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I want to find \begin{align}(1-p)\log\left [\int^{\infty}_{0} [f(x)]^pdx\right],\end{align} given that $\alpha,\beta, \theta$ and $p$ are constants and \begin{align} f(x)=\dfrac{ \frac{ \theta\alpha \beta}{x^2}\left( 1+\frac{ \beta}{x} \right) ^{-(1-\alpha)} }{\Big\{ 1-(1-\theta)\left[1-\left( 1+\frac{ \beta}{x} \right) ^{-\alpha} \right] \Big\}^2},\;x\in\Bbb{R}\end{align}

MY TRIAL

By substitution, let $u=1+\frac{ \beta}{x}, $ then $du=-[(u-1)^2/\beta]dx$ \begin{align} f(u)&=\dfrac{ \frac{ \theta\alpha }{\beta}(u-1)^2u ^{-(1-\alpha)} }{\Big\{ 1-(1-\theta)\left[1-u ^{-\alpha} \right] \Big\}^2}\\&=\dfrac{ \frac{ \theta\alpha }{\beta}(u-1)^2u ^{-(1-\alpha)} }{ 1-2(1-\theta)\left[1-u ^{-\alpha} \right]+(1-\theta)^2\left[1-u ^{-\alpha} \right] ^2}\end{align} I'm stuck here, please, how do I continue?

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  • $\begingroup$ What leads you to consider this horror? $\endgroup$ – Did Jan 8 at 8:42
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    $\begingroup$ Smiles... My statistical friend was going about his research work but got stuck here. He has no love for integration, so his only choice was to send it across to me. As you said, it seems like an horror! Smiles again... $\endgroup$ – Omojola Micheal Jan 8 at 8:47

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