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I want to compute:

$$I = \int_{0}^{\infty} x^2 [K_1(x)]^2 dx$$

where

$K_1(x)$ is the Bessel Modified Function given by:

$$ 2xK_1(x) = \int_{-\infty}^{\infty} \frac{e^{isx}}{(1+s^2)^{3/2}} ds $$

The Fourier Transform in cosines of $f(t) = e^{-at}$ is given by $F(w) = \sqrt{\frac{2}{\pi}} \frac{a}{a^2 + w^2}$

So to deal with the squares I was thinking on applying Parseval's Theorem which states that:

$$ \int_{-\infty}^{\infty}|F(w)|^2 dw = \int_{-\infty}^{\infty} |f(t)|^2 dt $$

But I don't know how to proceed. Tips are appreciated.

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AE_X12_31
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  • Seems to me that your idea will work fine since your result says that $xK_1(x)$ is the inverse Fourier transform of some nice function. So its squared $L^2$ norm is a multiple of the squared $L^2$ norm of that function. (However you will want to be careful about the fact that your desired integral is only on the half line. Can you write it as a multiple of an integral on the full line?) – Ian Jun 25 '20 at 04:23
  • Let $f(s) = (s^2 + 1)^{-3/2}$. For the FT with the kernel $e^{i x s}$, $$\mathcal Ff = 2 |x| K_1(|x|), \ | f |_2^2 = \frac 1 {2 \pi} | \mathcal F[f] |_2^2.$$ – Maxim Jun 25 '20 at 13:16
  • Thanks, I actually ended up with $ f(s) = \sqrt{\frac{\pi}{2}} (s^2 + 1)^{-3/2} $ through the invertion theorem. Since the functions are even, I managed to extend $ I $ to the line and apply Parseval. Finally ended up with $$ \frac{\pi}{2} \int_{0}^{\infty} \frac{1}{(1+s^2)^{3}} ds = \frac{3\pi^{2}}{32} $$ – AE_X12_31 Jun 26 '20 at 02:27

1 Answers1

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Since the OP stated in the comments that they successfully applied Parseval's/Plancherel's theorem, I'm going to evaluate a generalization.

Similar to my answer here, we can use the integral representation $$K_{\alpha}(x) K_{\beta}(x) = 2\int_0^\infty K_{\alpha + \beta} (2x \cosh t) \cosh \left((\alpha - \beta)t \right) \, \mathrm dt, \quad x >0,$$ to evaluate the Mellin transform $$I(s,\alpha, \beta) = \int_0^\infty x^{2s-1} K_{2 \alpha}(x) K_{2 \beta}(x) \, \mathrm dx \, , \quad \left( \alpha, \beta \ge 0, \, s > \alpha + \beta \right). $$

All we need is the following integral formula, where $s > |\nu|$:

$$\begin{align} \int_{0}^{\infty} \frac{\cosh(2 \nu t)}{\cosh^{2s}(t)} \, \mathrm dt &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cosh(2 \nu t)}{\cosh^{2s}(t)} \, \mathrm dt \\ &= 2^{2(s-1)} \int_{0}^{\infty} \frac{u^{2 \nu+ 2 s-1} + u^{-2 \nu+ 2s -1}}{\left(u^{2}+1 \right)^{2s}} \, \mathrm du \\ &= \frac{2^{2(s-1)}}{2} \int_{0}^{\infty} \frac{w^{\nu+ s-1} + w^{- \nu +s -1}}{(w+1)^{2 s}} \, \mathrm dw \\ &\overset{(1)}{=} \frac{2^{2(s-1)}}{2} \, \left(B(s+\nu, s- \nu) + B(s-\nu, s+ \nu) \right) \\ &= 2^{2(s-1)} \, B (s+ \nu, s-\nu). \end{align} $$

Then $$ \begin{align} I(s,\alpha, \beta) &= \int_0^\infty x^{2s-1} K_{2 \alpha}(x) K_{2 \beta}(x) \, \mathrm dx \\ &= 2 \int_0^\infty x^{2s-1} \int_{0}^{\infty} K_{2(\alpha+ \beta)}(2x \cosh t) \cosh \left(2(\alpha -\beta)t \right) \, \mathrm dt \, \mathrm dx \\ &\overset{(2)}{=} 2 \int_0^\infty x^{2s-1} \int_0^\infty \cosh \left(2(\alpha - \beta)t \right) \int_0^\infty e^{-2x \cosh (t) \cosh (v)} \cosh \left(2(\alpha + \beta)v \right) \, \mathrm dv \, \mathrm dt \, \mathrm dx \\ &= 2 \int_0^\infty \cosh \left(2(\alpha + \beta)v \right) \int_0^\infty \cosh \left(2(\alpha - \beta)t \right) \int_0^\infty x^{2s-1}e^{-2x \cosh (t) \cosh (v)} \, \mathrm dx \, \mathrm dt \, \mathrm dv \\ &\overset{(3)}{=} \frac{\Gamma(2s)}{2^{2s-1}} \left(\int_0^\infty \frac{\cosh \left(2(\alpha + \beta)v \right)}{\cosh^{2s}(v)} \, \mathrm dv\right) \left( \int_0^\infty \frac{\cosh \left(2(\alpha - \beta)t \right)}{\cosh^{2s}(t)} \, \mathrm dt \right) \\ &= \frac{ \Gamma(2s)}{2^{2s-1}} \, \left(2^{2s-2} \right)^2 \, B \left(s + \alpha + \beta, s- \alpha - \beta \right) B \left(s+ \alpha - \beta, s - \alpha + \beta \right) \\ &= \frac{2^{2s-3}}{\Gamma(2s)} \, \Gamma \left(s+ \alpha + \beta \right) \Gamma \left(s- \alpha - \beta \right) \Gamma \left(s+ \alpha - \beta \right) \Gamma \left(s- \alpha + \beta \right). \end{align}$$

For $s= \frac{3}{2}$ and $\alpha = \beta =\frac{1}{2} $, we have $$\int_{0}^{\infty} x^{2}K_{1}(x)^{2} \, \mathrm dx = \frac{1}{\Gamma(3)} \, \Gamma \left(\frac{5}{2} \right) \Gamma \left(\frac{1}{2} \right) \Gamma \left(\frac{3}{2} \right)^{2} = \frac{1}{2} \, \left(\frac{3\pi^{2}}{16} \right) = \frac{3 \pi^{2}}{32}. $$

$(1)$ https://en.wikipedia.org/wiki/Beta_function#Other_identities_and_formulas

$(2)$ https://dlmf.nist.gov/10.32.E9

$(3)$ $\int_{0}^{\infty} x^{s-1} e^{-ax} \, \mathrm dx = a^{-s} \, \Gamma(s), \quad a>0 $

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