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How to compute the following integral

\begin{equation} \int_{n_0}^{\infty} \exp(- c x^\lambda) dx, \end{equation}

when $0 < \lambda \leq 1.$

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Hint: $~\displaystyle\int_0^\infty\exp\Big(-c~x^\lambda\Big)~dx=\frac{\bigg(\dfrac1\lambda\bigg)!}{\sqrt[\lambda]c}~,~$ see $\Gamma$ function for more information. But since your

lower integration limit is $n_0$ instead of $0$, you will have to use the incomplete $\Gamma$ function. One can

also write it as $~\dfrac n\lambda\cdot\text{E}\bigg(1-\dfrac1\lambda~,~c~n^\lambda\bigg),~$ see exponential integral for more information.

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The computer algebra system maxima tells us: $$ \int \exp\left(-c\,x^\lambda\right) dx = -\dfrac{\Gamma\left(\frac{1}{\lambda},c\,x^\lambda\right)}{\lambda\,c^\frac{1}{\lambda}} $$ Where $\Gamma\left(s,x\right)$ is the upper incomplete gamma function, so: $$ \int_{n_{0}}^\infty \exp\left(-c\,x^\lambda\right) dx = \lim_{x\rightarrow\infty}-\dfrac{\Gamma\left(\frac{1}{\lambda},c\,x^\lambda\right)}{\lambda\,c^\frac{1}{\lambda}}+\dfrac{\Gamma\left(\frac{1}{\lambda},c\,n_0^\lambda\right)}{\lambda\,c^\frac{1}{\lambda}} = \dfrac{\Gamma\left(\frac{1}{\lambda},c\,n_0^\lambda\right)}{\lambda\,c^\frac{1}{\lambda}} $$

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    $\begingroup$ Why the minus sign? $\endgroup$ – grdgfgr Jun 1 '15 at 10:12
  • $\begingroup$ @grdgfgr the minus sign at the end is wrong because there is one missing in the beginning - thank you! $\endgroup$ – Lykos Jun 1 '15 at 10:16

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