Limit of an integral involving the floor function I am trying to obtain an asymptotic expansion of
$$\int_t^\infty \lfloor x\rfloor \frac {x}{\sqrt{x^2-t^2}} \ \Bbb d x$$
for $t \to \infty$ and $\lfloor x \rfloor $ denoting the floor function. I tried to reduce it to well known integrals that relate the floor function with the zeta function or the Euler's gamma constant, but without success. After several calculations, I obtained the expansion $$ t^2/8 + t \log (t)/2 -t+O (\log (t)) $$ but would like to have confirmation of this. Also, I believe that a more accurate estimate could be obtained, probably identifying a coefficient for the $\log (t) $ term and arriving to a $O(1 ) $ error term.
 A: First we shall calculate the indefinite integral:
$$\int \lfloor x \rfloor \frac {x}{\sqrt{x^2 -t^2}}dx$$
We can use a trick here known as implied integration. What an implied integral is, is an alternate version of the antiderivative that holds piecewise constant functions fixed and varies by piecewise constant functions. Therefore we can use implied integration and make the following two substitutions:
$$\lfloor x \rfloor = a$$
$$x^2 - t^2 = u$$
$$2x = du$$
$$x = \frac {du}2$$
Therefore, continuing on:
$$\int a \frac {1}{2\sqrt{u}}dx = a \int \frac {1}{2\sqrt{u}}dx = a \sqrt{u} = \lfloor x \rfloor \sqrt{x^2 - t^2} + C(x)$$
Of course, this is merely the implied integral. This isn't the true indefinite integral (hence the C(x) term). We want to find the function C(x) such that the particular implied integral is a continuous indefinite integral. There is a known formula for forms involving floor(x) and it is given below:
$$C(x) = -\sum_{n = 0}^{\lfloor x \rfloor} (lim_{a \to n^+} \lfloor a \rfloor \sqrt{a^2 - t^2} - lim_{a \to n^-} \lfloor a \rfloor \sqrt{a^2 - t^2})$$
$$C(x) = -\sum_{n = 0}^{\lfloor x \rfloor} (n \sqrt{n^2 - t^2} - (n - 1) \sqrt{n^2 - t^2})$$
$$C(x) = -\sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}$$
I am not sure how to proceed from here in terms of reducing the summation; however, we do not need to reduce it to continue.
$$\int_t^{\inf} \lfloor x \rfloor \frac {x}{\sqrt{x^2 -t^2}}dx = \lim_{x \to \inf} (\lfloor x \rfloor \sqrt{x^2 - t^2} - \sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}) - (\lfloor t \rfloor \sqrt{t^2 - t^2} - \sum_{n = 0}^{\lfloor t \rfloor} \sqrt{n^2 - t^2}) = \lim_{x \to \inf} (\lfloor x \rfloor \sqrt{x^2 - t^2} - \sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}) - (\lfloor t \rfloor \sqrt{0} - 0) = \lim_{x \to \inf} (\lfloor x \rfloor \sqrt{x^2 - t^2} - \sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}) = ln(e^{\lfloor x \rfloor \sqrt{x^2 - t^2} - \sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}}) = ln(\frac {e^{\lfloor x \rfloor \sqrt{x^2 - t^2}}} {e^{\sum_{n = 0}^{\lfloor x \rfloor} \sqrt{n^2 - t^2}}})) = ???$$
I simply lack the knowledge in series to continue onward. I believe L'hopital's rule would come into play, but the piecewise constant terms will just break it severely as they (mostly) evaluate as 0 under the derivative. I believe it is negatively divergent, but there is no way to tell with my skills.
