Integration of Bessel function of the second kind I need to calculate the below integration;
$$\int Y_0(x) \, dx,$$ where $ Y_0(x) $ is a zero order Bessel function of second kind. 
This seems like a simple integration, however I could nowhere find a solution for this. I tried in Wolform alpha alpha solution, answer comes with Struve functions which I am not familiar with.
Can anyone shed light on this matter.
 A: If you ask Wolfram Alpha to differentiate what it claims to be an antiderivative of $Y_{0}(x)$, it leaves the result in a messy form.
But you can show that indeed $$\int \text{Y}_{0}(x) \, dx = \frac{\pi x}{2} \Big(Y_{0}(x){\bf H}_{-1}(x)+Y_{1}(x) {\bf H}_{0}(x)\Big) + C \ $$
by using the identities
$$ \frac{d}{dx} {\bf H}_{\alpha}(x) = -\frac{\alpha}{x} \, {\bf H}_{\alpha}(x) + {\bf H}_{\alpha -1}(x) \tag{1}$$
and
$$\frac{d}{dx} {\bf H}_{\alpha}(x) = \frac{\alpha}{x} \, {\bf H}_{\alpha}(x) - {\bf H}_{\alpha +1}(x) + \frac{(\frac{x}{2})^{\alpha}}{\sqrt{\pi} \, \Gamma \left(\alpha + \frac{3}{2} \right)}  \tag{2}$$
which come from adding and subtracting the two recurrence relations here.
We also need the fact that for the Bessel function of the second kind (and similarly for the Bessel function of the first kind),
$$\frac{d}{dx} \, Y_{0}(x) = - Y_{1}(x) $$
and
$$ \frac{d}{dx} \, x  Y_{1}(x) = x  Y_{0}(x) .$$
Using $(2)$,
$$ \begin{align} &\frac{d}{dx}  \Big[ x Y_{0}(x){\bf H}_{-1}(x) \Big] \\ &= Y_{0}(x) {\bf H}_{-1}(x) -x Y_{1}(x) {\bf H}_{-1}(x)+ x Y_{0}(x) \left(- \frac{{\bf H}_{-1}(x)}{x} - {\bf H}_{0}(x)+ \frac{(\frac{2}{x})}{\sqrt{\pi} \, \Gamma \left( \frac{1}{2}\right)} \right) \\ &= -xY_{1}(x) {\bf H}_{-1}(x) - x Y_{0}(x) {\bf H}_{0}(x) + \frac{2 Y_{0}(x)}{\pi}. \end{align}$$
And using $(1)$,
$$ \begin{align} \frac{d}{dx} \Big[x \, Y_{1}(x) {\bf H}_{0}(x) \Big] &= xY_{0}(x){\bf H}_{0}(x) + x Y_{1}(x) \Big(0 + {\bf H}_{-1}(x) \Big) \\ &= xY_{0}(x) {\bf H}_{0}(x) + x Y_{1}(x) {\bf H}_{-1}(x) .  \end{align} $$
Therefore, 
$$ \frac{d}{dx} \left[\frac{\pi x}{2} \Big(Y_{0}(x){\bf H}_{-1}(x)+Y_{1}(x) {\bf H}_{0}(x)\Big) \right] = \frac{\pi}{2} \left(\frac{2 Y_{0}}{\pi} \right) = Y_{0}(x).$$
