Need help with integral with e: int[0,+infty] [x^n e^{-ax}] dx

[csleeab]

I saw the integral below (particularly the second case below [being the one in the red box in the attached graphic) and could not understand how the answer is arrived:

. . . . .\(\displaystyle \displaystyle \int_0^{+\infty}\, x^n\, e^{-ax}\, dx\, =\, \begin{cases}\dfrac{\Gamma(n\, +\, 1)}{a^{n+1}}&\mbox{for }\, n\, >\, -1,\, a\, >\, 0\\ \\ \dfrac{n!}{a^{n+1}}&\mbox{for }\, n\, =\, 0,\, 1,\, 2,\, ...,\, a\, >\, 0\end{cases}\)

Would anyone please explain to me? Many thanks!

 

[Dr.Peterson]

I saw the integral below (particularly the second case below [being the one in the red box in the attached graphic) and could not understand how the answer is arrived:

. . . . .\(\displaystyle \displaystyle \int_0^{+\infty}\, x^n\, e^{-ax}\, dx\, =\, \begin{cases}\dfrac{\Gamma(n\, +\, 1)}{a^{n+1}}&\mbox{for }\, n\, >\, -1,\, a\, >\, 0\\ \\ \dfrac{n!}{a^{n+1}}&\mbox{for }\, n\, =\, 0,\, 1,\, 2,\, ...,\, a\, >\, 0\end{cases}\)

Would anyone please explain to me? Many thanks!

Where did you see it?

The fact is, the first part of this supposedly piecewise answer could apply to all values of n other than negative integers (for which the gamma function is not defined). And the second part is just a special case, due to the fact that the gamma function Γ(x) is defined in such a way that Γ(n+1) = n! for non-negative integers n. The two are actually equal. So this is a strange way to state it. What they are showing is, in effect, the definition of gamma.

How much do you know about the gamma function?

 

[HallsofIvy]

Those are, in fact, the same answer. For n a positive integer, \(\displaystyle \Gamma(n)= (n-1)!\). The way this is written is a little strange. Normally when you have brackets like that it means that one or the other form is to be used. Here, the first line can be used for any x (not a negative integer), the lower form only for x a positive integer.