Reduction Formula and factorials: I_mn =int(x^m (1 - x)^n, x=0..1) = m!n!/(m+n+1)!

[Bronn]

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I_mn =int( x^m (1 - x)ⁿ, x=0..1) = m!n!/(m+n+1)!



really having trouble with this one.

first thing I tried was just integrating by parts as it is, which I couldn't make work.

I tried changing the expression to:

I_mn = int( (x^m-1) (1-x)ⁿ (1+x-1), x=0..1)

which simplified to:

I_mn = I_m-1,n+1 - I_m-1,n


which then integrating by parts on this new expression didn't have much luck either.

is there any hints on how I should approach this question?

thanks

 

[HallsofIvy]

Did you consider "proof by induction"? For fixed m do an induction on n. If n= 0, we have \(\displaystyle \int_0^1 x^m dx= \left[\frac{x^{m+1}}{m+1}\right]_0^1= \frac{1}{m+1}= \frac{m!0!}{(m+0+1)!}\).

Now, suppose that, for n= k, \(\displaystyle \int_0^1 x^m(1- x)^k dx= \frac{m!k!}{(m+k+1)!}\). Then \(\displaystyle \int_0^1 x^m(1- x)^{k+1}dx= \int_0^1 x^m(1- x)^k (1- x) dx\)
\(\displaystyle = \int_0^1 x^m(1- x)^k dx- \int_0^1 x^{m+1}(1- x)^k dx= \frac{m!k!}{(m+k+1)!}- \frac{(m+1)!k!}{((m+1)+k+1)!}\).

Can you finish that?

 

[Bronn]

Im not great with induction.
How did you go from 1/(m+1) to m!0!/(m+0+1)! ?

edit# ok nevermind, you multiplied by m!/m!






as for finishing your last step.


m!k!/(m+k+1)! - (m+1)!k!/((m+1)+k+1)!

= m!k!/(m+k+1)! - (m+1)m!k!/ (m+k+2)(m+k+1)!

=m!k!/(m+k+1)! * (1 - (m+1)/(m+k+2))

=(m + k + 2 - (m+1)) / (m+k+2) * m!k!/(m+k+1)!

= ((k+1) / (m+(k+1)+1)) * m!k!/(m+k+1)!

= m!(k+1)! / (m+(k+1)+1)!



so then you have to prove for m and its complete?

so proving m and n is true separately means they will be true together?


I can't seem to prove for n using the same method. Any clues for that?