How to calculate the definite integral $\int_0^\infty e^{-ax} x^4 dx$ How to calculate the following definite integral:
$$\int_0^\infty e^{-ax} x^4 \text{dx}$$
(Assuming $a > 0$.)
I know the solution is supposed to be $24a^{-5}$ but I don't even know where to start if I don't want to integrate by parts four times...
 A: Much easier than integration by parts since this is a "nice" definite integral - use Feynman's trick of differentiating under the integral sign.
Let $\displaystyle I(a) = \int_0^{\infty}e^{-ax}dx = -\frac 1ae^{-ax} \Biggr|_0^{\infty}=a^{-1}$
Now note that $\displaystyle \int_0^{\infty}x^4e^{-ax}dx = \frac{d^4}{da^4}I(a) = \frac{d^4}{da^4}a^{-1} = 24a^{-5}$
A: Integration by parts is, as suggested, the right approach, but this does not mean you have to integrate by parts four times to compute this integral. 
You can note generally that $$\int_{\mathbb{R}^{+}} e^{-x}x^n \ dx = \overbrace{\lim_{x \to +\infty}[-e^{-x}x^n]} ^{=0}+ n\int_{\mathbb{R}^{+}} e^{-x}x^{n-1} dx$$ and thus by induction it's easy to see that $\int_{\mathbb{R}^{+}} e^{-x}x^n \ dx = n!$
With this, we can evaluate $$\int_{\mathbb{R}^{+}} e^{-\alpha x} x^{n} \ dx = \alpha^{-n-1} \int_{\mathbb{R}^{+}} e^{-u} u^n \ du = \alpha^{-n-1} n!$$
Taking $n=4$ yields the desired result. 
A: Hint: use integration by parts
Substitute $u=x^4$ and $e^{-ax}$ and you should notice a recursive pattern, allowing you to easily cancel terms and get the desired answer of $\dfrac{24}{a^5}.$
