A tough integral : $\int_{0}^{\infty }\frac{\sin x \text{ or} \cos x}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$ Recently, I found these two interesting integrals in Handbook of special functions page 141.
$$\mathcal{I}=\int_{0}^{\infty }\frac{\sin(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$
$$\mathcal{J}=\int_{0}^{\infty }\frac{\cos(ax)}{\sqrt{x^{2}+z^{2}}}\ln\left ( x^{2}+z^{2} \right )\mathrm{d}x$$
In this book, it gives the answer below
$$\mathcal{I}=\frac{\pi }{2}\left (\ln\frac{z}{2a}-\gamma   \right )\left [ I_0\left ( az \right )- \mathbf{L}_0\left ( az \right )\right ]+\frac{1}{4\pi }G_{24}^{32}\left ( \frac{a^{2}z^{2}}{4}\middle|\begin{matrix}
 \dfrac{1}{2},\dfrac{1}{2} \\ 
0,0,\dfrac{1}{2},\dfrac{1}{2}
\end{matrix} \right )~~~,~~~\left (a,\Re z>0  \right )$$
$$\mathcal{J}=\left ( \ln\frac{z}{2a}-\gamma  \right )K_0\left ( az \right )~~~,~~~\left ( a,\Re z>0 \right )$$
where $I_0(\cdot)$ is modified bessel function of the first kind, $\mathbf{L}_0(\cdot)$ is modified struve function, $G_{pq}^{mn}(\cdot)$ is meijer-G function and $K_0(\cdot)$ is bessel function of rhe second kind.
So, I tried to figure out how to get the answer.

My attempt:
Let $x=z\tan t$, we have
\begin{align*}
\mathcal{I}&=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( z\sec t \right )\sec t\, \mathrm{d}t\\
&=2\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\ln\left ( \sec t \right )\sec t\, \mathrm{d}t+2\ln z\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec t\, \mathrm{d}t
\end{align*}
Hence, define
$$\mathcal{I}\left ( m \right )=\int_{0}^{\frac{\pi }{2}}\sin\left ( az\tan t \right )\sec^mt\, \mathrm{d}t$$
then using the taylor series of  $\sin x$ we get
\begin{align*}
\mathcal{I}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k+1}t\sec^mt\, \mathrm{d}t \\
&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k+1}}{\left ( 2k+1 \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k+1}t\cos^{-2k-m-1}t\, \mathrm{d}t
\end{align*}
By using the same way we get
$$\begin{align*}
\mathcal{J}\left ( m \right )&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\tan^{2k}t\sec^mt\, \mathrm{d}t \\
&=\sum_{k=0}^{\infty }\left ( -1 \right )^{k}\frac{\left ( az \right )^{2k}}{\left ( 2k \right )!}\int_{0}^{\frac{\pi }{2}}\sin^{2k}t\cos^{-2k-m}t\, \mathrm{d}t
\end{align*}$$
But how to evaluate the last integral, it seems can't be expressed by Beta function.
If I'm doing the wrong way, is there another way to solve the problem.
Any help will be appreciated!
 A: For the second integral 
Note that 
$$ K_\nu(az)=\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos at }{(t^2+z^2)^{\nu+1/2}} dt$$
By differentiation with respect to $\nu$
\begin{align}
\frac{\partial K_\nu(az)}{ \partial \nu} &=(\Gamma'(\nu+1/2)+\log(2z/a) )K_\nu(az)\\&-\frac{\Gamma(\nu+1/2)(2z/a)^\nu}{\sqrt{\pi}}\int_0^\infty\frac{\cos a t  }{(t^2+z^2)^{\nu+1/2}} \log(x^2+z^2)dt
\end{align}
Note that in 
$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$
This implies 
$$\int_0^\infty\frac{\cos a t  }{\sqrt{t^2+z^2)}} \log(x^2+z^2)\,dt = (\Gamma'(1/2)+\log(2z/a) )K_0(az)  $$
Note that 
$$\Gamma'(1/2)+\log(2z/a) = -\gamma -2\log(2)+\log(2)+\log(z/a) =\log(z/2a) -\gamma$$
Hence 
$$\mathcal{J}=\left ( \log (z/2a)-\gamma  \right )K_0\left ( az \right )$$

Addendum
$$K_\nu (z) = \int^\infty_0 e^{-z\cosh t} \cosh(\nu t)\,dt$$
This implies 
$$\frac{\partial K_\nu(z)}{ \partial \nu}  = \int^\infty_0t e^{-z\cosh t} \sinh(\nu t)\,dt$$
Hence we have 
$$\left|\frac{\partial K_\nu(z)}{ \partial \nu} \right|_{\nu=0} = 0$$
