Integral Representation of Modified Bessel Function of the Second Kind From DLMF, formula 10.32.7
$$K_{\nu }(x)=\sec
   \left(\frac{\nu  \pi
   }{2}\right) \int_0^{\infty
   } \cos (x \sinh (t)) \cosh
   (\nu  t) \, dt$$
for $x>0$ and $| \Re(\nu )| <1$.
A derivation is given on pp. 182–183 of G. N. Watson's A Treatise on the Theory of Bessel Functions which involves rather intricate manipulations with contour integrals. I am loking for an alternate derivation based on other integral representations for $K_{\nu }(x)$ along the real line.
 A: The derivation given by Watson is a bit difficult to follow, but we can simply use a quarter-circle contour in the first quadrant if we restrict $\nu$ to the interval $(0, 1)$.

Assume that $x>0$ and $0 < \nu < 1$.
(The case $\nu =0$ is discussed here.)
Notice that
$$ \begin{align} I &= \int_{0}^{\infty} \cos(x \sinh t) \cosh(\nu t) \, \mathrm dt \\&=  \frac{1}{2} \int_{-\infty}^{\infty} \cos \left(x \, \frac{e^{t}-e^{-t}}{2} \right) \left(\frac{e^{\nu t}+e^{-\nu t}}{2} \right) \, \mathrm dt \\&= \frac{1}{4} \left(\int_{-\infty}^{\infty} \cos \left(x \, \frac{e^{t}-e^{-t}}{2} \right)e^{\nu t} \, \mathrm dt + \int_{-\infty}^{\infty}\cos \left(x \, \frac{e^{t}-e^{-t}}{2} \right) e^{- \nu t} \, \mathrm dt \right)\\ &= \frac{1}{4} \left( \int_{0}^{\infty}\cos \left(x \, \frac{u-u^{-1}}{2} \right) u^{\nu-1} \, \mathrm du - \int_{\infty}^{0} \cos \left(x \, \frac{w^{-1}-w}{2} \right)w^{\nu-1} \, \mathrm dw \right) \\ &= \frac{1}{4} \left( \int_{0}^{\infty}\cos \left(x \, \frac{u-u^{-1}}{2} \right) u^{\nu-1} \, \mathrm du + \int_{0}^{\infty} \cos \left(x \, \frac{w-w^{-1}}{2} \right)w^{\nu-1} \, \mathrm dw \right) \\&= \frac{1}{2} \int_{0}^{\infty}\cos \left(x \, \frac{u-u^{-1}}{2} \right) u^{\nu-1} \, \mathrm du \\ &= \frac{1}{2} \,  \Re\int_{0}^{\infty} \exp \left[\frac{ix}{2} \left(u - \frac{1}{u} \right) \right] u^{\nu-1} \, \mathrm du. \end{align}$$
Now let's integrate the function $$f(z)=  \exp \left(\frac{ixz}{2} \right) \exp \left(-\frac{ix}{2z} \right) z^{\nu-1}  $$ around quarter-circle contour in the first quadrant that is indented at the origin.
Since $x>0$, the magnitude of $\exp \left(- \frac{ix}{2z} \right)$ never exceeds $1$ in upper half-plane.  We can thus appeal to Jordan's lemma to argue that the integral vanishes on the big arc as the radius of the arc goes to infinity.
And since $\nu >0$, the contribution from the small quarter-circle at the origin vanishes as the radius of the quarter-circle goes to zero.
Therefore, since there are no singularities inside the contour, we get $$\int_{0}^{\infty}\exp \left[\frac{ix}{2} \left(t-\frac{1}{t}\right)\right] t^{\nu-1} \, \mathrm dt + \int_{\infty}^{0}\exp \left[\frac{ix}{2} \left(it-\frac{1}{it}\right)\right] (te^{i\pi/2})^{\nu-1} \, (i \, \mathrm dt )  = 0 ,  $$ from which we can conclude that $$ \begin{align}\int_{0}^{\infty}\exp \left[\frac{ix}{2} \left(t-\frac{1}{t}\right)\right] t^{\nu-1} \, \mathrm du &=  e^{i\pi \nu /2} \int_{0}^{\infty} \exp \left[-\frac{x}{2} \left(t + \frac{1}{t} \right)\right] t^{\nu -1} \, \mathrm dt \\ &= e^{i\pi \nu /2} \int_{-\infty}^{\infty} \exp \left(-\frac{x}{2} \left(e^{y}+e^{-y} \right) \right) e^{\nu y} \, \mathrm dy \\ &= e^{i\pi \nu /2} \int_{-\infty}^{\infty} e^{-x \cosh y} e^{\nu y} \, \mathrm dy \\ &= e^{i\pi \nu /2} \left( \int_{-\infty}^{0} e^{-x \cosh y} e^{\nu y} \, \mathrm dy + \int_{0}^{\infty}  e^{-x \cosh y} e^{\nu y} \, \mathrm dy \right) \\ &=  e^{i\pi \nu /2} \left( -\int_{\infty}^{0} e^{-x \cosh z} e^{-\nu z} \, \mathrm dz + \int_{0}^{\infty}  e^{-x \cosh y} e^{\nu y} \, \mathrm dy \right) \\&= e^{i \pi \nu /2} \left( \int_{0}^{\infty} e^{-x \cosh z} e^{-\nu z} \, \mathrm dz + \int_{0}^{\infty}  e^{-x \cosh y} e^{\nu y} \, \mathrm dy\right) \\ &= 2 e^{i \pi \nu  /2} \int_{0}^{\infty} e^{- x \cosh y} \cosh(\nu y) \, \mathrm dy \\  &= 2  e^{i\pi \nu /2} K_{\nu}(x). \end{align} $$
Therefore, $$\int_{0}^{\infty} \cos(x \sinh t) \cosh(\nu t) \, \mathrm dt =  \frac{1}{2} \, \Re \left(2 e^{i\pi \nu /2} K_{\nu}(x) \right) = \cos \left(\frac{\pi \nu}{2} \right)K_{\nu}(x). $$
Since $K_{-\nu}(x) = K_{\nu}(x)$, the result holds for $\nu \in (-1, 1)$.

The same approach also shows that $$\int_{0}^{\infty} \sin(x \sinh t) \sinh(\nu t) \, \mathrm dt =  \frac{1}{2} \,  \Im \int_{0}^{\infty} \exp \left[\frac{ix}{2} \left(u - \frac{1}{u} \right) \right] u^{\nu-1} \, \mathrm du= \sin \left(\frac{\pi \nu}{2} \right) K_{\nu}(x)$$ for $x>0$ and $\nu \in (-1,1)$.
