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!
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.