Find $(1-p)logleft [int^{infty}_{0} [f(x)]^pdxright]$
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I want to find begin{align}(1-p)logleft [int^{infty}_{0} [f(x)]^pdxright],end{align}
given that $alpha,beta, theta$ and $p$ are constants and begin{align} f(x)=dfrac{ frac{ thetaalpha 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},;xinBbb{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{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{Big{ 1-(1-theta)left[1-u ^{-alpha} right] Big}^2}\&=dfrac{ frac{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{ 1-2(1-theta)left[1-u ^{-alpha} right]+(1-theta)^2left[1-u ^{-alpha} right] ^2}end{align}
I'm stuck here, please, how do I continue?
integration analysis probability-distributions
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
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add a comment |
$begingroup$
I want to find begin{align}(1-p)logleft [int^{infty}_{0} [f(x)]^pdxright],end{align}
given that $alpha,beta, theta$ and $p$ are constants and begin{align} f(x)=dfrac{ frac{ thetaalpha 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},;xinBbb{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{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{Big{ 1-(1-theta)left[1-u ^{-alpha} right] Big}^2}\&=dfrac{ frac{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{ 1-2(1-theta)left[1-u ^{-alpha} right]+(1-theta)^2left[1-u ^{-alpha} right] ^2}end{align}
I'm stuck here, please, how do I continue?
integration analysis probability-distributions
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
$endgroup$
$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
1
$begingroup$
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
add a comment |
$begingroup$
I want to find begin{align}(1-p)logleft [int^{infty}_{0} [f(x)]^pdxright],end{align}
given that $alpha,beta, theta$ and $p$ are constants and begin{align} f(x)=dfrac{ frac{ thetaalpha 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},;xinBbb{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{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{Big{ 1-(1-theta)left[1-u ^{-alpha} right] Big}^2}\&=dfrac{ frac{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{ 1-2(1-theta)left[1-u ^{-alpha} right]+(1-theta)^2left[1-u ^{-alpha} right] ^2}end{align}
I'm stuck here, please, how do I continue?
integration analysis probability-distributions
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
$endgroup$
I want to find begin{align}(1-p)logleft [int^{infty}_{0} [f(x)]^pdxright],end{align}
given that $alpha,beta, theta$ and $p$ are constants and begin{align} f(x)=dfrac{ frac{ thetaalpha 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},;xinBbb{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{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{Big{ 1-(1-theta)left[1-u ^{-alpha} right] Big}^2}\&=dfrac{ frac{ thetaalpha }{beta}(u-1)^2u ^{-(1-alpha)} }{ 1-2(1-theta)left[1-u ^{-alpha} right]+(1-theta)^2left[1-u ^{-alpha} right] ^2}end{align}
I'm stuck here, please, how do I continue?
integration analysis probability-distributions
integration analysis probability-distributions
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
edited Jan 8 at 8:37
Omojola Micheal
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
asked Jan 8 at 7:57
Omojola MichealOmojola Micheal
1,999424
1,999424
$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
1
$begingroup$
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
add a comment |
$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
1
$begingroup$
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
$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
1
1
$begingroup$
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
$begingroup$
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
add a comment |
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$begingroup$
What leads you to consider this horror?
$endgroup$
– Did
Jan 8 at 8:42
1
$begingroup$
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