What function $f(x)$ would the limit
$$\lim_{n\rightarrow{0^+}}\left(-\int_{-\infty}^{-x-n}\frac{xK_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da+\int_{-x+n}^{x-n}\frac{xK_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-\int_{x+n}^\infty\frac{xK_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
be equal to?
I tried doing numerical integration to get an idea of what the function would be, but didn't get to try enough of a variety of rectangle lengths or spreads of $a$ to see if it was really converging on anything. Also I know that simply subtracting $\infty$ from $\infty$ is undefined, which is the reason I expressed it as a limit.
Also $K_n(x)$ denotes the modified bessel function of the second kind.
The final answer is $$2\ \text{Chi}(x)\cosh(x) - 2\ \text{Ci}(x)\cos(x) - 2\ \text{Shi}(x)\sinh(x) - 2\ \text{Si}(x)\sin(x)+ \pi\sin(x)$$ for $x > 0$. $\text{Ci}(x)$ is the cosine integral, $\text{Si}(x)$ is the sine integral, $\text{Chi}(x)$ is the hyperbolic cosine integral, and $\text{Shi}(x)$ is the hyperbolic sine integral.
$x$ can be factored out to get:
$$x\lim_{n\rightarrow{0^+}}\left(-\int_{-\infty}^{-x-n}\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da+\int_{-x+n}^{x-n}\frac{K_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-\int_{x+n}^\infty\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
Since the first integral is equal to the third, and the integrand in the second integral is even, this simplifies to: $$x\lim_{n\rightarrow{0^+}}\left(2\int_{0}^{x-n}\frac{K_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-2\int_{x+n}^\infty\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
Using the integral definition $\big($formula $(5)\big)$ of $K_1(z)$, this becomes $$x\lim_{n\rightarrow{0^+}}\left(2\int_{0}^{x-n}\int_0^{\infty} \frac{\cos(t)}{(t^2+x^2-a^2)^{3/2}}dtda-2\int_{x+n}^\infty\int_0^{\infty} \frac{\cos(t)}{(t^2+a^2-x^2)^{3/2}}dtda\right)$$
Then switching the order of integration, this becomes $$x\lim_{n\rightarrow{0^+}}\left(2\int_0^{\infty}\int_{0}^{x-n} \frac{\cos(t)}{(t^2+x^2-a^2)^{3/2}}dadt-2\int_0^{\infty}\int_{x+n}^\infty \frac{\cos(t)}{(t^2+a^2-x^2)^{3/2}}dadt\right)$$
This simplifies to $$x\lim_{n\rightarrow{0^+}}\left(2\int_0^{\infty}\frac{\cos(t)(x-n)}{(t^2+x^2)\sqrt{2nx-n^2+t^2}}dt-2\int_0^{\infty}\cos(t)\left(\frac{1}{t^2-x^2}-\frac{n+x}{(t^2-x^2)\sqrt{n^2+t^2+2nx}}\right)dt\right)$$
Combining everything together into one integral yields $$2x\lim_{n\rightarrow{0^+}}\left(\int_0^{\infty}\cos(t)\left(\frac{x-n}{(t^2+x^2)\sqrt{2nx-n^2+t^2}}-\frac{1}{t^2-x^2}+\frac{n+x}{(t^2-x^2)\sqrt{n^2+t^2+2nx}}\right)dt\right)$$
Since this converges uniformly, $n = 0$ can be plugged in to get $$2x\left(\int_0^{\infty}\cos(t)\left(\frac{x}{t(t^2+x^2)}-\frac{1}{t^2-x^2}+\frac{x}{t(t^2-x^2)}\right)dt\right)$$
The integrand can be simplified $$2x\left(\int_0^{\infty}\cos(t)\left(\frac{x}{t(t^2+x^2)}-\frac{1}{t(t+x)}\right)dt\right)$$
Using integration by parts makes this $$f(x) = \int_{0}^{\infty}\ln\left(1+\frac{2xt}{x^{2}+t^{2}}\right)\sin\left(t\right)dt$$
Differentiating under the integral yields $$f'(x) = \int_0^{\infty}\left( \frac{2}{t+x}-\frac{2x}{t^2+x^2} \right) \sin(t) dt$$
Evaluating (with Mathematica), yields that $$f'(x) = 2\text{Ci}(x)\sin(x) + 2\text{Chi}(x)\sinh(x) - 2\text{Shi}(x)\cosh(x) - 2\text{Si}(x)\cos(x) + \pi\cos(x)$$ where $\text{Ci}(x)$ denotes the cosine integral, $\text{Si}(x)$ denotes the sine integral, $\text{Chi}(x)$ denotes the hyperbolic cosine integral, and $\text{Shi}(x)$ denotes the hyperbolic sine integral.
The antiderivative can be computed analytically, but it was somewhat long, so I will just skip over it.
Then since $f(0) = 0$, $f(x) = \int_0^x f'(u)du$, which evaluates to $$2\ \text{Chi}(x)\cosh(x) - 2\ \text{Ci}(x)\cos(x) - 2\ \text{Shi}(x)\sinh(x) - 2\ \text{Si}(x)\sin(x)+ \pi\sin(x)$$
I checked this numerically and it seems to hold for all positive $x$.