111
reputation
2

Pj30

I love this proof- Consider $$\int_0^\infty e^{-Ax} dx = \frac{1}{A} \space \space A>0 \\ \text{differentiate both sides w.r.t $A$} \\ \frac{d}{dA}\left(\int_0^\infty e^{-Ax} dx \right )=\frac{d}{dA}\left(\frac{1}{A}\right) \\ = \int_0^\infty \frac{\partial}{\partial A}(e^{-Ax}) dx = \frac{-1}{A^2} \\ = \int_0^\infty (-x)e^{-Ax} dx=\frac{-1}{A^2} \\ \text{by differentiating again w.r.t $A$,} \\ \int_0^\infty (-x)^2e^{-Ax} dx=\frac{(-1)(-2)}{A^3}\\ \text{so by taking n derivatives we get} \\ \int_0^\infty (-x)^n e^{-Ax} dx = \frac{(-1)^n n!}{A^n} \\ \text{so by cancelling out $(-1)^n$ and putting $A$ =1 we get} \\ \int_0^\infty x^n e^{-x} dx = n! \\ \text{hence, we get out desired result by replacing $n$ with $n-1$ } \\ \int_0^\infty x^{n-1}e^{-x} dx = \Gamma({n})=(n-1)!$$ honestly, I found this proof very simple, elegant and beautiful (though here it is shown only for natural $n$).

0
answers
1
question
~172
people reached
  • Member for 4 months
  • 0 profile views
  • Last seen Aug 14 at 13:15

Top tags (4)

Score 0
Posts 1
Posts % 100
Score 0
Posts 1
Score 0
Posts 1
Score 0
Posts 1

Top posts (1)

Badges (2)

Gold

Silver

Bronze

2

Rarest