2} \right) \, \mathrm{d}x\mathrm{d}y. \end{aligned}$ The final integral can be computed with various tools ranging from brutal-force to multivariate normal distribution. Here, we will apply the polar-coordinates change. Substituting $(x, y) = (r\cos\theta, r\sin\theta)$ , $\begin{aligned} \sum_{r=0}^{\infty}\frac{r!^2}{(2r+1)!} &= \int_{0}^{\pi/2}\int_{0}^{\infty} 2r \left( e^{-r^2(1-\cos\theta\sin\theta)} - e^{-r^2(1+\cos\theta\sin\theta)} \right) \, \mathrm{d}r\mathrm{d}\theta \\ &= \int_{0}^{\pi/2} \left( \frac{1}{1-\cos\theta\sin\theta} - \frac{1}{1+\cos\theta\sin\theta} \right) \, \mathrm{d}\theta. \end{aligned}$ Substituting $t=\tan\theta$ , $\begin{aligned} \sum_{r=0}^{\infty}\frac{r!^2}{(2r+1)!} &= \int_{0}^{\infty} \left( \frac{1}{1-t+t^2} - \frac{1}{1+t+t^2} \right) \, \mathrm{d}t \\ &= \frac{4\pi}{3\sqrt{3}} - \frac{2\pi}{3\sqrt{3}} \\ &= \frac{2\pi}{3\sqrt{3}}. \end{aligned}$ Therefore $(a, b) = (2, 3)$ and their sum is $5$ .

We can see that

$\begin{aligned} \sum_{r=0}^\infty \dfrac{(r!)^2}{(2r+1)!}& =\sum_{r=0}^\infty B(r+1,r+1)=\sum_{r=0}^\infty 2 \int_0^{\tfrac{\pi}{2}} (\sin\theta\times \cos\theta)^{2r+1} d\theta \\ & = \sum_{r=0}^\infty \int_0^{\tfrac{\pi}{2}} \dfrac{(\sin2\theta)^{2r+1}}{2^{2r}} d\theta =\int_0^{\tfrac{\pi}{2}} \sum_{r=0}^\infty \dfrac{(\sin2\theta)^{2r+1}}{2^{2r}} d\theta \\ & =\int_0^{\tfrac{\pi}{2}} \dfrac{(\sin2\theta)}{1-\dfrac{(\sin2\theta)^2}{2^2}} d\theta =4 \int_0^{\tfrac{\pi}{2}}\dfrac{(\sin2\theta)}{3+{(\cos2\theta)^2}} d\theta \\ \end{aligned}$

let $\cos2\theta=t$ , then $dt=-2\times sin2\theta d\theta$

$\begin{aligned} &=-2 \int_1^{-1}\dfrac{dt}{3+{t^2}} =\Big[\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{t}{\sqrt3}\right)\Big]_{-1}^1=\dfrac{2}{\sqrt{3}} \left( \tan^{-1}\dfrac{1}{\sqrt{3}}-\tan^{-1}\dfrac{-1}{\sqrt{3}}\right)=\dfrac{2\pi}{3\sqrt{3}}\\ \end{aligned}$

Thus, $a+b=2+3=\boxed{5}$