Trick to calculate integrals such as $\int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x}dx$?

Question

Besides integration by parts, is there any trick to calculate integrals such a$$\int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x}dx=1 $$

I got $K=4$ by using integration by parts multiple times, but I think there should be some sort of formula to solve this kind of integrals.

Answer 1

You could use the Gamma function,

$$ \Gamma(z) = \int_0^{+\infty}{\rm d}x~x^{z-1}e^{-x} $$

such that your integral becomes

$$ \frac{2}{3K} \int_0^{+\infty}{\rm d}x~x^{3}e^{-x} = \frac{2}{3K}\color{red}{\Gamma(4)} = \frac{2}{3K}\color{red}{3!} $$

where I've used the property

$$ \Gamma(n + 1) = n! $$

Answer 2

HINT: the indefinite integral is given by $$-2/3\,{\frac { \left( {x}^{3}+3\,{x}^{2}+6\,x+6 \right) {{\rm e}^{-x}} }{K}} +C$$

Answer 3

Let $$I_n=\int_\limits{0}^\infty x^ne^{-x}dx$$

Then applying by-parts, $$I_n=\left[-x^ne^{-x}\ \right]_0^\infty-\int_\limits 0^\infty n.x^{n-1}.(-e^{-x}\ )dx$$ $$\Rightarrow I_n=0+nI_{n-1}=n.I_{n-1}$$

This is the Gamma function

Answer 4

$$ \int_0^\infty \left(\frac {2x^3}{3K}\right) e^{-x} \, dx = \frac 2 {3K} \int_0^\infty x^3 e^{-x} \, dx. $$ The integral here is $\displaystyle\int_0^\infty x^3 e^{-x}\, dx;$ the fraction $\dfrac 2 {3K}$ can be attended to after evaluating the integral.

You say you did integration by parts multiple times. Now suppose you had had instead $$\int_0^\infty x^{50} e^{-x} \, dx.$$ Then you might think you'd have to do integration by parts a much larger number of times. However, all of those iterations of integration by parts are the same, thus: \begin{align} \int_0^\infty x^{50} \Big( e^{-x} \, dx\Big) & = \int u\, dv = uv - \int v\, du \\[10pt] & = \left. x^{50} (-e^{-x}) \vphantom{\frac \sum1}\, \right|_0^\infty - \int_0^\infty (-e^{-x}) \Big( 50 x^{49}\, dx\Big) \\[10pt] & = 50\int_0^\infty x^{49} e^{-x}\, dx. \end{align} You're back to the same integral, except that the exponent has been reduced from $50$ to $49,$ and you also have a factor of $50$ in front of the integral. So doing it again reduces the exponent from $49$ to $48$ and you get a factor of $49$ in front of the integral: $$ 50 \cdot 49 \int_0^\infty x^{48} e^{-x}\,dx. $$ Then do it again: $$ 50\cdot49\cdot48\int_0^\infty x^{47} e^{-x}\,dx. $$ And so on: $$ 50\cdot49\cdot48\cdot47\cdot46\cdots\cdots3\cdot2\cdot1\int_0^\infty x^0 e^{-x}\, dx $$ and you don't need integration by parts for that last one.

So once you see the pattern, you've got it.