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How to simplify the integral given below? $$ \int_{0}^{\infty}\frac{xe^{-\beta x}}{k^{x+1}}\mathrm{d}x,\quad k=1+e^{-a},\ a\in\mathbb{R},\;\beta>0 $$ Is there an explicit form, or can it be expressed with summation symbol? Do i need to use gamma function? if so, how?

($k$ is confusing. if $k=1$ it simply equals to $\beta^{-2}$.)

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    $\begingroup$ $k^{x+1} = \exp( \ln ( k^{x+1})) = \exp( (x+1) \ln(k)) = \exp( x \ln(k) ) \exp(\ln(k))$ $\endgroup$
    – Hetebrij
    Oct 26, 2015 at 13:24

3 Answers 3

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Shifting things around a bit you have this.

$$\int_0^\infty {xe^{-\beta x}\,dx \over k^{x + 1}} = {1\over k} \int_0^\infty {xe^{-\beta x}\,dx\over e^{x\log(k)}} = {1\over k}\int_0^\infty{xe^{x(\beta + \log(k))}}\, dx$$

The rest can easily be managed by an integration by parts and a $u$-sub.

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Using integration by parts, $$\int_0^\infty x e^{-\alpha x} dx = \frac{1}{\alpha^2}$$

\begin{align}\int_0^\infty \frac{x e^{-\beta x}}{k^{x+1}} dx &= \int_0^\infty x e^{-\beta x - (1+x)\log k} dx \\ &= e^{-\log k}\int_0^\infty x e^{-(\beta + \log k) x} dx \\ &= \frac{1}{k(\beta + \log k)^2}\end{align}

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  • $\begingroup$ Thanks all. i can't believe i didn't see that:) $\endgroup$
    – mert
    Oct 26, 2015 at 13:40
  • $\begingroup$ @mert It relies on the fact that $\beta + \log k > 0$ though $\endgroup$ Oct 26, 2015 at 13:42
  • $\begingroup$ actually $\beta>0$ is given. but i didn't add that:) i will edit the question. $\endgroup$
    – mert
    Oct 26, 2015 at 13:51
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$$\int_0^\infty\frac{xe^{-\beta x}}{k^{x+1}}\ dx = \frac 1k\int_0^\infty x\ \left(k\cdot e^{\beta}\right)^{-x}\ dx = \frac 1k\int_0^\infty xc^{-x}\ dx$$ Where $c = ke^\beta$ If we allow $u = x,\ dv = c^{-x}$, then $\dots$

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