I am trying to prove that $K_1(x) = - \frac{dK_0(x)}{dx}$. 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: $K_0(x) = \int_0^\infty \frac{cos(xu) du}{\sqrt{u^2 + 1}} \Longleftrightarrow \frac{dK_0(x)}{dx} = \int_0^\infty \frac{\partial }{\partial x}\bigg[\frac{cos(xu) du}{\sqrt{u^2 + 1}}\bigg] = -\int_0^\infty \frac{u}{\sqrt{u^2 + 1}} sin(xu) du$
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. Can anyone help me prove this?
Using that integral representation for $K_{0}(x)$ is problematic because you end up with a divergent integral if you assume you can differentiate under the integral sign.
You could instead use the integral representation $$K_{v}(x) = \int_{0}^{\infty} e^{-x \cosh t} \cosh (vt) \, \mathrm dt \, , \quad x>0.$$
Then $$K_{0}(x) = \int_{0}^{\infty} e^{-x \cosh t} \, \mathrm dt \, , \quad x>0, $$ which is absolutely integrable.
And if we assume that $0 < a \le x <\infty$, then
$$ \left| \frac{\partial}{\partial x} e^{-x \cosh t} \right| = e^{-x \cosh t} \cosh (t) $$ is dominated on $[0, \infty)$ by the integrable function $e^{-a \cosh t} \cosh(t)$.
Therefore, the dominated convergence theorem permits us to differentiate under the integral sign, and we can conclude that $$ \begin{align} \frac{\mathrm d}{\mathrm d x} K_{0}(x) &= \frac{\mathrm d}{\mathrm d x} \int_{0}^{\infty} e^{-x \cosh t} \, \mathrm dt \\&= \int_{0}^{\infty} \frac{\partial}{\partial x} e^{-x \cosh t} \, \mathrm dt \\ &= - \int_{0}^{\infty}e^{-x \cosh t} \cosh(t) \, \mathrm dt \\ &= - K_{1}(x) \end{align} $$ for $x>0$.
In general, $$\begin{align} \frac{\mathrm d}{\mathrm dx} K_{v}(x) &= - \int_{0}^{\infty} e^{-x \cosh t} \cosh(vt) \cosh(t) \, \mathrm dt \\ &= - \frac{1}{2} \int_{0}^{\infty} e^{-x \cosh t} \left(\cosh \left( (v+1)x\right) + \cosh(\left(v-1)x \right)\right) \mathrm dt \\ &= - \frac{1}{2} \left( K_{v+1}(x)+ K_{v-1}(x) \right) \\ &= - \frac{1}{2} \left( K_{v+1}(x)+ K_{|v-1|}(x) \right). \end{align} $$