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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!

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  • $\begingroup$ If $J(a,n)=\displaystyle\int_0^\infty K_0(ax)~I_1^n(bx)~dx~$ then your integral is $-J_1^{(1,1)}(a,1)$. $\endgroup$ – Lucian Feb 4 '15 at 15:05
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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.

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  • $\begingroup$ Thanks for your reply. I would say that eq.(4) is entirely unclear. At least, it's not explained enough. It seems to me, it's wrong. Eqs (5) are useless simply because there's no interest of the region of large $x$. $\endgroup$ – Nikita Feb 4 '15 at 16:32
  • $\begingroup$ @Nikita: it is more or less the same stated by Lucian. Equation $(5)$ is not useless: we have to be sure that $x\,K_1(ax)\,I_1(bx)$ is integrable, and the integral is took over $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Feb 4 '15 at 16:50

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