Is there a closed form solution for this integral?
$${\int_0^{\infty} x(1+{\frac{x}{\lambda}})^{(-\alpha-1)} dx}$$
where $$\alpha>1$$ $$\lambda>0$$
It's in Pareto statistical distribution family and I'm trying to find an analytical solution.
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1$\begingroup$ Does the$\int_{0}^{\infty} xdx$ converge? $\endgroup$– dmtriSep 20, 2019 at 20:08
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1$\begingroup$ Good point! I forgot the top-half of the equation. I've edited it now $\endgroup$– ChrisSep 20, 2019 at 20:12
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1$\begingroup$ Thanks for pointing that out. It's fixed now $\endgroup$– ChrisSep 20, 2019 at 20:15
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$\begingroup$ Welcome, You may edit the title of the equation also ) ; $\endgroup$– dmtriSep 20, 2019 at 20:26
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2 Answers
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The integral in the denominator looks like the expected value of a Lomax distribution. (some constants missing) I read that you need $\alpha >1$, otherwise undefined. Looking at the Wikipedia material, It appears the solution is
$$\int_0^{\infty} x\left(1+{\frac{x}{\lambda}}\right)^{(-\alpha-1)}dx = \frac{\lambda^2}{\alpha(\alpha -1)}$$
but verify.
answered Sep 20, 2019 at 20:23
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$$I=\int_{0}^{\infty} \frac{x}{(1+x/b)^{(1+a)}} dx =b^{(1+a)} \int_{0}^{\infty} \left( \frac{1}{(b+x)^a} - \frac{b}{(b+x)^{1+a}} \right)=b^{1+a} \left( \frac{(b+x)^{1-a}}{1-a}+b \frac{(b+x)^{-a}}{a} \right)_{0}^{\infty}=\frac{b^2}{a(a-1)},~ a > 1$$
