Improper integral involving modified bessel functions and and its logarithm I am trying to solve the following integral:
$$\int_0^\infty x \, K_1(ax) \,I_1(bx) \,\log \big[ I_1(bx) \big] \mathrm{d} x$$
where $a$ and $b$ are positive real numbers, $I_1(x)$ and $K_1(x)$ are the modified Bessel functions of order one of the first and second kind, respectively.
Are there any ideas how to overcome this problem?
Thanks!
 A: We can notice that for any $x\geq 0$:
$$ \int_{0}^{+\infty}\!\!\!\!\frac{\cos(xz)}{(1+z^2)^{3/2}}\,dz = x\cdot K_1(x) \tag{1}$$
$$ \frac{1}{\pi}\int_{0}^{\pi}\exp(x\cos\theta)\cos\theta\,d\theta = I_1(x),\tag{2} $$
$$ K_1(x) = \int_{0}^{+\infty}\exp(-x\cosh t)\cosh t\,dt\tag{3}$$
and our integral is given by:
$$ \left.\frac{d}{d\lambda}\int_{0}^{+\infty} \!\!x\,K_1(ax)\,I_1(bx)^{\lambda}\,dx\,\right|_{\lambda=1}.\tag{4}$$
Notice however that, as long as $z\to +\infty$,
$$ I_1(z)\approx e^z\sqrt{\frac{1}{2\pi z}},\qquad K_1(z)\approx e^{-z}\sqrt{\frac{\pi}{2z}} \tag{5}$$
so $x\,K_1(ax)\,I_1(bx)$ cannot belong to $L^1(\mathbb{R})$ if $\Re(b)>\Re(a)$. 
Assuming $\Re(a-b)>0$ and defining $f(\lambda)=\int_{0}^{+\infty} \!\!x\,K_1(ax)\,I_1(bx)^{\lambda}\,dx$ we have:
$$ f(1)=\int_{0}^{+\infty}x\,K_1(ax)\,I_1(bx)\,dx = \frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi}\frac{\cosh t\,\cos\theta}{(b\cos \theta-a\cosh t)^2}\,d\theta\,dt\tag{6}$$
or:
$$ f(1) = \int_{0}^{+\infty}\frac{b \cosh t}{(a^2\cosh^2 t-b^2)^{3/2}}\,dt = \int_{0}^{+\infty}\frac{b\,dz}{((a^2-b^2)+a^2 z^2)^{3/2}}=\color{red}{\frac{b}{a(a^2-b^2)}}\tag{7}$$
The next step is to derive a functional equation for $f(\lambda)$ in order to compute $f'(1)$. Continues.
