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About Rayleigh's formula

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?

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

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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.

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.

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.

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