Proving $\int_0^1 \exp \left(x-\frac{1}{x}\right) \, dx=\frac{1}{2} (1+\pi \pmb{H}_{-1}(2)+\pi Y_1(2))$ and generalize How to prove
$$\int_0^1 \exp \left(x-\frac{1}{x}\right) \, dx=\frac{1}{2} (1+\pi  \pmb{H}_{-1}(2)+\pi  Y_1(2))$$
Where $\pmb{H}, Y$ denotes Struve and Bessel function respectively? Any help will be appreciated.

Let $u=x-\frac{1}{x} \implies dx = \frac{1}{2}\left(1+\frac{u}{\sqrt{u^2+4}}\right)du$. We get the integral
$$\frac{1}{2}\int_{-\infty}^0 e^u\left(1+\frac{u}{\sqrt{u^2+4}}\right)du = \frac{1}{2} - \int_0^\infty \frac{t}{\sqrt{t^2+1}}e^{-2t}\:dt$$
by the further substitution $u=-2t$. Now using integration by parts on the remaining integral
$$-\sqrt{t^2+1}e^{-2t}\Bigr|_0^\infty - 2\int_0^\infty \sqrt{t^2+1}e^{-2t}\:dt = 1-2\int_0^\infty(1+t^2)^{1-\frac{1}{2}}e^{-2t}\:dt$$
we also have that
$$\int_0^\infty (1+t^2)^{\alpha-\frac{1}{2}}e^{-xt}\:dt = \frac{\sqrt{\pi}\Gamma\left(\alpha+\frac{1}{2}\right)}{2\left(\frac{x}{2}\right)^\alpha}K_\alpha(x)$$
from the integral representation of $K_\alpha(x)$, the Struve function of the second kind. Plugging in, the integral becomes
$$1-2\left(\frac{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}{2}K_1(2)\right)=1-\frac{\pi}{2}K_1(2)$$
leaving the final answer as
$$\frac{3}{2}-\frac{\pi}{2}K_1(2) \equiv \frac{1}{2}(3 - \pi H_1(2) + \pi Y_1(2)) \approx 0.31594$$
which is numerically equivalent to your form of the answer.
