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I've encountered this integral many, many times (exams, exercises) and always end up wasting a bit of time calculating it for different $\rm a,b$'s.

Is there any way to calculate the following integral?

$$\rm \int x^ae^{-bx}dx.\quad a,b \in \Bbb Z_{\geq 0}.$$

YoTengoUnLCD
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3 Answers3

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Hint

There is a representation in terms of the Gamma function if you want this for general $a$ and $b$... For integer $a$ and $b$, integration by parts will suffice $$ \int x^ae^{-bx}dx$$ $$= -\frac{x^ae^{-bx}}{b}+\frac{a-1}{b}\int x^{a-1}e^{-bx}\,dx$$ Now just recursively apply this rule to the final integral... you'll soon find a pattern emerging $$= -\frac{x^ae^{-bx}}{b}+\frac{a-1}{b}\left[-\frac{x^{a-1}e^{-bx}}{b}+\frac{a-2}{b}\int x^{a-2}e^{-bx}\,dx\right]$$ $$= -\frac{x^ae^{-bx}}{b} -\frac{x^{a-1}e^{-bx}(a-1)}{b^2}+\frac{(a-1)(a-2)}{b^2}\int x^{a-2}e^{-bx}\,dx$$ $$= -\frac{x^ae^{-bx}}{b} -\frac{x^{a-1}e^{-bx}(a-1)}{b^2}+\frac{(a-1)(a-2)}{b^2}\left[-\frac{x^{a-2}e^{-bx}}{b}+\frac{a-3}{b}\int x^{a-3}e^{-bx}\,dx\right]$$ $$=-\frac{x^ae^{-bx}}{b} -\frac{x^{a-1}e^{-bx}(a-1)}{b^2}-\frac{x^{a-2}e^{-bx}(a-1)(a-2)}{b^3}+\frac{(a-1)(a-2)(a-3)}{b^3}\int x^{a-3}e^{-bx}\,dx$$

Pattern matching, the rule appears to be, $\forall \{a,b\} \in \Bbb N$: $$\int x^ae^{-bx}dx = \color{red}{-\frac{e^{-bx}}{a}\sum_{k=0}^{a} \frac{x^{a-k}(a)_{k+1}}{b^{k+1}}}$$ That last sum might be a little off as I was quickly looking, so feel free to check me as an exercise. You should get the basic idea though.

Brevan Ellefsen
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  • What does this indefinite integral have to do with the $\Gamma$ function? – Alex M. Dec 22 '15 at 17:23
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    @AlexM. First note that, by the Fundamental Theorem of Calculus, $$\frac{d}{dx} \left[\int_{g(x)}^{\infty} h(t) ,dt\right] = -h(g(x))g'(x)$$ Also, the definition of the Incomplete Upper Gamma Function is $$\Gamma(a,x)=\int_x^{\infty} t^{a-1}e^{-t},dt $$ Using these facts, we see that $$\frac{d}{dx} \Gamma(a+1,bx) = \frac{d}{dx} \int_{bx}^{\infty} t^{a}e^{-t},dt = -(bx)^ae^{-bx}b = -b^{a+1}x^ae^{-bx} ;[\forall x > 0]$$ Therefore, we get that $$\int x^ae^{-bx}dx = \int b^{a+1}x^ae^{-bx}b^{-a-1} = -b^{-a-1}\int\left[\frac{d}{dx} \Gamma(a+1,bx)\right]dx = \color{red}{-b^{-a-1}\Gamma(a+1,bx)}$$ – Brevan Ellefsen Dec 23 '15 at 08:03
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    @AlexM. I would also like to point out that I hit the character limit *exactly*, first try in my comment above simply pasting all this in from notepad (without checking length along the way). It's a Christmas miracle... I was really thinking I was going to go over the limit a bit. Hopefully that clears things up a bit. The Incomplete Gamma Function isn't something I mind using in the slightest (I love special functions), but some mathematicians much smarter than I am disagree with me, so I figured I would just leave the answer as it was :) – Brevan Ellefsen Dec 23 '15 at 08:06
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In general, the answer is $\mathbf{no}$.

In particular, take $b=-1$ and $a=-1$. Then you are left with the integral:

$$ \int \frac{e^x}{x} dx$$

And it is well known that this integral has no "closed form" solution.

Claude gave an example of a solution that uses the incomplete gamma function, but this is generally seen as a "nonelementary" function.

Even if we require $a>0,b>0$, we can still find an example. Pick $a=\frac{1}{2}$ and $b=1$. Then we have: $$ \int \sqrt{x}e^{-x}dx$$

With the substitution $x=u^2$, this gives us: $$ 2\int u^2e^{-u^2}du$$ And one iteration of iteration by parts leaves us with an integral of the form: $$ \int e^{-u^2}du$$ Which is known to have no closed form.

ASKASK
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For $a \in \mathbb N$ you can develop a finite length iterative reduction formula and collect it into a compact sum with sigma notation. I believe this is that the OP intended for his/her/-insert appropriate gender pronoun here- answer, as he/she/something did state non-negative integral a and b.

Jack Tiger Lam
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