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