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how to evaluate this integral: $$l(y)=\int\limits_\beta^\infty \theta\exp(-y\theta)\alpha\exp(-\alpha\theta) \, d\theta$$ where $\alpha,\beta,\theta,y>0.$ Because I find it infinity! Can anyone help me to evaluate this integral? Thank you.$$$$ I find this solution :$$\left(\left.\frac{-1}{(\alpha+y)^2}\exp(-(\alpha+y)\theta)\right)\right|_{\beta}^\infty-\left.\frac{\theta}{(\alpha+y)}\exp(-(\alpha+y)\theta)\right|_{\beta}^\infty.$$ $$$$ in which in the second term, I obtain infinity value!

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Try substituting $t = exp(-\alpha \theta)$ . – Dilawar Nov 13 '12 at 8:14
Hint: Note that $\ell(y) = - \dfrac{\partial}{\partial y} \int_\beta^\infty \alpha \exp(-(y+\alpha) \theta)\ d\theta$ – Robert Israel Nov 13 '12 at 8:16
@Dilawar Or not. – Did Nov 13 '12 at 9:53

The product $$ \begin{align} x\exp(-x) &=x/\exp(x)\\ &=x/(1+x+x^2/2+x^3/6+\dots)\\ &=1/(1/x+1+x/2+x^2/6+\dots) \end{align} $$ $1/x\to0$ as $x\to\infty$ . Sum of other terms in the denominator $\to\infty$ as $x\to\infty$. So the product $x\exp(-x)\to0$ as $x\to\infty$.

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It makes answers easier to read if you use MathJax. – robjohn Jan 16 '13 at 1:18

You're wrong, your 2nd term is not infinite ! Indeed exponential growth is much stronger than polynomial one, so $\lim_{x \to +\infty} x.e^{-x} = 0$.

Beware of the $\alpha$ factor inside the integral and signs, I found for my part: $$ l(y) = \left( \frac\alpha{(y + \alpha)^2} - \frac{\alpha \beta}{y + \alpha} \right) e^{-(y + \alpha) \beta} $$

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