Question

Diff eq question! (round 3...)

Answer 1

\[\Gamma(p + 1) = \int\limits_{0}^{\infty} e ^{-x} x ^{p}dx\]

Show that, for p > 0,\[\Gamma(p+1) = p \Gamma(p)\]

Answer 2

You can start by writing what
\[ p\Gamma(p) \]
would look like, based on the definition above.

Answer 3

Wellp, I'm bored, here ya go:
\[ p\Gamma(p) = p \cdot \int_0^\infty e^{-x}x^{p-1} dx \]
\[ = \int_0^\infty e^{-x} \left( p x^{p-1} dx \right) \]
Now we integrate by parts to get:
\[ =\left. e^{-x}\cdot x^p \right|_0^\infty - \int_0^\infty\left(- e^{-x}\right) x^p dx \]
\[ = 0 + \int_0^\infty e^{-x} x^p dx = \Gamma(p+1) \]