Ceiling Integral

Consider the function $f(x) = \left \lceil \sqrt{x+\frac 14} + \frac 12 \right \rceil$ for $x \ge - \frac 14$ and note that

$\begin{array} {rll} f(x) = 1 & \rm when & -\frac 14 \le x \le 0 \\ f(x) = 2 & \rm when & 0 < x \le 2 \\ f(x) = 3 & \rm when & 2 < x \le 6 \\ f(x) = 4 & \rm when & 6 < x \le 12 \\ \cdots & \cdots & \cdots \\ \implies f(x) = n & \rm when & (n-1)(n-2) < x \le n(n-1) \text{ for }n \ge 0 \end{array}$

Then we have

$\begin{aligned} I & = \int_{-\frac 14}^\infty e^{1-\left \lceil \sqrt{x+\frac 14} + \frac 12 \right \rceil} dx \\ & = \int_{-\frac 14}^0 e^0 dx + \int_0^2 e^{-1} dx + \int_2^6 e^{-2} dx + \int_6^{12} e^{-3} dx + \cdots \\ & = 0 - \left(-\frac 14\right) + \frac {2-0}e + \frac {6-2}{e^2} + \frac {12-6}{e^3} + \cdots \\ & = \frac 14 + \frac 2e + \frac 4{e^2} + \frac 6{e^3} + \frac 8{e^4} + \cdots \\ & = \frac 14 + \frac 2e \left(1 + \frac 2e + \frac 3{e^2} + \frac 4{e^3} + \cdots \right) \\ & = \frac 14 + \frac 2e \cdot \frac d{dx} \bigg[ x+x^2 + x^3 + x^4 + \cdots \bigg]_{x =1/e} \\ & = \frac 14 + \frac 2e \cdot \frac d{dx} \left[\frac x{1-x} \right]_{x =1/e} & \small \blue{\text{for }|x| < 1} \\ & = \frac 14 + \frac 2e \left[\frac 1{(1-x)^2} \right]_{x =1/e} \\ & = \frac 14 + \frac {2e}{(e-1)^2} \approx \boxed{2.09} \end{aligned}$

Step 1: Find discontinuities

Because the expression contains a ceiling function, we know that the integrand is discontinuous whenever $\sqrt{x + \frac{1}{4}} + \frac{1}{2}$ is an integer. Therefore, for any $n \in \mathbb{Z}$ , the function is discontinuous for any $x$ satisfying:

$n = \sqrt{x + \frac{1}{4}} + \frac{1}{2}$

Solving for x:

$x = n^2 - n$ , for any non-negative $n \in \mathbb{Z}$

Step 2: Rewrite the integral

From here, we can rewrite the integral, but first it's easier to split it up as $- \frac{1}{4}$ isn't an integer.

$\displaystyle\int_{-\frac{1}{4}}^{\infty}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx = \displaystyle\int_{-\frac{1}{4}}^{0}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx + \displaystyle\int_{0}^{\infty}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx$

We can rewrite the second integral as a sum now that we know the general form for the x values which make the function discontinuous:

$\displaystyle\int_{0}^{\infty} e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx = \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{n^{2} - n}^{n^2 + n}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx$

We can simplify this even further, as we know on the interval $(n^{2} - n, n^{2} + n)$ , $\lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil = n + 1$

$\displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{n^{2} - n}^{n^2 + n}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx = \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{n^{2} - n}^{n^2 + n}\ e^{-n} dx$

Our final expression becomes:

$\displaystyle\int_{-\frac{1}{4}}^{\infty}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx = \displaystyle\int_{-\frac{1}{4}}^{0}\ e^{0} dx + \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{n^{2} - n}^{n^2 + n}\ e^{-n} dx$

Step 3: Evaluate the integral

$\displaystyle\int_{-\frac{1}{4}}^{0}\ 1 dx = \frac{1}{4}$

$\displaystyle\int_{n^{2} - n}^{n^2 + n}\ e^{-n} dx = (\frac{1}{e})^{n}x\big|_{n^{2}-n}^{n^{2}+n} = (\frac{1}{e})^{n}(n^{2} + n) - (\frac{1}{e})^{n}(n^{2} - n) = 2n(\frac{1}{e})^{n}$

$\frac{1}{4} + 2\displaystyle\sum_{n=1}^{\infty}n(\frac{1}{e})^{n}$

Step 4: Evaluate the sum

The sum

$\displaystyle\sum_{n=1}^{\infty}n(\frac{1}{e})^{n}$

can be evaluated as:

$S = 1(\frac{1}{e})^{1} + 2(\frac{1}{e})^{2} + 3(\frac{1}{e})^{3} + 4(\frac{1}{e})^{4} + ...$

$-\frac{1}{e}S = -1(\frac{1}{e})^{2} - 2(\frac{1}{e})^{3} - 3(\frac{1}{e})^{4} - 4(\frac{1}{e})^{5} + ...$

$(1-\frac{1}{e})S = (\frac{1}{e})^{1} + (\frac{1}{e})^{2} + (\frac{1}{e})^{3} + (\frac{1}{e})^{4} + ...$

$(1-\frac{1}{e})S = \frac{\frac{1}{e}}{1 - \frac{1}{e}}$

$(\frac{e-1}{e})S = \frac{1}{e-1}$

$S = \frac{e}{(e-1)^{2}}$

Therefore,

$\displaystyle\int_{-\frac{1}{4}}^{\infty}\ e^{1 - \lceil \sqrt{x + \frac{1}{4}} + \frac{1}{2} \rceil} dx = \frac{1}{4} + \frac{2e}{(e-1)^{2}} = 2.091$