The Spacelike Propagator

Question

I am having trouble showing that the spacelike correlation function (Peskin and Schroeder exercise 2.3) is a function of $K_1$, as presented in the solutions manual. This manual mentions that we should use contour integration. A post regarding this exercise can be found here (Bessel function representation of spacelike KG propagator) However, I am not familiar with contour integration and complex analysis, and thus cannot tell how once arrives at (3) (see the link) from the equation right before (3). To that end, I tried to avoid contour integration entirely. I found that (letting $u = \frac{p}{m}$ and $du = \frac{dp}{m}$)

$D(x - y) = \frac{1}{mr(2\pi)^2} \int_0^\infty \frac{mu \times sin(mru) mdu}{\sqrt{u^2 + 1}} = \frac{m}{(2\pi)^2 r}\int_0^\infty \frac{u}{\sqrt{u^2 + 1}} sin(mru) du$

From this link (https://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html) I found that

$K_\nu(x) = \frac{\Gamma(\nu + \frac{1}{2}) (2x)^\nu}{\sqrt{\pi}} \int_0^\infty \frac{cos(t) dt}{(t^2 + x^2)^{\nu + \frac{1}{2}}} = \frac{\Gamma(\nu + \frac{1}{2}) (2x)^\nu}{\sqrt{\pi}} \int_0^\infty \frac{cos(t) dt}{x^{2\nu + 1}((\frac{t}{x})^2 + 1)^{\nu + \frac{1}{2}}}$.

From this I found: $K_0(x) = \int_0^\infty \frac{cos(xu) du}{\sqrt{u^2 + 1}} \Longleftrightarrow \frac{dK_0(x)}{dx} = \int_0^\infty \frac{\partial }{\partial x}\bigg[\frac{cos(xu) du}{\sqrt{u^2 + 1}}\bigg] = -\int_0^\infty \frac{u}{\sqrt{u^2 + 1}} sin(xu) du$

and from this I can conclude: $D(x - y) = - \frac{m}{(2\pi)^2 r} \frac{dK_0(x)}{dx}\bigg|_{x = mr}$.

Now I need to prove that $-\frac{dK_0(x)}{dx} = K_1(x)$. To this end I found:

Next we note that $\Gamma (\frac{3}{2}) = \frac{\sqrt{\pi}}{2}$. We also note that the integration by parts formula is $\int_0^\infty u dv = uv|_{0}^\infty - \int_0^\infty v du$. In the integral that follows, we let $u = cos(t)$ and $dv = \frac{x}{(t^2 + x^2)^{3/2}} dt$. From this we see that $du = -sin(t) dt$. Now let $t = xtan(\beta)$; thus $dt = x sec^2(\beta) d\beta$ and $t^2 + x^2 = x^2 sec^2(\beta)$. Hence we see $dv = \frac{x dt}{(t^2 + x^2)^{3/2}} = \frac{1}{x} cos(\beta) d\beta$. From this we see $v = \frac{sin(\beta)}{x} = \frac{t}{x\sqrt{x^2 + t^2}}$. Thus we see:

\begin{align} K_1(z) &= \int_0^\infty \frac{x}{(t^2 + x^2)^{3/2}}cos(t) dt = \bigg[\frac{t cos(t)}{x \sqrt{x^2 + t^2}} \bigg]_0^\infty + \int_0^\infty \frac{t}{x\sqrt{x^2 + t^2}} sin(t) dt \end{align}

This is close to what I want, but the term in the big square bracket diverges.

My Questions: Is there I way I can use the integral formula to finish my proof? If not, how can I arrive at equation (3) --- see the stack exchange link I provided--- from the equation right above (3)?