As the title says, I would like to know if there is a closed form for the integral:
\begin{align*} \int_{0}^{\infty}\ln(z)z^{\lambda - 1}\exp\left(-\frac{w}{2}\left(z + \frac{1}{z}\right)\right)\mathrm{d}z \end{align*}
I am particularly interested in references on the subject.
Yes, it has a closed form in terms of trascendental functions. What you are looking for is the derivative of this expression:
$$ \int_{0}^{\infty}x^{s-1}\exp\left(-\alpha x^h-\beta x^{-h}\right)dx = \frac{2}{h}\left(\frac{\beta}{\alpha}\right)^{\frac{s}{2h}}K_{\frac{s}{h}}(2\alpha^{\frac{1}{2}} \beta^{\frac{1}{2}}), \quad h>0,\Re a>0, \Re b>0$$
Where $K_v(x)$ is the Basset function, also called: the modified Bassel function of the third kind, Bassel's function of the second kind of imaginary parameter, Mcdonald's function or the modified Hankel function.
The reference is Erdelyi, A., Ed. (1954) Tables of Integral Transforms. Volume 1, page 313.
In the same collection Erdélyi A., Higher Transcendental Functions. Volume 2, page 5 contains a good discussion of the Basset function and similar functions.
If we start by taking $ z \rightarrow e^u$ to get $$ \int^{\infty}_{-\infty} u e^{\lambda u - w\cosh(u)} du = \Omega(w,\lambda) $$ We can then find $$ \int \Omega(w,\lambda) d\lambda = \int^{\infty}_{-\infty} e^{\lambda u - w\cosh(u)} du $$ We can then use the following identities $$ H^{(1)}_{\nu}(z) = \frac{e^{-i\frac{\pi}{2}\nu}}{i\pi}\int^{\infty}_{-\infty}e^{iz\cosh(t)-\nu t}dt $$ $$ H^{(1)}_{-\nu}(z) = e^{i\pi\nu}H^{(1)}_{\nu}(z) $$ for z in the upper half complex plane where $H^{(1)}_{\nu}$ is the Hankel function of the first kind, using the identity $$ \int \Omega(w,\lambda) d\lambda = i\pi e^{i\frac{\pi}{2}\lambda}H^{(1)}_{\lambda}(iw) $$ We can then using the following identity $$ K_{\nu}(z) = \frac{\pi}{2} i^{\nu+1}H^{(1)}_{\nu}(iz) \, , \, -\pi < arg(z) \leq \frac{\pi}{2} $$ where $K_{\nu}$ is the modified bessel function of the second kind, to then give us that... $$ \int \Omega(w,\lambda) d\lambda = 2K_{\lambda}(w) \, , \, 0 \leq \Re(w) $$ Hence $$ \int^{\infty}_{0}ln(z)z^{\lambda-1}e^{\frac{w}{2}(z+1/z)}dz = 2 \frac{\partial}{\partial \lambda} K_{\lambda}(w) \, , \, 0 \leq \Re(w) $$ Unfortunately I'm unaware of any good closed form expression for the derivitive with respect to order of the modified bessel function of the second kind. As an additional tid bit I also found this in my exploration of the problem... $$ \int^{\infty}_{0}ln(z)z^{\lambda-1}e^{\frac{w}{2}(z+1/z)}dz = \frac{\sqrt{e}ln(w)}{w^{\lambda}}e^{-\frac{w^{2}}{2}} + \sum^{\infty}_{n=-\infty} J_{n}(w) \frac{\Gamma(\lambda+n)\psi(\lambda+n)}{w^{\lambda+n}} $$ ...which when taken as a identity for the order wise derivitive of the modified bessel function is not something I've seen before.
