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I want to know if $\lim_{T-> \infty}$ of this integral

$$ \frac{\sigma^{4}C_{H,K}^{2}}{4 T^{4HK}e^{2\theta T }}\\ \times \int\limits_{[0,T]^{4}}e^{\theta(t_{1}-s_{1})}e^{\theta(t_{2}-s_{2})}\left\vert(t_{1}t_{2})^{2H-1}(t_{1}^{2H}+t_{2}^{2H})^{K-2}+(2HK-1)\vert t_1-t_2\vert^{2HK-2} \right\vert\ \\ \times\left\vert(s_{1}s_{2})^{2H-1}(s_{1}^{2H}+s_{2}^{2H})^{K-2}+(2HK-1)\vert s_{1}-s_{2}\vert^{2HK-2}\right\vert dt_{1}ds_{1}dt_{2}ds_{2}$$ with 2HK>1 and H,K $\in$ (0,1) and $\theta>0$

is finite. Thanks to help me

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    $\begingroup$ Have you already tried to multiply each of the four variables with $T$ and keep the domain of integration fixed as $[0,1]^4$? This should solve your problem. I am not sure if this really a research level question, and therefore recommend that you post questions like this on math.stackexchange. $\endgroup$
    – Sebastian Goette
    Commented Mar 12, 2016 at 17:07
  • $\begingroup$ Your integral seems to be connected to vector fractional brownian motion and its Laplace transform. Would David Nulart's famous book "fractional brownian motion stochastic calculus and applications" be helpful? Since the integrand is nonnegative and is a product of the same function evaluated at two different pairs of arguments, bounding one gives a bound on the whole integral. You can use Tonelli theorem first, and isolate the singularity first. $H,K$ play the critical role. But since Sebastian has recommended it be posted on another site, maybe you should do so! $\endgroup$
    – Chee
    Commented Mar 13, 2016 at 3:13
  • $\begingroup$ Depending on the value of $H,K$, you may see phase transition! $\endgroup$
    – Chee
    Commented Mar 13, 2016 at 3:15

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