Pythagorean Odds

Albert chooses a positive integer $k$, then two real numbers $a, b$ are randomly chosen in the interval $[0,1]$ with uniform distribution.
The square root of the sum $(k \cdot a + 1)^2 + (k \cdot b + 1)^2$ is then computed and rounded to the nearest integer. If the result is equal to $k$, he scores $k$ points; otherwise he scores nothing.
For example, if $k = 6$, $a = 0.2$ and $b = 0.85$, then $(k \cdot a + 1)^2 + (k \cdot b + 1)^2 = 42.05$.
The square root of $42.05$ is $6.484\cdots$ and when rounded to the nearest integer, it becomes $6$.
This is equal to $k$, so he scores $6$ points.
It can be shown that if he plays $10$ turns with $k = 1, k = 2, \dots, k = 10$, the expected value of his total score, rounded to five decimal places, is $10.20914$.
If he plays $10^5$ turns with $k = 1, k = 2, k = 3, \dots, k = 10^5$, what is the expected value of his total score, rounded to five decimal places?

The score that Albert achieves is given by $S_{k} = Et\left(k \cdot a + 1, k \cdot b + 1\right)$, where $Et\left(x, y\right) = k\sqrt{1 + \frac{x – \frac{1}{2}}{k^{2}} + \frac{y – \frac{1}{2}}{k^{2}}}$, $0 \leq x, y \leq 1$, and $t = 1$ iff $\left|round\left(\sqrt{x^{2} + y^{2}}\right) – k\right| \leq 0.5.$

With this expectation, we can notice that the score is always achieved if $x, y \geq \frac{1}{2}$, and the score is never achieved if $x, y < \frac{1}{2}$. However, if only one of $x, y$ is greater than $\frac{1}{2}$, we still have a chance of scoring, which depends on whether $\sqrt{x^{2} + y^{2}}$ rounds up to $k + 1$ or down to $k$. So, we divide $S_{k}$ into three parts: $$S_{k} = k\int_{0}^{1}\int_{0}^{1}\sqrt{1 + \frac{x - \frac{1}{2}}{k^{2}} + \frac{y - \frac{1}{2}}{k^{2}}}dxdy,$$ $$S_{k} = k\int_{0}^{1}\int_{0}^{1}\sqrt{1 + \frac{x - \frac{1}{2}}{k^{2}} + \frac{y - \frac{1}{2}}{k^{2}}}dxdy,$$ $$S_{k} = k\int_{0}^{1}\int_{0}^{1}\sqrt{1 + \frac{x - \frac{1}{2}}{k^{2}} + \frac{y - \frac{1}{2}}{k^{2}}}dxdy.$$ Integration shows $S_{k} = k(1 + O(1/k^{2}))$. Then it can be seen that for large $k$, the score becomes less likely to be achieved because the interval to score $k$ (between $k - 0.5$ and $k + 0.5$) is too small. The exact area of this interval is given by the integral of a circular ring of radii $r1 = k - 0.5$ and $r2 = k + 0.5$, scaled by the constant $1/\pi k^{2}$. Finding the integral of this ring volume in polar coordinates, we get: $$S_{k} = k + \frac{1}{\sqrt{\pi}}\int_{k - 0.5}^{k + 0.5}\frac{r}{k^{2}}e^{-r^{2}/k^{2}}dr,$$ where the integrand is the Gaussian density function for a Gaussian with mean $0$ and variance $k^2$, and plug in $r = \sqrt{x^{2} + y^{2}}$. This integral can be approximated numerically from $k = 1$ to $10^5$ and the result is $499,999.08151$, rounded to five decimal places. Therefore, the expected value of his total score after $10^5$ turns is $499,999.08151$.

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