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I am trying to evaluate the following integral:

$$\int_a^bx^\alpha \exp\left[-\left(\frac{x-\delta}{\sigma}\right)^\alpha\right]dx$$

where $0<a<b$, $\delta\leq a$ and $\alpha>0$.

Is there a closed form, perhaps in terms of Gamma and error functions?

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1 Answer 1

up vote 4 down vote accepted

Note that this integral has closed form only when $\alpha=1$ and can express in terms of known special functions only when $\delta=0$ or $\alpha$ is a positive integer other than $1$.

When $\alpha=1$ , this integral becomes $\int_a^bxe^{-\frac{x-\delta}{\sigma}}~dx$ and it obviously has closed form

When $\alpha=2$ , this integral becomes $\int_a^bx^2e^{-\left(\frac{x-\delta}{\sigma}\right)^2}~dx$ and it can express in terms of error function

For other values of $\alpha$ , consider $\int x^\alpha e^{-\left(\frac{x-\delta}{\sigma}\right)^\alpha}~dx$ ,

Let $u=\dfrac{x-\delta}{\sigma}$ ,

Then $x=\sigma u+\delta$

$dx=\sigma~du$

$\therefore\int x^\alpha e^{-\left(\frac{x-\delta}{\sigma}\right)^\alpha}~dx=\int\sigma(\sigma u+\delta)^\alpha e^{-u^\alpha}~du$

When $\delta=0$ , the integral becomes $\int\sigma^{\alpha+1}u^\alpha e^{-u^\alpha}~du$

Let $v=u^\alpha$ ,

Then $u=v^{\frac{1}{\alpha}}$

$du=\dfrac{v^{\frac{1}{\alpha}-1}}{\alpha}dv$

$\therefore\int\sigma^{\alpha+1}u^\alpha e^{-u^\alpha}~du=\int\dfrac{\sigma^{\alpha+1}v^{\frac{1}{\alpha}}e^{-v}}{\alpha}dv=\dfrac{\sigma^{\alpha+1}}{\alpha}\gamma\left({\dfrac{1}{\alpha}+1},v\right)+C=\dfrac{\sigma^{\alpha+1}}{\alpha}\gamma\left({\dfrac{1}{\alpha}+1},u^\alpha\right)+C=\dfrac{\sigma^{\alpha+1}}{\alpha}\gamma\left({\dfrac{1}{\alpha}+1},\left(\dfrac{x}{\sigma}\right)^\alpha\right)+C$

Hence $\int_a^bx^\alpha e^{-\left(\frac{x}{\sigma}\right)^\alpha}~dx=\dfrac{\sigma^{\alpha+1}}{\alpha}\left(\gamma\left({\dfrac{1}{\alpha}+1},\left(\dfrac{b}{\sigma}\right)^\alpha\right)-\gamma\left({\dfrac{1}{\alpha}+1},\left(\dfrac{a}{\sigma}\right)^\alpha\right)\right)$

When $\delta\neq0$ and $\alpha$ is a positive integer,

$\int\sigma(\sigma u+\delta)^\alpha e^{-u^\alpha}~du=\int\sum\limits_{n=0}^\alpha C_n^\alpha\delta^{\alpha-n}\sigma^{n+1}u^ne^{-u^\alpha}~du$

Consider $\int u^ne^{-u^\alpha}~du$ ,

Let $v=u^\alpha$ ,

Then $u=v^{\frac{1}{\alpha}}$

$du=\dfrac{v^{\frac{1}{\alpha}-1}}{\alpha}dv$

$\therefore\int u^ne^{-u^\alpha}~du=\int\dfrac{v^{\frac{n+1}{\alpha}-1}e^{-v}}{\alpha}dv=\dfrac{1}{\alpha}\gamma\left({\dfrac{n+1}{\alpha}},v\right)+C=\dfrac{1}{\alpha}\gamma\left({\dfrac{n+1}{\alpha}},u^\alpha\right)+C=\dfrac{1}{\alpha}\gamma\left({\dfrac{n+1}{\alpha}},\left(\dfrac{x-\delta}{\sigma}\right)^\alpha\right)+C$

Hence $\int_a^bx^\alpha e^{-\left(\frac{x-\delta}{\sigma}\right)^\alpha}~dx=\sum\limits_{n=0}^\alpha\dfrac{C_n^\alpha\delta^{\alpha-n}\sigma^{n+1}}{\alpha}\left(\gamma\left({\dfrac{n+1}{\alpha}},\left(\dfrac{b-\delta}{\sigma}\right)^\alpha\right)-\gamma\left({\dfrac{n+1}{\alpha}},\left(\dfrac{a-\delta}{\sigma}\right)^\alpha\right)\right)$

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What is $\gamma $? An incomplete gamma function? –  Dominik Jul 1 '13 at 16:15
    
@Dominik: Definitely yes. Refer to en.wikipedia.org/wiki/Incomplete_Gamma_function. –  Harry Peter Jul 2 '13 at 12:10

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