pjmathematician
• Member for 4 years, 5 months
• Last seen more than 1 year ago
Stats
313
reputation
9k
reached
2
3
questions
Communities
View all

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 $$n$$).

This user doesn’t have any gold badges yet.
4
12
4
Score
2
Posts
40
Posts %
4
Score
1
Posts
20
Posts %
4
Score
1
Posts
20
Posts %
0
Score
2
Posts
40
Posts %
0
Score
1
Posts
20
Posts %
0
Score
1
Posts
20
Posts %
Top posts
4
May 29, 2020
4
Jun 11, 2020
2
Apr 5, 2020
1
Mar 31, 2020
0
Mar 31, 2020
Top network posts
View all network posts