Consider the integral $$ \int_{0}^{\infty} \left\{\frac{1}{\sqrt{t}}\right\}e^{-t/a} \mathrm dt $$ where $ \{x\} =x-[x] $ is the fractional part of $x\in \Bbb R$. Using the representation of the gamma function plus analytic continuation
$$ \Gamma (s) = \int_{0}^{\infty}t^{s-1} \mathrm e^{-t}\mathrm dt $$ or is there an straight calculation to this integral ?? thanks.
I wonder whether a nice expression of this integral exists (that is you know it), or you don't know. The substitution $t = y^{-2}$ gives us an integral $$ 2\int_0^\infty \frac{\{y\}}{y^3}\mathrm e^{-1/ay^2}\mathrm dy = 2\sum_{k=0}^\infty\int_k^{k+1} \frac{y-k}{y^3}\mathrm e^{-1/ay^2}\mathrm dy. $$ The latter integral can be evaluated e.g. in Mathematica for $a>0$: $$ \int_k^{k+1} \frac{y-k}{y^3}\mathrm e^{-1/ay^2}\mathrm dy = \frac{k}{2a}\left(\mathrm e^{-a/k^2} - \mathrm e^{-a/(k+1)^2}\right) + \sqrt{\frac{\pi}{2a}}\left(\Phi\left(\frac{\sqrt a}{k} \right)- \Phi\left(\frac{\sqrt a}{k+1}\right)\right) $$ where $\Phi$ is the error function $\mathrm{Erf}$. As a result, the second part of the sum is telescopic and gives you $$ \sum_{k=0}^\infty \sqrt{\frac{\pi}{2a}}\left(\Phi\left(\frac{\sqrt a}{k} \right)- \Phi\left(\frac{\sqrt a}{k+1}\right)\right) = \sqrt\frac{\pi}{2a} $$ while for the first part you have $$ \sum_{k=0}^\infty \frac{k}{2a}\left(\mathrm e^{-a/k^2} - \mathrm e^{-a/(k+1)^2}\right) $$ which as well may be further simplified - I am not sure at the moment.
Updated: It obviously holds that $$ \sum_{k=0}^\infty k (f_k -f_{k+1}) = f_1 - f_2 +2f_2 - 2f_3 + 3f_3 -3f_4 + \dots $$ $$ =f_1+f_2+f_3+\dots = \sum_{k=1}^\infty f_k. $$ If I am not mistaken, then $$ \sum_{k=0}^\infty \frac{k}{2a}\left(\mathrm e^{-a/k^2} - \mathrm e^{-a/(k+1)^2}\right) = \frac1{2a} \sum_{k=1}^\infty \mathrm e^{-a/k^2} =\infty $$ and so the original integral is infinite.