I'm looking for a way of evaluating $$\int_0^\pi\sin x \exp(-a/\sin x)dx$$ to get a second order Bickley function $K_2(a)$, which is basically the same integral, but $\cos x$ instead of $\sin x$ and the limits change from $0$ to $\pi/2$, which is understandable.
I'm a bit lost what kind of variable substitution could I do. Any suggestions? I have tried $-a/\sin x = u$, but that doesnt seem to give reasonable results. Thanks in advance!
Elaborating on my comment.
Though I now consider the limits to be from $0$ to $\pi/2$, so the final function would have to be multiplied by $2$.
Substitution: $$u=\frac{1}{\sin x}$$
makes the integral:
$$f(a)=\int_1^\infty e^{-a u} \frac{du}{u^2 \sqrt{u^2-1}}=\int_0^\infty \frac{dt}{\cosh^2 t}e^{-a \cosh t}$$
This makes it obvious that:
$$f''(a)=K_0(a)$$
Where the initial conditions can be easily found from the original integral:
$$f(0)=1 \\ f'(0)=-\frac{\pi}{2}$$
Does $f(a)$ have a closed form? Probably not a nice one. We can always use numerical methods.